Given the following data frame:
a = pd.DataFrame({'A': [1,2], 'B': [4,0], 'C': [1,2]})
a
A B C
0 1 4 1
1 2 0 2
I would like to create a new column D containing the non-null values (per row) separated by columns. Like this:
A B C D
0 1 4 1 1,4,1
1 2 0 2 1,0,2
In reality, I will have many columns. Thanks in advance!
An alternative:
a['D'] = a.apply(lambda row: ','.join(row.dropna()
.astype(int).astype(str)), axis=1)
print(a)
A B C D
0 1 4 1 1,4,1
1 2 0 2 2,0,2
apply with a lambda and a loop?# example data with NaN values
a = pd.DataFrame({'A': [np.nan,2], 'B': [4,np.nan], 'C': [1,2]})
a
A B C
0 NaN 4.0 1
1 2.0 NaN 2
# make new column with non-null values
a['D'] = a.apply(lambda x: [val for val in x if not np.isnan(val)], axis=1)
a
A B C D
0 NaN 4.0 1 [4.0, 1.0]
1 2.0 NaN 2 [2.0, 2.0]
a = pd.DataFrame({'A': [1,2], 'B': [4,0], 'C': [1,2]}, dtype=float)?ValueError: Wrong number of items passed 3, placement implies 1. I knocked it around a bit just now but couldn't figure out why it barfs when all values aren't NaN. I'll have a look later. Good observation, any ideas?You can do something along the lines of the following:
combVals = []
a = a.T
for col in a.columns:
combVals.append(str(a[col].dropna().astype(int).tolist())[1:-1])
a = a.T
a['D'] = combVals
print(a)
A B C D
0 1 4 1 1, 4, 1
1 2 0 2 2, 0, 2
You can remove the spaces in column D by doing: a['D'] = a['D'].str.replace(' ','')
df.apply(lambda x :','.join(x.astype(str)),axis=1)