efficient way to get the adjacency list of a network?











up vote
0
down vote

favorite












Consider this simple example



pd.DataFrame({'id' : [1,1,2,3,4],
'place' : ['bar','pool','bar','kitchen','bar']})

Out[4]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
4 4 bar


Here the network structure is such that a given id is connected to another id if they went to the same place.



For instance, here 1 is connected to 2 and 4 because they are at the bar.



1 and 3 are NOT connected because 1 went to bar and pool which does not include kitchen (the only place 3 went to)



My real data is huge, about 500k. What is the most efficient way to proceed to get the adjacency list? Here this is just a string with the format source target target like in https://networkx.github.io/documentation/networkx-1.10/reference/readwrite.adjlist.html



adjacency_list
1 2 4
2 1 4
4 1 2


Can we avoid loops and use Pandas tricks?



Thanks!










share|improve this question
























  • so if more than 2 id went to the same place, do we place them under the same edge_df row?
    – Rocky Li
    Nov 12 at 2:22










  • thanks for the comment. I meant the adjacency list indeed. let me edit the question
    – ℕʘʘḆḽḘ
    Nov 12 at 2:26










  • for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
    – W-B
    Nov 12 at 2:30










  • @W-B I dont think so. 1 only went to be and c while 2 went to d
    – ℕʘʘḆḽḘ
    Nov 12 at 2:37










  • df.groupby('place').id.unique()?
    – W-B
    Nov 12 at 2:39















up vote
0
down vote

favorite












Consider this simple example



pd.DataFrame({'id' : [1,1,2,3,4],
'place' : ['bar','pool','bar','kitchen','bar']})

Out[4]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
4 4 bar


Here the network structure is such that a given id is connected to another id if they went to the same place.



For instance, here 1 is connected to 2 and 4 because they are at the bar.



1 and 3 are NOT connected because 1 went to bar and pool which does not include kitchen (the only place 3 went to)



My real data is huge, about 500k. What is the most efficient way to proceed to get the adjacency list? Here this is just a string with the format source target target like in https://networkx.github.io/documentation/networkx-1.10/reference/readwrite.adjlist.html



adjacency_list
1 2 4
2 1 4
4 1 2


Can we avoid loops and use Pandas tricks?



Thanks!










share|improve this question
























  • so if more than 2 id went to the same place, do we place them under the same edge_df row?
    – Rocky Li
    Nov 12 at 2:22










  • thanks for the comment. I meant the adjacency list indeed. let me edit the question
    – ℕʘʘḆḽḘ
    Nov 12 at 2:26










  • for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
    – W-B
    Nov 12 at 2:30










  • @W-B I dont think so. 1 only went to be and c while 2 went to d
    – ℕʘʘḆḽḘ
    Nov 12 at 2:37










  • df.groupby('place').id.unique()?
    – W-B
    Nov 12 at 2:39













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider this simple example



pd.DataFrame({'id' : [1,1,2,3,4],
'place' : ['bar','pool','bar','kitchen','bar']})

Out[4]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
4 4 bar


Here the network structure is such that a given id is connected to another id if they went to the same place.



For instance, here 1 is connected to 2 and 4 because they are at the bar.



1 and 3 are NOT connected because 1 went to bar and pool which does not include kitchen (the only place 3 went to)



My real data is huge, about 500k. What is the most efficient way to proceed to get the adjacency list? Here this is just a string with the format source target target like in https://networkx.github.io/documentation/networkx-1.10/reference/readwrite.adjlist.html



adjacency_list
1 2 4
2 1 4
4 1 2


Can we avoid loops and use Pandas tricks?



Thanks!










share|improve this question















Consider this simple example



pd.DataFrame({'id' : [1,1,2,3,4],
'place' : ['bar','pool','bar','kitchen','bar']})

Out[4]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
4 4 bar


Here the network structure is such that a given id is connected to another id if they went to the same place.



For instance, here 1 is connected to 2 and 4 because they are at the bar.



