Fastest way to perform complex search on pandas dataframe
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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
add a comment
|
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
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 aboutdf['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 infromNum, fromCode
expected output, that's what makes this question complex imo.
– Erfan
May 28 at 14:26
add a comment
|
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
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
python pandas binary-search-tree
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 aboutdf['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 infromNum, fromCode
expected output, that's what makes this question complex imo.
– Erfan
May 28 at 14:26
add a comment
|
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 aboutdf['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 infromNum, 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
add a comment
|
2 Answers
2
active
oldest
votes
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
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
|
show 5 more comments
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
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
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
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
|
show 5 more comments
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
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
|
show 5 more comments
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
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
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
|
show 5 more comments
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
|
show 5 more comments
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
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
add a comment
|
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
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
add a comment
|
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
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
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
add a comment
|
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
add a comment
|
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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