Fastest way to perform complex search on pandas dataframe





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15

















I am trying to figure out the fastest way to perform search and sort on a pandas dataframe. Below are before and after dataframes of what I am trying to accomplish.



Before:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC DEF 123 456 8000 8000
DEF XYZ 456 893 9999 9999
AAA BBB 473 917 5555 5555
BBB CCC 917 341 5555 5555


After search/sort:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC XYZ 123 893 8000 9999
AAA CCC 473 341 5555 5555


In this example I am essentially trying to filter out 'flights' that exist in between end destinations. This should be done by using some sort of drop duplicates method but what leaves me confused is how to handle all of the columns. Would a binary search be the best way to accomplish this? Hints appreciated, trying hard to figure this out.



possible edge case:



What if the data is switched up and our end connections are in the same column?



flight1  flight2      1Num    2Num     1Code   2Code
ABC DEF 123 456 8000 8000
XYZ DEF 893 456 9999 9999


After search/sort:



flight1  flight2      1Num    2Num     1Code   2Code
ABC XYZ 123 893 8000 9999


This case logically shouldn't happen. After all how can you go DEF-ABC and DEF-XYZ? You can't, but the 'endpoints' would still be ABC-XYZ










share|improve this question





























  • Are the connecting flights always adjacent in the data frame?

    – Mike
    May 28 at 14:14











  • np.where(condition)

    – Dadu Khan
    May 28 at 14:14











  • how about df['flightFrom'].shift() != df['fightTo']?

    – IanS
    May 28 at 14:17













  • @Mike the information can be completely random in the DataFrame

    – MaxB
    May 28 at 14:18






  • 1





    @IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

    – Erfan
    May 28 at 14:26




















15

















I am trying to figure out the fastest way to perform search and sort on a pandas dataframe. Below are before and after dataframes of what I am trying to accomplish.



Before:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC DEF 123 456 8000 8000
DEF XYZ 456 893 9999 9999
AAA BBB 473 917 5555 5555
BBB CCC 917 341 5555 5555


After search/sort:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC XYZ 123 893 8000 9999
AAA CCC 473 341 5555 5555


In this example I am essentially trying to filter out 'flights' that exist in between end destinations. This should be done by using some sort of drop duplicates method but what leaves me confused is how to handle all of the columns. Would a binary search be the best way to accomplish this? Hints appreciated, trying hard to figure this out.



possible edge case:



What if the data is switched up and our end connections are in the same column?



flight1  flight2      1Num    2Num     1Code   2Code
ABC DEF 123 456 8000 8000
XYZ DEF 893 456 9999 9999


After search/sort:



flight1  flight2      1Num    2Num     1Code   2Code
ABC XYZ 123 893 8000 9999


This case logically shouldn't happen. After all how can you go DEF-ABC and DEF-XYZ? You can't, but the 'endpoints' would still be ABC-XYZ










share|improve this question





























  • Are the connecting flights always adjacent in the data frame?

    – Mike
    May 28 at 14:14











  • np.where(condition)

    – Dadu Khan
    May 28 at 14:14











  • how about df['flightFrom'].shift() != df['fightTo']?

    – IanS
    May 28 at 14:17













  • @Mike the information can be completely random in the DataFrame

    – MaxB
    May 28 at 14:18






  • 1





    @IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

    – Erfan
    May 28 at 14:26
















15












15








15


9






I am trying to figure out the fastest way to perform search and sort on a pandas dataframe. Below are before and after dataframes of what I am trying to accomplish.



Before:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC DEF 123 456 8000 8000
DEF XYZ 456 893 9999 9999
AAA BBB 473 917 5555 5555
BBB CCC 917 341 5555 5555


After search/sort:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC XYZ 123 893 8000 9999
AAA CCC 473 341 5555 5555


In this example I am essentially trying to filter out 'flights' that exist in between end destinations. This should be done by using some sort of drop duplicates method but what leaves me confused is how to handle all of the columns. Would a binary search be the best way to accomplish this? Hints appreciated, trying hard to figure this out.



possible edge case:



What if the data is switched up and our end connections are in the same column?



flight1  flight2      1Num    2Num     1Code   2Code
ABC DEF 123 456 8000 8000
XYZ DEF 893 456 9999 9999


After search/sort:



flight1  flight2      1Num    2Num     1Code   2Code
ABC XYZ 123 893 8000 9999


This case logically shouldn't happen. After all how can you go DEF-ABC and DEF-XYZ? You can't, but the 'endpoints' would still be ABC-XYZ










share|improve this question

















I am trying to figure out the fastest way to perform search and sort on a pandas dataframe. Below are before and after dataframes of what I am trying to accomplish.



