Rank groups within a grouped sequence of TRUE/FALSE and NA





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10















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question




















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    15 hours ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    15 hours ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    15 hours ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    15 hours ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    15 hours ago




















10















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question




















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    15 hours ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    15 hours ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    15 hours ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    15 hours ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    15 hours ago
















10












10








10








I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question
















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!







r dplyr data.table rank






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edited 15 hours ago







Humpelstielzchen

















asked 17 hours ago









HumpelstielzchenHumpelstielzchen

1,3901318




1,3901318








  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    15 hours ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    15 hours ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    15 hours ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    15 hours ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    15 hours ago
















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    15 hours ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    15 hours ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    15 hours ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    15 hours ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    15 hours ago










1




1





you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

– user20650
15 hours ago







you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

– user20650
15 hours ago















that is a really funny solution. Very good job!

– Humpelstielzchen
15 hours ago





that is a really funny solution. Very good job!

– Humpelstielzchen
15 hours ago













In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

– smci
15 hours ago





In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

– smci
15 hours ago













No, when group A stops so stops the sequence for group A.

– Humpelstielzchen
15 hours ago







No, when group A stops so stops the sequence for group A.

– Humpelstielzchen
15 hours ago















But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

– smci
15 hours ago







But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

– smci
15 hours ago














4 Answers
4






active

oldest

votes


















7














Another data.table approach:



library(data.table)
setDT(dt)
dt[, cr := rleid(criterium)][
(criterium), goal := rleid(cr), by=.(group)]





share|improve this answer



















  • 1





    Tried with rleid but didn't get it to work. (+1)

    – markus
    15 hours ago











  • works for me. And seems to be the most elegant answer.

    – Humpelstielzchen
    15 hours ago



















6














Maybe I have over-complicated this but one way with dplyr is



library(dplyr)

df %>%
mutate(temp = replace(criterium, is.na(criterium), FALSE),
temp1 = cumsum(!temp)) %>%
group_by(temp1) %>%
mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
group_by(group) %>%
mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
select(-temp, -temp1)

# group criterium goal
# <fct> <lgl> <int>
# 1 A NA NA
# 2 A TRUE 1
# 3 A TRUE 1
# 4 A TRUE 1
# 5 A FALSE NA
# 6 A FALSE NA
# 7 A TRUE 2
# 8 A TRUE 2
# 9 A FALSE NA
#10 A TRUE 3
#11 A TRUE 3
#12 A TRUE 3
#13 B NA NA
#14 B FALSE NA
#15 B TRUE 1
#16 B TRUE 1
#17 B TRUE 1
#18 B FALSE NA


We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






share|improve this answer































    4














    A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



    f1 <- function(x) {
    x[is.na(x)] <- FALSE
    rle1 <- rle(x)
    y <- rle1$values
    rle1$values[!y] <- 0
    rle1$values[y] <- cumsum(rle1$values[y])
    return(inverse.rle(rle1))
    }


    do.call(rbind,
    lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
    i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
    i}))


    Of course, If you want you can apply it via dplyr, i.e.



    library(dplyr)

    df %>%
    group_by(group) %>%
    mutate(goal = f1(criterium),
    goal = replace(goal, is.na(criterium)|!criterium, NA))


    which gives,




    # A tibble: 18 x 3
    # Groups: group [2]
    group criterium goal
    <fct> <lgl> <dbl>
    1 A NA NA
    2 A TRUE 1
    3 A TRUE 1
    4 A TRUE 1
    5 A FALSE NA
    6 A FALSE NA
    7 A TRUE 2
    8 A TRUE 2
    9 A FALSE NA
    10 A TRUE 3
    11 A TRUE 3
    12 A TRUE 3
    13 B NA NA
    14 B FALSE NA
    15 B TRUE 1
    16 B TRUE 1
    17 B TRUE 1
    18 B FALSE NA






