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!
python pandas networkx
|
show 2 more comments
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!
python pandas networkx
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 theadjacency 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
|
show 2 more comments
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!
python pandas networkx
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
python pandas networkx
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 theadjacency 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
|
show 2 more comments
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 theadjacency 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
|
show 2 more comments
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
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
add a comment |
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
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
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
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
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
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
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
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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