1 and 3 are NOT connected because 1 went to bar and pool which does not include kitchen (the only place 3 went to)



My real data is huge, about 500k. What is the most efficient way to proceed to get the adjacency list? Here this is just a string with the format source target target like in https://networkx.github.io/documentation/networkx-1.10/reference/readwrite.adjlist.html



adjacency_list
1 2 4
2 1 4
4 1 2


Can we avoid loops and use Pandas tricks?



Thanks!







python pandas networkx






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 2:57

























asked Nov 12 at 2:17









ℕʘʘḆḽḘ

6,74794199




6,74794199












  • so if more than 2 id went to the same place, do we place them under the same edge_df row?
    – Rocky Li
    Nov 12 at 2:22










  • thanks for the comment. I meant the adjacency list indeed. let me edit the question
    – ℕʘʘḆḽḘ
    Nov 12 at 2:26










  • for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
    – W-B
    Nov 12 at 2:30










  • @W-B I dont think so. 1 only went to be and c while 2 went to d
    – ℕʘʘḆḽḘ
    Nov 12 at 2:37










  • df.groupby('place').id.unique()?
    – W-B
    Nov 12 at 2:39


















  • so if more than 2 id went to the same place, do we place them under the same edge_df row?
    – Rocky Li
    Nov 12 at 2:22










  • thanks for the comment. I meant the adjacency list indeed. let me edit the question
    – ℕʘʘḆḽḘ
    Nov 12 at 2:26










  • for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
    – W-B
    Nov 12 at 2:30










  • @W-B I dont think so. 1 only went to be and c while 2 went to d
    – ℕʘʘḆḽḘ
    Nov 12 at 2:37










  • df.groupby('place').id.unique()?
    – W-B
    Nov 12 at 2:39
















so if more than 2 id went to the same place, do we place them under the same edge_df row?
– Rocky Li
Nov 12 at 2:22




so if more than 2 id went to the same place, do we place them under the same edge_df row?
– Rocky Li
Nov 12 at 2:22












thanks for the comment. I meant the adjacency list indeed. let me edit the question
– ℕʘʘḆḽḘ
Nov 12 at 2:26




thanks for the comment. I meant the adjacency list indeed. let me edit the question
– ℕʘʘḆḽḘ
Nov 12 at 2:26












for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
– W-B
Nov 12 at 2:30




for example , 1 --b 1--c 2--d 3--c 3--d will 1 and 2 connected ?
– W-B
Nov 12 at 2:30












@W-B I dont think so. 1 only went to be and c while 2 went to d
– ℕʘʘḆḽḘ
Nov 12 at 2:37




@W-B I dont think so. 1 only went to be and c while 2 went to d
– ℕʘʘḆḽḘ
Nov 12 at 2:37












df.groupby('place').id.unique()?
– W-B
Nov 12 at 2:39




df.groupby('place').id.unique()?
– W-B
Nov 12 at 2:39












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Using unique then switch the column 0 to 1 and column 1 to 0 concat the both df together



adj=pd.DataFrame(df.groupby('place').id.unique().loc[lambda x : x.str.len()>1].tolist())
pd.concat([adj,adj.rename(columns={0:1,1:0})])
Out[810]:
0 1
0 1 2
0 2 1


Update :



newdf=df.merge(df,on='place')
x=nx.from_pandas_dataframe(newdf,'id_x','id_y') # using merge to get the connect for all id by link columns place.
[list(itertools.permutations(x, len(x)) for x in list(nx.connected_components(x))] # using permutations get the all combination for each connected_components in networkx
Out[821]: [[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]]


Data input



df
Out[822]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 bar





share|improve this answer























  • nice but would that work with multiple targets for the same source?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:44










  • @ℕʘʘḆḽḘ check the update
    – W-B
    Nov 12 at 2:53










  • sounds promising. do you mind explaining a bit whats going on with the update? thanks!
    – ℕʘʘḆḽḘ
    Nov 12 at 2:55










  • @ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
    – W-B
    Nov 12 at 2:59












  • well done bro, well done
    – ℕʘʘḆḽḘ
    Nov 12 at 3:10


















up vote
1
down vote













What about:



>>> df
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
>>> df.groupby('place').id.nunique().value_counts()
1 2
2 1
Name: id, dtype: int64





share|improve this answer





















  • not sure I understand whats going on here?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:52










