How to make a column with lists from columns of list elements in a pandas dataframe?
1
I have a pandas dataframe like
test = pd.DataFrame([[['P','N'], ['Z', 'P']],[['N','N'], ['Z', 'P']]],
columns=['c1', 'c2'])
I want to add another column c3 to test whose elements are
['PZ', 'NP']
['NZ', 'NP']
How can I do this?
python-3.x pandas dataframe
edited Nov 14 '18 at 20:19
Malik Asad
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
add a comment |
1
I have a pandas dataframe like
test = pd.DataFrame([[['P','N'], ['Z', 'P']],[['N','N'], ['Z', 'P']]],
columns=['c1', 'c2'])
I want to add another column c3 to test whose elements are
['PZ', 'NP']
['NZ', 'NP']
How can I do this?
python-3.x pandas dataframe
edited Nov 14 '18 at 20:19
Malik Asad
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
add a comment |
1
1
1
I have a pandas dataframe like
test = pd.DataFrame([[['P','N'], ['Z', 'P']],[['N','N'], ['Z', 'P']]],
columns=['c1', 'c2'])
I want to add another column c3 to test whose elements are
['PZ', 'NP']
['NZ', 'NP']
How can I do this?
python-3.x pandas dataframe
edited Nov 14 '18 at 20:19
Malik Asad
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
I have a pandas dataframe like
test = pd.DataFrame([[['P','N'], ['Z', 'P']],[['N','N'], ['Z', 'P']]],
columns=['c1', 'c2'])
I want to add another column c3 to test whose elements are
['PZ', 'NP']
['NZ', 'NP']
How can I do this?
python-3.x pandas dataframe
python-3.x pandas dataframe
edited Nov 14 '18 at 20:19
Malik Asad
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
edited Nov 14 '18 at 20:19
Malik Asad
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
edited Nov 14 '18 at 20:19
Malik Asad
316112
edited Nov 14 '18 at 20:19
Malik Asad
316112
edited Nov 14 '18 at 20:19
Malik Asad
316112
316112
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
asked Nov 14 '18 at 16:45
anotheroneanotherone
411417
411417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
Use assign:
df = test.assign(c3 = [[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
Or:
df = test.assign(c3 = list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
print(df)
c1 c2 c3
0 [P, N] [Z, P] [PZ, NP]
1 [N, N] [Z, P] [NZ, NP]
print([[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
[['PZ', 'NP'], ['NZ', 'NP']]
print(list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
[['PZ', 'NP'], ['NZ', 'NP']]
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Use assign:
df = test.assign(c3 = [[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
Or:
df = test.assign(c3 = list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
print(df)
c1 c2 c3
0 [P, N] [Z, P] [PZ, NP]
1 [N, N] [Z, P] [NZ, NP]
print([[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
[['PZ', 'NP'], ['NZ', 'NP']]
print(list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
[['PZ', 'NP'], ['NZ', 'NP']]
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
add a comment |
2
Use assign:
df = test.assign(c3 = [[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
Or:
df = test.assign(c3 = list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
print(df)
c1 c2 c3
0 [P, N] [Z, P] [PZ, NP]
1 [N, N] [Z, P] [NZ, NP]
print([[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
[['PZ', 'NP'], ['NZ', 'NP']]
print(list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
[['PZ', 'NP'], ['NZ', 'NP']]
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
add a comment |
2
2
2
Use assign:
df = test.assign(c3 = [[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
Or:
df = test.assign(c3 = list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
print(df)
c1 c2 c3
0 [P, N] [Z, P] [PZ, NP]
1 [N, N] [Z, P] [NZ, NP]
print([[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
[['PZ', 'NP'], ['NZ', 'NP']]
print(list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
[['PZ', 'NP'], ['NZ', 'NP']]
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
Use assign:
df = test.assign(c3 = [[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
Or:
df = test.assign(c3 = list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
print(df)
c1 c2 c3
0 [P, N] [Z, P] [PZ, NP]
1 [N, N] [Z, P] [NZ, NP]
print([[x[0]+y[0], x[1]+y[1]] for x,y in test.values.tolist()])
[['PZ', 'NP'], ['NZ', 'NP']]
print(list(map(list,zip(test.c1.str[0]+test.c2.str[0],test.c1.str[1]+test.c2.str[1]))))
[['PZ', 'NP'], ['NZ', 'NP']]
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
edited Nov 15 '18 at 4:36
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
answered Nov 14 '18 at 16:49
Sandeep KadapaSandeep Kadapa
7,388831
7,388831
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
add a comment |
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
thanks! was just wondering if there is any sensible way to avoid the loop? maybe for better performance..
– anotherone
Nov 14 '18 at 20:26
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
@anotherone Check the update
– Sandeep Kadapa
Nov 15 '18 at 3:46
1
1
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
thanks again. interesting way...
– anotherone
Nov 15 '18 at 8:54
add a comment |
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