replace all values in range of columns based on condition - python-3.x

Question

How can I replace values in multiple columns based on a condition?
Suppose I have a df looking something like this:
df = pd.DataFrame({'A': [1,2,3,4], 'C': [1,2,3,4], 'B': [3,4,6,6]})
With numpy I can change the value of a column based on a condition like this:
df['A'] = np.where((df['B'] < 5), '-', df['A'])
But how can I change the value of many columns based on a condition? Thought I could do something like below but that's not working.
df[['A','C']] = np.where((df['B'] < 5), '-', df[['A', 'C']])
I could do a loop but that does not feel very pythonic/pands
cols = ['A', 'C']
for col in cols:
df[col] = np.where((df['B'] < 5), '-', df[col])

Answer 1

One idea is use DataFrame.mask:
df[['A','C']] = df[['A', 'C']].mask(df['B'] < 5, '-')
print (df)
A C B
0 - - 3
1 - - 4
2 3 3 6
3 4 4 6
Alternative solution with DataFrame.loc:
df.loc[df['B'] < 5, ['A','C']] = '-'
print (df)
A C B
0 - - 3
1 - - 4
2 3 3 6
3 4 4 6
Solution with numpy.where and broadcasting mask:
df[['A','C']] = np.where((df['B'] < 5)[:, None], '-', df[['A', 'C']])
Performance if mixed values - numeric with string -:
df = pd.DataFrame({'A': [1,2,3,4], 'C': [1,2,3,4], 'B': [3,4,6,6]})
#400k rows
df = pd.concat([df] * 100000, ignore_index=True)
In [217]: %timeit df[['A','C']] = df[['A', 'C']].mask(df['B'] < 5, '-')
171 ms ± 13.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [219]: %timeit df[['A','C']] = np.where((df['B'] < 5)[:, None], '-', df[['A', 'C']])
72.5 ms ± 11.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [221]: %timeit df.loc[df['B'] < 5, ['A','C']] = '-'
27.8 ms ± 533 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
Performace if replace by numeric:
df = pd.DataFrame({'A': [1,2,3,4], 'C': [1,2,3,4], 'B': [3,4,6,6]})
df = pd.concat([df] * 100000, ignore_index=True)
In [229]: %timeit df[['A','C']] = df[['A', 'C']].mask(df['B'] < 5, 0)
187 ms ± 4.24 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [231]: %timeit df[['A','C']] = np.where((df['B'] < 5)[:, None], 0, df[['A', 'C']])
20.8 ms ± 455 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [233]: %timeit df.loc[df['B'] < 5, ['A','C']] = 0
61.3 ms ± 1.06 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)