  • @ℕʘʘḆḽḘ, W-B's answer is good.
    – pygo
    Nov 12 at 2:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Using unique then switch the column 0 to 1 and column 1 to 0 concat the both df together



adj=pd.DataFrame(df.groupby('place').id.unique().loc[lambda x : x.str.len()>1].tolist())
pd.concat([adj,adj.rename(columns={0:1,1:0})])
Out[810]:
0 1
0 1 2
0 2 1


Update :



newdf=df.merge(df,on='place')
x=nx.from_pandas_dataframe(newdf,'id_x','id_y') # using merge to get the connect for all id by link columns place.
[list(itertools.permutations(x, len(x)) for x in list(nx.connected_components(x))] # using permutations get the all combination for each connected_components in networkx
Out[821]: [[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]]


Data input



df
Out[822]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 bar





share|improve this answer























  • nice but would that work with multiple targets for the same source?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:44










  • @ℕʘʘḆḽḘ check the update
    – W-B
    Nov 12 at 2:53










  • sounds promising. do you mind explaining a bit whats going on with the update? thanks!
    – ℕʘʘḆḽḘ
    Nov 12 at 2:55










  • @ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
    – W-B
    Nov 12 at 2:59












  • well done bro, well done
    – ℕʘʘḆḽḘ
    Nov 12 at 3:10















up vote
1
down vote



accepted










Using unique then switch the column 0 to 1 and column 1 to 0 concat the both df together



adj=pd.DataFrame(df.groupby('place').id.unique().loc[lambda x : x.str.len()>1].tolist())
pd.concat([adj,adj.rename(columns={0:1,1:0})])
Out[810]:
0 1
0 1 2
0 2 1


Update :



newdf=df.merge(df,on='place')
x=nx.from_pandas_dataframe(newdf,'id_x','id_y') # using merge to get the connect for all id by link columns place.
[list(itertools.permutations(x, len(x)) for x in list(nx.connected_components(x))] # using permutations get the all combination for each connected_components in networkx
Out[821]: [[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]]


Data input



df
Out[822]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 bar





share|improve this answer























  • nice but would that work with multiple targets for the same source?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:44










  • @ℕʘʘḆḽḘ check the update
    – W-B
    Nov 12 at 2:53










  • sounds promising. do you mind explaining a bit whats going on with the update? thanks!
    – ℕʘʘḆḽḘ
    Nov 12 at 2:55










  • @ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
    – W-B
    Nov 12 at 2:59












  • well done bro, well done
    – ℕʘʘḆḽḘ
    Nov 12 at 3:10













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Using unique then switch the column 0 to 1 and column 1 to 0 concat the both df together



adj=pd.DataFrame(df.groupby('place').id.unique().loc[lambda x : x.str.len()>1].tolist())
pd.concat([adj,adj.rename(columns={0:1,1:0})])
Out[810]:
0 1
0 1 2
0 2 1


Update :



newdf=df.merge(df,on='place')
x=nx.from_pandas_dataframe(newdf,'id_x','id_y') # using merge to get the connect for all id by link columns place.
[list(itertools.permutations(x, len(x)) for x in list(nx.connected_components(x))] # using permutations get the all combination for each connected_components in networkx
Out[821]: [[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]]


Data input



df
Out[822]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 bar





share|improve this answer














Using unique then switch the column 0 to 1 and column 1 to 0 concat the both df together



adj=pd.DataFrame(df.groupby('place').id.unique().loc[lambda x : x.str.len()>1].tolist())
pd.concat([adj,adj.rename(columns={0:1,1:0})])
Out[810]:
0 1
0 1 2
0 2 1


Update :



newdf=df.merge(df,on='place')
x=nx.from_pandas_dataframe(newdf,'id_x','id_y') # using merge to get the connect for all id by link columns place.
[list(itertools.permutations(x, len(x)) for x in list(nx.connected_components(x))] # using permutations get the all combination for each connected_components in networkx
Out[821]: [[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)]]


Data input



df
Out[822]:
id place
0 1 bar
1 1 pool
2 2 bar
3 3 bar






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 at 2:56

























answered Nov 12 at 2:42









W-B

97.5k73162




97.5k73162












  • nice but would that work with multiple targets for the same source?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:44










