Replacing empty values in a DataFrame with value of a column

Question

Say I have the following pandas dataframe:

df = pd.DataFrame([[3, 2, np.nan, 0],
                    [5, 4, 2, np.nan],
                    [7, np.nan, np.nan, 5],
                    [9, 3, np.nan, 4]],
                    columns=list('ABCD'))

which returns this:

   A    B    C    D
0  3  2.0  NaN  0.0
1  5  4.0  2.0  NaN
2  7  NaN  NaN  5.0
3  9  3.0  NaN  4.0

I'd like that if a np.nan is found, that the value is replaced by a value in the A column. So that would mean the result to be this:

   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0

I've tried multiple things, but I could not get anything to work. Can anyone help?

Answer 1

Here is necessary double transpose:

cols = ['B','C', 'D']
df[cols] = df[cols].T.fillna(df['A']).T
print(df)
   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0

because:

df[cols] = df[cols].fillna(df['A'], axis=1)
print(df)

NotImplementedError: Currently only can fill with dict/Series column by column

Another solution with numpy.where and broadcasting column A:

df = pd.DataFrame(np.where(df.isnull(), df['A'].values[:, None], df), 
                  index=df.index, 
                  columns=df.columns)
print (df)
     A    B    C    D
0  3.0  2.0  3.0  0.0
1  5.0  4.0  2.0  5.0
2  7.0  7.0  7.0  5.0
3  9.0  3.0  9.0  4.0

Thank you @pir for another solution:

df = pd.DataFrame(np.where(df.isnull(), df[['A']], df), 
                  index=df.index, 
                  columns=df.columns)

Answer 2

Currently, fillna doesn't allow for broadcasting a series across columns while aligning the indices.

pandas.DataFrame.mask

This functions exactly like what we'd want fillna to do. Finds the the nulls, fills it in with df.A along axis=0

df.mask(df.isna(), df.A, axis=0)

   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0

---

pandas.DataFrame.fillna using a dictionary

However, you can pass a dictionary to fillna that tells it what to do for each column.

df.fillna({k: df.A for k in df})

   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0

Answer 3

DO fillna with reindex

df.fillna(df[['A']].reindex(columns=df.columns).ffill(1))
Out[20]: 
   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0

Or combine_first

df.combine_first(df.fillna(0).add(df.A,0))
Out[35]: 
   A    B    C    D
0  3  2.0  3.0  0.0
1  5  4.0  2.0  5.0
2  7  7.0  7.0  5.0
3  9  3.0  9.0  4.0