Reputation: 4608
I have a pandas dataframe as follows.
a b c d e
a 0 1 0 1 1
b 1 0 1 6 3
c 0 1 0 1 2
d 5 1 1 0 8
e 1 3 2 8 0
I want to replace values that is below 6 <=5
with 0. So my output should be as follows.
a b c d e
a 0 0 0 0 0
b 0 0 0 6 0
c 0 0 0 0 0
d 0 0 0 0 8
e 0 0 0 8 0
I was trying to do this using the following code.
df['a'].replace([1, 2, 3, 4, 5], 0)
df['b'].replace([1, 2, 3, 4, 5], 0)
df['c'].replace([1, 2, 3, 4, 5], 0)
df['d'].replace([1, 2, 3, 4, 5], 0)
df['e'].replace([1, 2, 3, 4, 5], 0)
However, I am sure that there is a more easy way of doing this task in pandas.
I am happy to provide more details if needed.
Upvotes: 3
Views: 9069
Reputation: 402263
For performance, I recommend np.where
. You can assign the array back inplace using sliced assignment (df[:] = ...
).
df[:] = np.where(df < 6, 0, df)
df
a b c d e
a 0 0 0 0 0
b 0 0 0 6 0
c 0 0 0 0 0
d 0 0 0 0 8
e 0 0 0 8 0
Another option involves fillna
:
df[df>=6].fillna(0, downcast='infer')
a b c d e
a 0 0 0 0 0
b 0 0 0 6 0
c 0 0 0 0 0
d 0 0 0 0 8
e 0 0 0 8 0
Upvotes: 3
Reputation: 323226
Using mask
df=df.mask(df<=5,0)
df
Out[380]:
a b c d e
a 0 0 0 0 0
b 0 0 0 6 0
c 0 0 0 0 0
d 0 0 0 0 8
e 0 0 0 8 0
Upvotes: 5