pandas: replace values by row based on condition

Question

I have a pandas dataframe as follows:

df2
   amount  1  2  3  4
0   5      1  1  1  1
1   7      0  1  1  1
2   9      0  0  0  1
3   8      0  0  1  0
4   2      0  0  0  1

What I want to do is replace the 1s on every row with the value of the amount field in that row and leave the zeros as is. The output should look like this

   amount  1  2  3  4
0   5      5  5  5  5
1   7      0  7  7  7
2   9      0  0  0  9
3   8      0  0  8  0
4   2      0  0  0  2

I've tried applying a lambda function row-wise like this, but I'm running into errors

df2.apply(lambda x: x.loc[i].replace(0, x['amount']) for i in len(x), axis=1)

Any help would be much appreciated. Thanks

just multiply : 0 returns 0 and 1 returns the amount value: df.drop('amount',1).mul(df['amount'],axis=0) — anky
– anky, Commented Mar 22, 2020 at 17:12
but I'm running into errors Please provide the entire error message, as well as a minimal reproducible example. Asn aside, why you use 0/1 instead of actual boolean values? — AMC
– AMC, Commented Mar 22, 2020 at 17:43

Scott Boston · Accepted Answer · 2020-03-22 17:16:44Z

5

Let's use mask:

df2.mask(df2 == 1, df2['amount'], axis=0)

Output:

   amount  1  2  3  4
0       5  5  5  5  5
1       7  0  7  7  7
2       9  0  0  0  9
3       8  0  0  8  0
4       2  0  0  0  2

answered Mar 22, 2020 at 17:16

Scott Boston

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1 Comment

FBruzzesi Over a year ago

For a more general case, maybe use df2.mask(df2 != 0, df2['amount'], axis=0)

Jaroslav Bezděk · Accepted Answer · 2020-03-22 17:28:15Z

2

You can also do it wit pandas.DataFrame.mul() method, like this:

>>> df2.iloc[:, 1:] = df2.iloc[:, 1:].mul(df2['amount'], axis=0)
>>> print(df2)
   amount  1  2  3  4
0       5  5  5  5  5
1       7  0  7  7  7
2       9  0  0  0  9
3       8  0  0  8  0
4       2  0  0  0  2

answered Mar 22, 2020 at 17:28

Jaroslav Bezděk

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