    share|improve this answer

































      4














      A data.table option using rle



      library(data.table)
      DT <- as.data.table(dat)
      DT[, goal := {
      r <- rle(replace(criterium, is.na(criterium), FALSE))
      r$values <- with(r, cumsum(values) * values)
      out <- inverse.rle(r)
      replace(out, out == 0, NA)
      }, by = group]
      DT
      # group criterium goal
      # 1: A NA NA
      # 2: A TRUE 1
      # 3: A TRUE 1
      # 4: A TRUE 1
      # 5: A FALSE NA
      # 6: A FALSE NA
      # 7: A TRUE 2
      # 8: A TRUE 2
      # 9: A FALSE NA
      #10: A TRUE 3
      #11: A TRUE 3
      #12: A TRUE 3
      #13: B NA NA
      #14: B FALSE NA
      #15: B TRUE 1
      #16: B TRUE 1
      #17: B TRUE 1
      #18: B FALSE NA


      step by step



      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



      r
      #Run Length Encoding
      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


      We manipulate the values compenent in the following way



      r$values <- with(r, cumsum(values) * values)
      r
      #Run Length Encoding
      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
      # values : int [1:9] 0 1 0 2 0 3 0 4 0


      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



      out <- inverse.rle(r)
      out
      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


      This is almost what OP wants but we need to replace the 0s with NA



      replace(out, out == 0, NA)


      This is done for each group.



      data



      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
      -18L))





      share|improve this answer


























      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

        – Humpelstielzchen
        16 hours ago






      • 1





        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

        – markus
        16 hours ago













      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

        – Humpelstielzchen
        15 hours ago












      Your Answer






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer



















      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        15 hours ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        15 hours ago
















      7














      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer



















      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        15 hours ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        15 hours ago














      7












      7








      7







      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer













      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 15 hours ago









      chinsoon12chinsoon12

      9,93611420




      9,93611420








      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        15 hours ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        15 hours ago














      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        15 hours ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        15 hours ago








      1




      1





      Tried with rleid but didn't get it to work. (+1)

      – markus
      15 hours ago





      Tried with rleid but didn't get it to work. (+1)

      – markus
      15 hours ago













      works for me. And seems to be the most elegant answer.

      – Humpelstielzchen
      15 hours ago





      works for me. And seems to be the most elegant answer.

      – Humpelstielzchen
      15 hours ago













      6














      Maybe I have over-complicated this but one way with dplyr is



      library(dplyr)

      df %>%
      mutate(temp = replace(criterium, is.na(criterium), FALSE),
      temp1 = cumsum(!temp)) %>%
      group_by(temp1) %>%
      mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
      group_by(group) %>%
      mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
      select(-temp, -temp1)

      # group criterium goal
      # <fct> <lgl> <int>
      # 1 A NA NA
      # 2 A TRUE 1
      # 3 A TRUE 1
      # 4 A TRUE 1
      # 5 A FALSE NA
      # 6 A FALSE NA
      # 7 A TRUE 2
      # 8 A TRUE 2
      # 9 A FALSE NA
      #10 A TRUE 3
      #11 A TRUE 3
      #12 A TRUE 3
      #13 B NA NA
      #14 B FALSE NA
      #15 B TRUE 1
      #16 B TRUE 1
      #17 B TRUE 1
      #18 B FALSE NA


      We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






      share|improve this answer




























        6














        Maybe I have over-complicated this but one way with dplyr is



        library(dplyr)

        df %>%
        mutate(temp = replace(criterium, is.na(criterium), FALSE),
        temp1 = cumsum(!temp)) %>%
        group_by(temp1) %>%
        mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
        group_by(group) %>%
        mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
        select(-temp, -temp1)

        # group criterium goal
        # <fct> <lgl> <int>
        # 1 A NA NA
        # 2 A TRUE 1
        # 3 A TRUE 1
        # 4 A TRUE 1
        # 5 A FALSE NA
        # 6 A FALSE NA
        # 7 A TRUE 2
        # 8 A TRUE 2
        # 9 A FALSE NA
        #10 A TRUE 3
        #11 A TRUE 3
        #12 A TRUE 3
        #13 B NA NA
        #14 B FALSE NA
        #15 B TRUE 1
        #16 B TRUE 1
        #17 B TRUE 1
        #18 B FALSE NA


