Pandas replace column by value in row

Question

How can I vectorise a replace, by looking for a value in the row.

For a dataframe as follows:

df = pd.DataFrame([(1, 2, 3, 4, np.NaN, np.NaN, 4),
                   (1, 2, 3, 0, 0,      np.NaN, 0),
                   (1, 2, 3, 4, 5,      np.NaN, 5)],
                 columns = ['P0', 'P1', 'P2', 'P3', 'P4', 'P5',  'Last_not_NaN_value'],
                 index = ['row1', 'row2', 'row3'])

Output df:

        P0  P1  P2  P3  P4     P5   Last_not_NaN_value
row1    1   2   3   4   NaN    NaN  4
row2    1   2   3   0   0.0    NaN  0
row3    1   2   3   4   5.0    NaN  5

How can I do something like

df.replace(df['Last_not_NaN_value'], 0 ) <- which does nothing.

How can I look for where the last_not_NaN_value is in the df, and replace that with a 0, eg:

        P0  P1  P2  P3  P4     P5   Last_not_NaN_value
row1    1   2   3   *0* NaN    NaN  4
row2    1   2   3   0   *0*    NaN  0
row3    1   2   3   4   *0*    NaN  5

Answer 1

Vectorized, as requested. Perform broadcasted comparison, find the indices of replacement, and just replace accordingly. Afterwards, you can assign the result back using a neat df[:] = ... trick.

v = df.values
i = v[:, :-1]
j = v[:, -1]

v[np.arange(v.shape[0]), (i == j[:, None]).argmax(axis=1)] = 0
df[:] = v

df
       P0   P1   P2   P3   P4  P5  Last_not_NaN_value
row1  1.0  2.0  3.0  0.0  NaN NaN                 4.0
row2  1.0  2.0  3.0  0.0  0.0 NaN                 0.0
row3  1.0  2.0  3.0  4.0  0.0 NaN                 5.0

Answer 2

This is one solution, though not vectorised:

for i in range(6):
    df.loc[i==(df['Last_not_NaN_value']-1), 'P'+str(i)] = 0

#       P0  P1  P2  P3   P4  P5  Last_not_NaN_value
# row1   1   2   3   0  NaN NaN                   4
# row2   1   2   3   0  0.0 NaN                   0
# row3   1   2   3   4  0.0 NaN                   5