Before:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC DEF 123 456 8000 8000
DEF XYZ 456 893 9999 9999
AAA BBB 473 917 5555 5555
BBB CCC 917 341 5555 5555


After search/sort:



flightTo  flightFrom  toNum  fromNum  toCode  fromCode
ABC XYZ 123 893 8000 9999
AAA CCC 473 341 5555 5555


In this example I am essentially trying to filter out 'flights' that exist in between end destinations. This should be done by using some sort of drop duplicates method but what leaves me confused is how to handle all of the columns. Would a binary search be the best way to accomplish this? Hints appreciated, trying hard to figure this out.



possible edge case:



What if the data is switched up and our end connections are in the same column?



flight1  flight2      1Num    2Num     1Code   2Code
ABC DEF 123 456 8000 8000
XYZ DEF 893 456 9999 9999


After search/sort:



flight1  flight2      1Num    2Num     1Code   2Code
ABC XYZ 123 893 8000 9999


This case logically shouldn't happen. After all how can you go DEF-ABC and DEF-XYZ? You can't, but the 'endpoints' would still be ABC-XYZ







python pandas binary-search-tree






share|improve this question
















share|improve this question













share|improve this question




share|improve this question








edited May 28 at 18:40







MaxB

















asked May 28 at 14:07









MaxBMaxB

2031 silver badge12 bronze badges




2031 silver badge12 bronze badges
















  • Are the connecting flights always adjacent in the data frame?

    – Mike
    May 28 at 14:14











  • np.where(condition)

    – Dadu Khan
    May 28 at 14:14











  • how about df['flightFrom'].shift() != df['fightTo']?

    – IanS
    May 28 at 14:17













  • @Mike the information can be completely random in the DataFrame

    – MaxB
    May 28 at 14:18






  • 1





    @IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

    – Erfan
    May 28 at 14:26





















  • Are the connecting flights always adjacent in the data frame?

    – Mike
    May 28 at 14:14











  • np.where(condition)

    – Dadu Khan
    May 28 at 14:14











  • how about df['flightFrom'].shift() != df['fightTo']?

    – IanS
    May 28 at 14:17













  • @Mike the information can be completely random in the DataFrame

    – MaxB
    May 28 at 14:18






  • 1





    @IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

    – Erfan
    May 28 at 14:26



















Are the connecting flights always adjacent in the data frame?

– Mike
May 28 at 14:14





Are the connecting flights always adjacent in the data frame?

– Mike
May 28 at 14:14













np.where(condition)

– Dadu Khan
May 28 at 14:14





np.where(condition)

– Dadu Khan
May 28 at 14:14













how about df['flightFrom'].shift() != df['fightTo']?

– IanS
May 28 at 14:17







how about df['flightFrom'].shift() != df['fightTo']?

– IanS
May 28 at 14:17















@Mike the information can be completely random in the DataFrame

– MaxB
May 28 at 14:18





@Mike the information can be completely random in the DataFrame

– MaxB
May 28 at 14:18




1




1





@IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

– Erfan
May 28 at 14:26







@IanS check the values in fromNum, fromCode expected output, that's what makes this question complex imo.

– Erfan
May 28 at 14:26














2 Answers
2






active

oldest

votes


















14


















This is network problem , so we using networkx , notice , here you can have more than two stops , which means you can have some case like NY-DC-WA-NC



import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')

# create the nx object from pandas dataframe

l=list(nx.connected_components(G))

# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]

# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others

d={k: v for d in L for k, v in d.items()}

# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output

Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555


Installation networkx





  • Pip: pip install networkx


  • Anaconda: conda install -c anaconda networkx






share|improve this answer
























  • 2





    great answer! Looked into networkx couple times, will do more now!

