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Aggregate method and lambda function
2019 Community Moderator ElectionDifference between append vs. extend list methods in PythonUsing global variables in a functionUnderstanding Python super() with __init__() methodsHow to make a chain of function decorators?Does Python have a string 'contains' substring method?Adding new column to existing DataFrame in Python pandasHow can I replace all the NaN values with Zero's in a column of a pandas dataframe“Large data” work flows using pandasChange data type of columns in PandasGet list from pandas DataFrame column headers
I have a dataframe df
and would like to get the ID
list for each class:
import pandas as pd
list1 = [[1, ('A', 'B', 'C')], [2, ('M', 'E', 'F')], [3, ('H', 'A', 'D')], [4, ('H', 'A', 'D')]]
df = pd.DataFrame(list1, columns=['ID', 'Class'])
df
looks like
ID Class
1 (A, B, C)
2 (M, E, F)
3 (H, A, D)
4 (H, A, D)
The desired output is:
Class ID
(A, B, C) [1]
(M, E, F) [2]
(H, A, D) [3, 4]
I tried to use the code below, but it is not working. Thanks.
df.groupby('Class')['ID'].aggregate(lambda x: list(x))
python pandas
add a comment |
I have a dataframe df
and would like to get the ID
list for each class:
import pandas as pd
list1 = [[1, ('A', 'B', 'C')], [2, ('M', 'E', 'F')], [3, ('H', 'A', 'D')], [4, ('H', 'A', 'D')]]
df = pd.DataFrame(list1, columns=['ID', 'Class'])
df
looks like
ID Class
1 (A, B, C)
2 (M, E, F)
3 (H, A, D)
4 (H, A, D)
The desired output is:
Class ID
(A, B, C) [1]
(M, E, F) [2]
(H, A, D) [3, 4]
I tried to use the code below, but it is not working. Thanks.
df.groupby('Class')['ID'].aggregate(lambda x: list(x))
python pandas
1
lambda x: list(x)
is equivalent tolist
, but with more runtime overhead.
– chepner
Mar 6 at 16:34
add a comment |
I have a dataframe df
and would like to get the ID
list for each class:
import pandas as pd
list1 = [[1, ('A', 'B', 'C')], [2, ('M', 'E', 'F')], [3, ('H', 'A', 'D')], [4, ('H', 'A', 'D')]]
df = pd.DataFrame(list1, columns=['ID', 'Class'])
df
looks like
ID Class
1 (A, B, C)
2 (M, E, F)
3 (H, A, D)
4 (H, A, D)
The desired output is:
Class ID
(A, B, C) [1]
(M, E, F) [2]
(H, A, D) [3, 4]
I tried to use the code below, but it is not working. Thanks.
df.groupby('Class')['ID'].aggregate(lambda x: list(x))
python pandas
I have a dataframe df
and would like to get the ID
list for each class:
import pandas as pd
list1 = [[1, ('A', 'B', 'C')], [2, ('M', 'E', 'F')], [3, ('H', 'A', 'D')], [4, ('H', 'A', 'D')]]
df = pd.DataFrame(list1, columns=['ID', 'Class'])
df
looks like
ID Class
1 (A, B, C)
2 (M, E, F)
3 (H, A, D)
4 (H, A, D)
The desired output is:
Class ID
(A, B, C) [1]
(M, E, F) [2]
(H, A, D) [3, 4]
I tried to use the code below, but it is not working. Thanks.
df.groupby('Class')['ID'].aggregate(lambda x: list(x))
python pandas
python pandas
asked Mar 6 at 16:29
Steve PSteve P
686
686
1
lambda x: list(x)
is equivalent tolist
, but with more runtime overhead.
– chepner
Mar 6 at 16:34
add a comment |
1
lambda x: list(x)
is equivalent tolist
, but with more runtime overhead.
– chepner
Mar 6 at 16:34
1
1
lambda x: list(x)
is equivalent to list
, but with more runtime overhead.– chepner
Mar 6 at 16:34
lambda x: list(x)
is equivalent to list
, but with more runtime overhead.– chepner
Mar 6 at 16:34
add a comment |
2 Answers
2
active
oldest
votes
lot of solutions here one:
df = df.groupby('Class')['ID'].apply(list).reset_index()
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
add a comment |
You should using
df.groupby('Class').ID.agg(list).reset_index()
Out[473]:
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
To fix your code
df.groupby('Class')['ID'].aggregate(lambda x: x.tolist())
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
|
show 5 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
lot of solutions here one:
df = df.groupby('Class')['ID'].apply(list).reset_index()
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
add a comment |
lot of solutions here one:
df = df.groupby('Class')['ID'].apply(list).reset_index()
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
add a comment |
lot of solutions here one:
df = df.groupby('Class')['ID'].apply(list).reset_index()
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
lot of solutions here one:
df = df.groupby('Class')['ID'].apply(list).reset_index()
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
answered Mar 6 at 16:36
FrenchyFrenchy
1,3422414
1,3422414
add a comment |
add a comment |
You should using
df.groupby('Class').ID.agg(list).reset_index()
Out[473]:
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
To fix your code
df.groupby('Class')['ID'].aggregate(lambda x: x.tolist())
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
|
show 5 more comments
You should using
df.groupby('Class').ID.agg(list).reset_index()
Out[473]:
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
To fix your code
df.groupby('Class')['ID'].aggregate(lambda x: x.tolist())
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
|
show 5 more comments
You should using
df.groupby('Class').ID.agg(list).reset_index()
Out[473]:
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
To fix your code
df.groupby('Class')['ID'].aggregate(lambda x: x.tolist())
You should using
df.groupby('Class').ID.agg(list).reset_index()
Out[473]:
Class ID
0 (A, B, C) [1]
1 (H, A, D) [3, 4]
2 (M, E, F) [2]
To fix your code
df.groupby('Class')['ID'].aggregate(lambda x: x.tolist())
answered Mar 6 at 16:32
Wen-BenWen-Ben
117k83369
117k83369
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
|
show 5 more comments
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
Hi Wenben, I got some error message. When I run df.groupby('Class').ID.agg(list).reset_index(), it says TypeError: 'type' object is not iterable.
– Steve P
Mar 6 at 16:45
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
When I run df.groupby('Class')['ID'].aggregate(lambda x: x.tolist()), it says ValueError: Length of names must match number of levels in MultiIndex.
– Steve P
Mar 6 at 16:46
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
@SteveP make sure that is tuple or list? Incolumn class
– Wen-Ben
Mar 6 at 16:50
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
Hi Wen-Ben, column Class such as (A, B, C) is a tuple.
– Steve P
Mar 6 at 16:51
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
@SteveP if it is same like what you show to us , my query should work
– Wen-Ben
Mar 6 at 17:06
|
show 5 more comments
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1
lambda x: list(x)
is equivalent tolist
, but with more runtime overhead.– chepner
Mar 6 at 16:34