        We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






        share|improve this answer


























          6












          6








          6







          Maybe I have over-complicated this but one way with dplyr is



          library(dplyr)

          df %>%
          mutate(temp = replace(criterium, is.na(criterium), FALSE),
          temp1 = cumsum(!temp)) %>%
          group_by(temp1) %>%
          mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
          group_by(group) %>%
          mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
          select(-temp, -temp1)

          # group criterium goal
          # <fct> <lgl> <int>
          # 1 A NA NA
          # 2 A TRUE 1
          # 3 A TRUE 1
          # 4 A TRUE 1
          # 5 A FALSE NA
          # 6 A FALSE NA
          # 7 A TRUE 2
          # 8 A TRUE 2
          # 9 A FALSE NA
          #10 A TRUE 3
          #11 A TRUE 3
          #12 A TRUE 3
          #13 B NA NA
          #14 B FALSE NA
          #15 B TRUE 1
          #16 B TRUE 1
          #17 B TRUE 1
          #18 B FALSE NA


          We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






          share|improve this answer













          Maybe I have over-complicated this but one way with dplyr is



          library(dplyr)

          df %>%
          mutate(temp = replace(criterium, is.na(criterium), FALSE),
          temp1 = cumsum(!temp)) %>%
          group_by(temp1) %>%
          mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
          group_by(group) %>%
          mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
          select(-temp, -temp1)

          # group criterium goal
          # <fct> <lgl> <int>
          # 1 A NA NA
          # 2 A TRUE 1
          # 3 A TRUE 1
          # 4 A TRUE 1
          # 5 A FALSE NA
          # 6 A FALSE NA
          # 7 A TRUE 2
          # 8 A TRUE 2
          # 9 A FALSE NA
          #10 A TRUE 3
          #11 A TRUE 3
          #12 A TRUE 3
          #13 B NA NA
          #14 B FALSE NA
          #15 B TRUE 1
          #16 B TRUE 1
          #17 B TRUE 1
          #18 B FALSE NA


          We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 16 hours ago









          Ronak ShahRonak Shah

          46k104268




          46k104268























              4














              A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



              f1 <- function(x) {
              x[is.na(x)] <- FALSE
              rle1 <- rle(x)
              y <- rle1$values
              rle1$values[!y] <- 0
              rle1$values[y] <- cumsum(rle1$values[y])
              return(inverse.rle(rle1))
              }


              do.call(rbind,
              lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
              i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
              i}))


              Of course, If you want you can apply it via dplyr, i.e.



              library(dplyr)

              df %>%
              group_by(group) %>%
              mutate(goal = f1(criterium),
              goal = replace(goal, is.na(criterium)|!criterium, NA))


              which gives,




              # A tibble: 18 x 3
              # Groups: group [2]
              group criterium goal
              <fct> <lgl> <dbl>
              1 A NA NA
              2 A TRUE 1
              3 A TRUE 1
              4 A TRUE 1
              5 A FALSE NA
              6 A FALSE NA
              7 A TRUE 2
              8 A TRUE 2
              9 A FALSE NA
              10 A TRUE 3
              11 A TRUE 3
              12 A TRUE 3
              13 B NA NA
              14 B FALSE NA
              15 B TRUE 1
              16 B TRUE 1
              17 B TRUE 1
              18 B FALSE NA






              share|improve this answer






























                4














                A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                f1 <- function(x) {
                x[is.na(x)] <- FALSE
                rle1 <- rle(x)
                y <- rle1$values
                rle1$values[!y] <- 0
                rle1$values[y] <- cumsum(rle1$values[y])
                return(inverse.rle(rle1))
                }


                do.call(rbind,
                lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                i}))


                Of course, If you want you can apply it via dplyr, i.e.



