python dataframe how to convert set column to list
2
I tried to convert a set column to list in python dataframe, but failed. Not sure what's best way to do so. Thanks.
Here is the example:
I tried to create a 'c' column which convert 'b' set column to list. but 'c' is still set.
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = list(tdf['b'])
tdf
a b c
0 [1, 2, 3] {33, 11, 22} {33, 11, 22}
1 [2, 3, 4] {222, 111} {222, 111}
python list dataframe set
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
add a comment |
2
I tried to convert a set column to list in python dataframe, but failed. Not sure what's best way to do so. Thanks.
Here is the example:
I tried to create a 'c' column which convert 'b' set column to list. but 'c' is still set.
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = list(tdf['b'])
tdf
a b c
0 [1, 2, 3] {33, 11, 22} {33, 11, 22}
1 [2, 3, 4] {222, 111} {222, 111}
python list dataframe set
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
1
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46
add a comment |
2
2
2
I tried to convert a set column to list in python dataframe, but failed. Not sure what's best way to do so. Thanks.
Here is the example:
I tried to create a 'c' column which convert 'b' set column to list. but 'c' is still set.
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = list(tdf['b'])
tdf
a b c
0 [1, 2, 3] {33, 11, 22} {33, 11, 22}
1 [2, 3, 4] {222, 111} {222, 111}
python list dataframe set
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
I tried to convert a set column to list in python dataframe, but failed. Not sure what's best way to do so. Thanks.
Here is the example:
I tried to create a 'c' column which convert 'b' set column to list. but 'c' is still set.
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = list(tdf['b'])
tdf
a b c
0 [1, 2, 3] {33, 11, 22} {33, 11, 22}
1 [2, 3, 4] {222, 111} {222, 111}
python list dataframe set
python list dataframe set
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
asked Dec 28 '18 at 23:49
user1828513user1828513
10039
10039
1
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46
add a comment |
1
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46
1
1
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46
add a comment |
2 Answers
2
active
oldest
votes
1
You could do:
import pandas as pd
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = [list(e) for e in tdf.b]
print(tdf)
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
1
@U9-Forward if you seemap,apply,lambda,np.vectorize(a complete misnomer)... it's running in Python time.
– roganjosh
Dec 28 '18 at 23:56
add a comment |
1
Use apply:
tdf['c'] = tdf['b'].apply(list)
Because using list is doing to whole column not one by one.
Or do:
tdf['c'] = tdf['b'].map(list)
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
1
You could do:
import pandas as pd
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = [list(e) for e in tdf.b]
print(tdf)
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
1
@U9-Forward if you seemap,apply,lambda,np.vectorize(a complete misnomer)... it's running in Python time.
– roganjosh
Dec 28 '18 at 23:56
add a comment |
1
You could do:
import pandas as pd
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = [list(e) for e in tdf.b]
print(tdf)
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
1
@U9-Forward if you seemap,apply,lambda,np.vectorize(a complete misnomer)... it's running in Python time.
– roganjosh
Dec 28 '18 at 23:56
add a comment |
1
1
1
You could do:
import pandas as pd
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = [list(e) for e in tdf.b]
print(tdf)
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
You could do:
import pandas as pd
data = [{'a': [1,2,3], 'b':{11,22,33}},{'a':[2,3,4],'b':{111,222}}]
tdf = pd.DataFrame(data)
tdf['c'] = [list(e) for e in tdf.b]
print(tdf)
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
answered Dec 28 '18 at 23:51
Daniel MesejoDaniel Mesejo
16k21130
16k21130
1
@U9-Forward if you seemap,apply,lambda,np.vectorize(a complete misnomer)... it's running in Python time.
– roganjosh
Dec 28 '18 at 23:56
add a comment |
1
@U9-Forward if you seemap,apply,lambda,np.vectorize(a complete misnomer)... it's running in Python time.
– roganjosh
Dec 28 '18 at 23:56
1
1
@U9-Forward if you see
map, apply, lambda, np.vectorize (a complete misnomer)... it's running in Python time.– roganjosh
Dec 28 '18 at 23:56
@U9-Forward if you see
map, apply, lambda, np.vectorize (a complete misnomer)... it's running in Python time.– roganjosh
Dec 28 '18 at 23:56
add a comment |
1
Use apply:
tdf['c'] = tdf['b'].apply(list)
Because using list is doing to whole column not one by one.
Or do:
tdf['c'] = tdf['b'].map(list)
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
add a comment |
1
Use apply:
tdf['c'] = tdf['b'].apply(list)
Because using list is doing to whole column not one by one.
Or do:
tdf['c'] = tdf['b'].map(list)
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
add a comment |
1
1
1
Use apply:
tdf['c'] = tdf['b'].apply(list)
Because using list is doing to whole column not one by one.
Or do:
tdf['c'] = tdf['b'].map(list)
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
Use apply:
tdf['c'] = tdf['b'].apply(list)
Because using list is doing to whole column not one by one.
Or do:
tdf['c'] = tdf['b'].map(list)
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
answered Dec 28 '18 at 23:51
U9-ForwardU9-Forward
14.1k21337
14.1k21337
add a comment |
add a comment |
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1
This should never be a dataframe in the first place. You have non-scalar values in columns, you're making life harder for yourself. You will get no benefit of pandas at all through this structure.
– roganjosh
Dec 28 '18 at 23:51
Related meta: Generic “Don't Do It” Answer
– jpp
Dec 29 '18 at 0:46