    – Erfan
    May 28 at 14:21






  • 2





    @Erfan love the enthusiasm ;) same here(for networkx)

    – anky_91
    May 28 at 14:22






  • 2





    This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

    – Erfan
    May 28 at 14:24








  • 1





    @Erfan ok let me working on it

    – WeNYoBen
    May 28 at 14:24






  • 1





    Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

    – Prayson W. Daniel
    May 28 at 15:40



















6


















Here's a NumPy solution, which might be convenient in the case performance is relevant:



def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)




Which fo the proposed dataframe yields the expected output:



print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555


This approach does not assume any particular order in terms of the rows where the duplicate is, and the same applies to the columns (to cover the edge case described in the question). If we use for instance the following dataframe:



    flightTo flightFrom  toNum  fromNum  toCode  fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555





share|improve this answer





























  • Would this generalize to the case where the flights are randomly distributed over the dataframe?

    – Erfan
    May 28 at 14:38











  • I think the only problem is //2

    – WeNYoBen
    May 28 at 14:48













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









14


















This is network problem , so we using networkx , notice , here you can have more than two stops , which means you can have some case like NY-DC-WA-NC



import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')

# create the nx object from pandas dataframe

l=list(nx.connected_components(G))

# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]

# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others

d={k: v for d in L for k, v in d.items()}

# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output

Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555


Installation networkx





  • Pip: pip install networkx


  • Anaconda: conda install -c anaconda networkx






share|improve this answer
























  • 2





    great answer! Looked into networkx couple times, will do more now!

    – Erfan
    May 28 at 14:21






  • 2





    @Erfan love the enthusiasm ;) same here(for networkx)

    – anky_91
    May 28 at 14:22






  • 2





    This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

    – Erfan
    May 28 at 14:24








  • 1





    @Erfan ok let me working on it

    – WeNYoBen
    May 28 at 14:24






  • 1





    Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

    – Prayson W. Daniel
    May 28 at 15:40
















14


















This is network problem , so we using networkx , notice , here you can have more than two stops , which means you can have some case like NY-DC-WA-NC



import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')

# create the nx object from pandas dataframe

l=list(nx.connected_components(G))

# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]

# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others

d={k: v for d in L for k, v in d.items()}

# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output

Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555


Installation networkx





  • Pip: pip install networkx


  • Anaconda: conda install -c anaconda networkx






share|improve this answer
























  • 2





    great answer! Looked into networkx couple times, will do more now!

    – Erfan
    May 28 at 14:21






  • 2





    @Erfan love the enthusiasm ;) same here(for networkx)

    – anky_91
    May 28 at 14:22






  • 2





    This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

    – Erfan
    May 28 at 14:24








  • 1





    @Erfan ok let me working on it

    – WeNYoBen
    May 28 at 14:24






  • 1





    Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

    – Prayson W. Daniel
    May 28 at 15:40














14














14










14









This is network problem , so we using networkx , notice , here you can have more than two stops , which means you can have some case like NY-DC-WA-NC



import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')

# create the nx object from pandas dataframe

l=list(nx.connected_components(G))

# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]

# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others

d={k: v for d in L for k, v in d.items()}

# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output

Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555


Installation networkx





  • Pip: pip install networkx


  • Anaconda: conda install -c anaconda networkx






share|improve this answer
















This is network problem , so we using networkx , notice , here you can have more than two stops , which means you can have some case like NY-DC-WA-NC



import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')

# create the nx object from pandas dataframe

l=list(nx.connected_components(G))

# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]

# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others

d={k: v for d in L for k, v in d.items()}

# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output

Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555


Installation networkx





  • Pip: pip install networkx


  • Anaconda: conda install -c anaconda networkx







share|improve this answer















share|improve this answer




share|improve this answer








edited May 28 at 18:18

























answered May 28 at 14:19









WeNYoBenWeNYoBen

162k9 gold badges57 silver badges92 bronze badges




162k9 gold badges57 silver badges92 bronze badges











  • 2





    great answer! Looked into networkx couple times, will do more now!

    – Erfan
    May 28 at 14:21






  • 2





    @Erfan love the enthusiasm ;) same here(for networkx)

    – anky_91
    May 28 at 14:22






  • 2





    This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

    – Erfan
    May 28 at 14:24








  • 1





    @Erfan ok let me working on it

    – WeNYoBen
    May 28 at 14:24






  • 1





    Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

    – Prayson W. Daniel
    May 28 at 15:40














  • 2





    great answer! Looked into networkx couple times, will do more now!

