How to implement sql coalesce in pandas

Question

I have a data frame like

df = pd.DataFrame({\"A\":[1,2,np.nan],\"B\":[np.nan,10,np.nan], \"C\":[5,10,7]})
     A     B   C
0  1.0   NaN   5
1  2.0  10.0  10         


        

Answer 1

Another way is to explicitly fill column D with A,B,C in that order.

df['D'] = np.nan
df['D'] = df.D.fillna(df.A).fillna(df.B).fillna(df.C)

Answer 2

Another approach is to use the combine_first method of a pd.Series. Using your example df,

>>> import pandas as pd
>>> import numpy as np
>>> df = pd.DataFrame({"A":[1,2,np.nan],"B":[np.nan,10,np.nan], "C":[5,10,7]})
>>> df
     A     B   C
0  1.0   NaN   5
1  2.0  10.0  10
2  NaN   NaN   7

we have

>>> df.A.combine_first(df.B).combine_first(df.C)
0    1.0
1    2.0
2    7.0

We can use reduce to abstract this pattern to work with an arbitrary number of columns.

>>> from functools import reduce
>>> cols = [df[c] for c in df.columns]
>>> reduce(lambda acc, col: acc.combine_first(col), cols)
0    1.0
1    2.0
2    7.0
Name: A, dtype: float64

Let's put this all together in a function.

>>> def coalesce(*args):
...     return reduce(lambda acc, col: acc.combine_first(col), args)
...
>>> coalesce(*cols)
0    1.0
1    2.0
2    7.0
Name: A, dtype: float64

Answer 3

I think you need bfill with selecting first column by iloc:

df['D'] = df.bfill(axis=1).iloc[:,0]
print (df)
     A     B   C    D
0  1.0   NaN   5  1.0
1  2.0  10.0  10  2.0
2  NaN   NaN   7  7.0

same as:

df['D'] = df.fillna(method='bfill',axis=1).iloc[:,0]
print (df)
     A     B   C    D
0  1.0   NaN   5  1.0
1  2.0  10.0  10  2.0
2  NaN   NaN   7  7.0

Answer 4

There is already a method for Series in Pandas that does this:

df['D'] = df['A'].combine_first(df['C'])

Or just stack them if you want to look up values sequentially:

df['D'] = df['A'].combine_first(df['B']).combine_first(df['C'])

This outputs the following:

>>> df
     A     B   C    D
0  1.0   NaN   5  1.0
1  2.0  10.0  10  2.0
2  NaN   NaN   7  7.0

Answer 5

option 1
pandas

df.assign(D=df.lookup(df.index, df.isnull().idxmin(1)))

     A     B   C    D
0  1.0   NaN   5  1.0
1  2.0  10.0  10  2.0
2  NaN   NaN   7  7.0

option 2
numpy

v = df.values
j = np.isnan(v).argmin(1)
df.assign(D=v[np.arange(len(v)), j])

     A     B   C    D
0  1.0   NaN   5  1.0
1  2.0  10.0  10  2.0
2  NaN   NaN   7  7.0

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naive time test
over given data

over larger data