I have a question regards about np.where()
Currently, I have 2 columns, each column contains Null values and categorical values. Values from each column are distinct and will not overlap.
For now, I want to apply all the Non-Null values from these 2 columns into the new column and fill the NaN value in the new column as a categorical value.
My idea is using np.where()
Basic idea is if df[‘A’]==’A’, fill the value A into new column fist, elif df[‘B’]==’B’, fill the value B into new column as well, Else fill the value ‘C’ for all the NaN values.
However, a syntax error returned.
ValueError: operands could not be broadcast together with shapes (544,) () (3,)
Thanks for the help always!
A B C Desired col C user1 Null Null user1 user1 Null Null user1 user1 Null Null user1 user1 Null Null user1 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null user2 Null Null Null user3 Null Null Null user3 Null Null Null user3 Null Null Null user3
Assuming your initial df is only cols A, B, and C:
# convert value you don't want to NaNs df = df.where(df != 'Null') # temporary list lst =  # iterate row-wise for r in df.iterrows(): # test if all values in row are the same (1 = no) if r.nunique() == 1: # if different, find the one that is the string and append to list a,b,c = r # *this is specific to your example with three cols* for i in [a,b,c]: if isinstance(i,str): lst.append(i) else: # if same append specified value to list lst.append('user3') df['D'] = lst
It’s verbose and will be bit slow for very large dfs, but it produces your expected result. And it’s readable.
It would be cleaner if you didn’t have the rows with all nulls. Then a cleaner, one-liner would be more feasible df.where(), .apply(lambda), or masked array approach easier.