                library(dplyr)

                df %>%
                group_by(group) %>%
                mutate(goal = f1(criterium),
                goal = replace(goal, is.na(criterium)|!criterium, NA))


                which gives,




                # A tibble: 18 x 3
                # Groups: group [2]
                group criterium goal
                <fct> <lgl> <dbl>
                1 A NA NA
                2 A TRUE 1
                3 A TRUE 1
                4 A TRUE 1
                5 A FALSE NA
                6 A FALSE NA
                7 A TRUE 2
                8 A TRUE 2
                9 A FALSE NA
                10 A TRUE 3
                11 A TRUE 3
                12 A TRUE 3
                13 B NA NA
                14 B FALSE NA
                15 B TRUE 1
                16 B TRUE 1
                17 B TRUE 1
                18 B FALSE NA






                share|improve this answer




























                  4












                  4








                  4







                  A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                  f1 <- function(x) {
                  x[is.na(x)] <- FALSE
                  rle1 <- rle(x)
                  y <- rle1$values
                  rle1$values[!y] <- 0
                  rle1$values[y] <- cumsum(rle1$values[y])
                  return(inverse.rle(rle1))
                  }


                  do.call(rbind,
                  lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                  i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                  i}))


                  Of course, If you want you can apply it via dplyr, i.e.



                  library(dplyr)

                  df %>%
                  group_by(group) %>%
                  mutate(goal = f1(criterium),
                  goal = replace(goal, is.na(criterium)|!criterium, NA))


                  which gives,




                  # A tibble: 18 x 3
                  # Groups: group [2]
                  group criterium goal
                  <fct> <lgl> <dbl>
                  1 A NA NA
                  2 A TRUE 1
                  3 A TRUE 1
                  4 A TRUE 1
                  5 A FALSE NA
                  6 A FALSE NA
                  7 A TRUE 2
                  8 A TRUE 2
                  9 A FALSE NA
                  10 A TRUE 3
                  11 A TRUE 3
                  12 A TRUE 3
                  13 B NA NA
                  14 B FALSE NA
                  15 B TRUE 1
                  16 B TRUE 1
                  17 B TRUE 1
                  18 B FALSE NA






                  share|improve this answer















                  A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                  f1 <- function(x) {
                  x[is.na(x)] <- FALSE
                  rle1 <- rle(x)
                  y <- rle1$values
                  rle1$values[!y] <- 0
                  rle1$values[y] <- cumsum(rle1$values[y])
                  return(inverse.rle(rle1))
                  }


                  do.call(rbind,
                  lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                  i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                  i}))


                  Of course, If you want you can apply it via dplyr, i.e.



                  library(dplyr)

                  df %>%
                  group_by(group) %>%
                  mutate(goal = f1(criterium),
                  goal = replace(goal, is.na(criterium)|!criterium, NA))


                  which gives,




                  # A tibble: 18 x 3
                  # Groups: group [2]
                  group criterium goal
                  <fct> <lgl> <dbl>
                  1 A NA NA
                  2 A TRUE 1
                  3 A TRUE 1
                  4 A TRUE 1
                  5 A FALSE NA
                  6 A FALSE NA
                  7 A TRUE 2
                  8 A TRUE 2
                  9 A FALSE NA
                  10 A TRUE 3
                  11 A TRUE 3
                  12 A TRUE 3
                  13 B NA NA
                  14 B FALSE NA
                  15 B TRUE 1
                  16 B TRUE 1
                  17 B TRUE 1
                  18 B FALSE NA







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 14 hours ago

























                  answered 16 hours ago









                  SotosSotos

                  31.4k51741




                  31.4k51741























                      4














                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer


























                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        16 hours ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        16 hours ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        15 hours ago
















                      4














                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer


























                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        16 hours ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        16 hours ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        15 hours ago














                      4












                      4








                      4







                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer















                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 12 hours ago

























                      answered 16 hours ago









                      markusmarkus

                      15.4k11336




                      15.4k11336













                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        16 hours ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        16 hours ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        15 hours ago



















                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        16 hours ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        16 hours ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        15 hours ago

















                      Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                      – Humpelstielzchen
                      16 hours ago





                      Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                      – Humpelstielzchen
                      16 hours ago




                      1




                      1





                      @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                      – markus
                      16 hours ago







                      @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                      – markus
                      16 hours ago















                      Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                      – Humpelstielzchen
                      15 hours ago





                      Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                      – Humpelstielzchen
                      15 hours ago


















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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029