    – Erfan
    May 28 at 14:21






  • 2





    @Erfan love the enthusiasm ;) same here(for networkx)

    – anky_91
    May 28 at 14:22






  • 2





    This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

    – Erfan
    May 28 at 14:24








  • 1





    @Erfan ok let me working on it

    – WeNYoBen
    May 28 at 14:24






  • 1





    Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

    – Prayson W. Daniel
    May 28 at 15:40








2




2





great answer! Looked into networkx couple times, will do more now!

– Erfan
May 28 at 14:21





great answer! Looked into networkx couple times, will do more now!

– Erfan
May 28 at 14:21




2




2





@Erfan love the enthusiasm ;) same here(for networkx)

– anky_91
May 28 at 14:22





@Erfan love the enthusiasm ;) same here(for networkx)

– anky_91
May 28 at 14:22




2




2





This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

– Erfan
May 28 at 14:24







This answer deserves to be broken down in more explanation :) (so I can learn from it hehe)

– Erfan
May 28 at 14:24






1




1





@Erfan ok let me working on it

– WeNYoBen
May 28 at 14:24





@Erfan ok let me working on it

– WeNYoBen
May 28 at 14:24




1




1





Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

– Prayson W. Daniel
May 28 at 15:40





Best answer I have read. Is it possible to edit variables, using information names, instead of letters, and expands the solution. Or best write a post/article on medium(or other place) explaining this methodology

– Prayson W. Daniel
May 28 at 15:40













6


















Here's a NumPy solution, which might be convenient in the case performance is relevant:



def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)




Which fo the proposed dataframe yields the expected output:



print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555


This approach does not assume any particular order in terms of the rows where the duplicate is, and the same applies to the columns (to cover the edge case described in the question). If we use for instance the following dataframe:



    flightTo flightFrom  toNum  fromNum  toCode  fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555





share|improve this answer





























  • Would this generalize to the case where the flights are randomly distributed over the dataframe?

    – Erfan
    May 28 at 14:38











  • I think the only problem is //2

    – WeNYoBen
    May 28 at 14:48
















6


















Here's a NumPy solution, which might be convenient in the case performance is relevant:



def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)




Which fo the proposed dataframe yields the expected output:



print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555


This approach does not assume any particular order in terms of the rows where the duplicate is, and the same applies to the columns (to cover the edge case described in the question). If we use for instance the following dataframe:



    flightTo flightFrom  toNum  fromNum  toCode  fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555





share|improve this answer





























  • Would this generalize to the case where the flights are randomly distributed over the dataframe?

    – Erfan
    May 28 at 14:38











  • I think the only problem is //2

    – WeNYoBen
    May 28 at 14:48














6














6










6









Here's a NumPy solution, which might be convenient in the case performance is relevant:



def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)




Which fo the proposed dataframe yields the expected output:



print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555


This approach does not assume any particular order in terms of the rows where the duplicate is, and the same applies to the columns (to cover the edge case described in the question). If we use for instance the following dataframe:



    flightTo flightFrom  toNum  fromNum  toCode  fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555





share|improve this answer
















Here's a NumPy solution, which might be convenient in the case performance is relevant:



def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)




Which fo the proposed dataframe yields the expected output:



print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555


This approach does not assume any particular order in terms of the rows where the duplicate is, and the same applies to the columns (to cover the edge case described in the question). If we use for instance the following dataframe:



    flightTo flightFrom  toNum  fromNum  toCode  fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555

remove_middle_dest(df)

flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555






share|improve this answer















share|improve this answer




share|improve this answer








edited May 29 at 12:35

























answered May 28 at 14:32









yatuyatu

35.7k6 gold badges27 silver badges58 bronze badges




35.7k6 gold badges27 silver badges58 bronze badges
















  • Would this generalize to the case where the flights are randomly distributed over the dataframe?

    – Erfan
    May 28 at 14:38











  • I think the only problem is //2

    – WeNYoBen
    May 28 at 14:48



















  • Would this generalize to the case where the flights are randomly distributed over the dataframe?

    – Erfan
    May 28 at 14:38











  • I think the only problem is //2

    – WeNYoBen
    May 28 at 14:48

















Would this generalize to the case where the flights are randomly distributed over the dataframe?

– Erfan
May 28 at 14:38





Would this generalize to the case where the flights are randomly distributed over the dataframe?

– Erfan
May 28 at 14:38













I think the only problem is //2

– WeNYoBen
May 28 at 14:48





I think the only problem is //2

– WeNYoBen
May 28 at 14:48



















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