  • @ℕʘʘḆḽḘ check the update
    – W-B
    Nov 12 at 2:53










  • sounds promising. do you mind explaining a bit whats going on with the update? thanks!
    – ℕʘʘḆḽḘ
    Nov 12 at 2:55










  • @ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
    – W-B
    Nov 12 at 2:59












  • well done bro, well done
    – ℕʘʘḆḽḘ
    Nov 12 at 3:10


















  • nice but would that work with multiple targets for the same source?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:44










  • @ℕʘʘḆḽḘ check the update
    – W-B
    Nov 12 at 2:53










  • sounds promising. do you mind explaining a bit whats going on with the update? thanks!
    – ℕʘʘḆḽḘ
    Nov 12 at 2:55










  • @ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
    – W-B
    Nov 12 at 2:59












  • well done bro, well done
    – ℕʘʘḆḽḘ
    Nov 12 at 3:10
















nice but would that work with multiple targets for the same source?
– ℕʘʘḆḽḘ
Nov 12 at 2:44




nice but would that work with multiple targets for the same source?
– ℕʘʘḆḽḘ
Nov 12 at 2:44












@ℕʘʘḆḽḘ check the update
– W-B
Nov 12 at 2:53




@ℕʘʘḆḽḘ check the update
– W-B
Nov 12 at 2:53












sounds promising. do you mind explaining a bit whats going on with the update? thanks!
– ℕʘʘḆḽḘ
Nov 12 at 2:55




sounds promising. do you mind explaining a bit whats going on with the update? thanks!
– ℕʘʘḆḽḘ
Nov 12 at 2:55












@ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
– W-B
Nov 12 at 2:59






@ℕʘʘḆḽḘ added , since if we need to create the network we need two column , you only have one column id and another is the link place ,we using the link to create another missing column , then we can using networkx
– W-B
Nov 12 at 2:59














well done bro, well done
– ℕʘʘḆḽḘ
Nov 12 at 3:10




well done bro, well done
– ℕʘʘḆḽḘ
Nov 12 at 3:10












up vote
1
down vote













What about:



>>> df
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
>>> df.groupby('place').id.nunique().value_counts()
1 2
2 1
Name: id, dtype: int64





share|improve this answer





















  • not sure I understand whats going on here?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:52










  • @ℕʘʘḆḽḘ, W-B's answer is good.
    – pygo
    Nov 12 at 2:54















up vote
1
down vote













What about:



>>> df
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
>>> df.groupby('place').id.nunique().value_counts()
1 2
2 1
Name: id, dtype: int64





share|improve this answer





















  • not sure I understand whats going on here?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:52










  • @ℕʘʘḆḽḘ, W-B's answer is good.
    – pygo
    Nov 12 at 2:54













up vote
1
down vote










up vote
1
down vote









What about:



>>> df
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
>>> df.groupby('place').id.nunique().value_counts()
1 2
2 1
Name: id, dtype: int64





share|improve this answer












What about:



>>> df
id place
0 1 bar
1 1 pool
2 2 bar
3 3 kitchen
>>> df.groupby('place').id.nunique().value_counts()
1 2
2 1
Name: id, dtype: int64






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 2:45









pygo

1,7361416




1,7361416












  • not sure I understand whats going on here?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:52










  • @ℕʘʘḆḽḘ, W-B's answer is good.
    – pygo
    Nov 12 at 2:54


















  • not sure I understand whats going on here?
    – ℕʘʘḆḽḘ
    Nov 12 at 2:52










  • @ℕʘʘḆḽḘ, W-B's answer is good.
    – pygo
    Nov 12 at 2:54
















not sure I understand whats going on here?
– ℕʘʘḆḽḘ
Nov 12 at 2:52




not sure I understand whats going on here?
– ℕʘʘḆḽḘ
Nov 12 at 2:52












@ℕʘʘḆḽḘ, W-B's answer is good.
– pygo
Nov 12 at 2:54




@ℕʘʘḆḽḘ, W-B's answer is good.
– pygo
Nov 12 at 2:54


















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