How to replace integer values in pandas in python?

Question

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.

BENY · Accepted Answer · 2019-02-19 02:30:12Z

5

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

answered Feb 19, 2019 at 2:30

BENY

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Comments

cs95 · Accepted Answer · 2019-02-19 02:44:53Z

3

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

edited Feb 19, 2019 at 2:44

answered Feb 19, 2019 at 2:37

cs95

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2 Comments

anky Over a year ago

Didn't know we can use np where on the entire df. Very nice :)

cs95 Over a year ago

@anky_91 Happy my post taught you something new!

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