Pandas - append column with sum of even row values
3
I have a dataframe and I want a new column with sum of other columns in the same row. Criteria is I want Nan if the sum is odd and the sum if the value is even.
df = pd.DataFrame([[1, 8, 7, 2],
[8, 5, 9, 4],
[1, -5, 3, -2]], columns=list('pqrs'))
df
p q r s
0 1 8 7 2
1 8 5 9 4
2 1 -5 3 -2
Expected output:
p q r s t
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
python pandas dataframe sum
edited Nov 28 '18 at 5:35
Anand
116
add a comment |
3
I have a dataframe and I want a new column with sum of other columns in the same row. Criteria is I want Nan if the sum is odd and the sum if the value is even.
df = pd.DataFrame([[1, 8, 7, 2],
[8, 5, 9, 4],
[1, -5, 3, -2]], columns=list('pqrs'))
df
p q r s
0 1 8 7 2
1 8 5 9 4
2 1 -5 3 -2
Expected output:
p q r s t
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
python pandas dataframe sum
edited Nov 28 '18 at 5:35
Anand
116
add a comment |
3
3
3
I have a dataframe and I want a new column with sum of other columns in the same row. Criteria is I want Nan if the sum is odd and the sum if the value is even.
df = pd.DataFrame([[1, 8, 7, 2],
[8, 5, 9, 4],
[1, -5, 3, -2]], columns=list('pqrs'))
df
p q r s
0 1 8 7 2
1 8 5 9 4
2 1 -5 3 -2
Expected output:
p q r s t
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
python pandas dataframe sum
edited Nov 28 '18 at 5:35
Anand
116
I have a dataframe and I want a new column with sum of other columns in the same row. Criteria is I want Nan if the sum is odd and the sum if the value is even.
df = pd.DataFrame([[1, 8, 7, 2],
[8, 5, 9, 4],
[1, -5, 3, -2]], columns=list('pqrs'))
df
p q r s
0 1 8 7 2
1 8 5 9 4
2 1 -5 3 -2
Expected output:
p q r s t
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
python pandas dataframe sum
python pandas dataframe sum
edited Nov 28 '18 at 5:35
Anand
116
edited Nov 28 '18 at 5:35
Anand
116
edited Nov 28 '18 at 5:35
Anand
116
edited Nov 28 '18 at 5:35
Anand
116
edited Nov 28 '18 at 5:35
Anand
116
116
asked Nov 28 '18 at 5:22
user10715091
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1 Answer
1
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oldest
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2
Using np.where
df['new']=np.where(df.sum(1)%2==0,df.sum(1),np.nan)
df
p q r s new
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
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
Using np.where
df['new']=np.where(df.sum(1)%2==0,df.sum(1),np.nan)
df
p q r s new
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
add a comment |
2
Using np.where
df['new']=np.where(df.sum(1)%2==0,df.sum(1),np.nan)
df
p q r s new
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
add a comment |
2
2
2
Using np.where
df['new']=np.where(df.sum(1)%2==0,df.sum(1),np.nan)
df
p q r s new
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
Using np.where
df['new']=np.where(df.sum(1)%2==0,df.sum(1),np.nan)
df
p q r s new
0 1 8 7 2 18.0
1 8 5 9 4 26.0
2 1 -5 3 -2 NaN
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
answered Nov 28 '18 at 5:25
Wen-BenWen-Ben
117k83469
117k83469
add a comment |
add a comment |
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