Python: Divide two columns, a/b, if b is 0 -> assign 0

Question

I need to make new column in df (apples per pupils) divide two columns (apples and pupils), and if the number of pupils is 0, I should just assign zero to that row.

Current values:

apples	pupils
5	2
6	3
5	0

Expected output:

apples	pupils	avg_apples_per_pupils
5	2	2.5
6	3	2.0
5	0	0.0

I tried using this:

if df[df['pupils']!=0]:
    df['avg_apples_per_pupils'] = df['apples'] / df['pupils']
else:
    df['avg_apples_per_pupils'] = df.assign(avg_apples_per_pupils='0')

but I get this error:

ValueError: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I’m completely open for other approaches

Asked By: Beginner in the house

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Source

Answer 1

Because you compere array and 0 (if df[df[‘pupils’]!=0)
you must specify which value from the df[‘pupils’] you want to compere to 0

if(df['pupils'] != 0).any():
df['avg_apples_per_pupils']=df['apples'] / df['pupils']

else:
df[‘avg_apples_per_pupils’] = 0

Answered By: Mohamed Sid Abdalla Rgagde

Answer 2

Use np.where condition:

df['avg_apples_per_pupils'] = np.where(df['pupils'].ne(0), df['apples'] / df['pupils'], df['pupils'])

   apples  pupils  avg_apples_per_pupils
0       5       2                    2.5
1       6       3                    2.0
2       5       0                    0.0

Answered By: RomanPerekhrest

Answer 3

You can use a mask:

df['avg_apples_per_pupils'] = (df['apples'].div(df['pupils'])
                               .mask(df['pupils'].eq(0), 0)
                               )

Or numpy.divide:

df['avg_apples_per_pupils'] = np.divide(df['apples'], df['pupils'],
                                        where=df['pupils'].ne(0))

Output:

   apples  pupils  avg_apples_per_pupils
0       5       2                    2.5
1       6       3                    2.0
2       5       0                    0.0

Answered By: mozway

Python: Divide two columns, a/b, if b is 0 -> assign 0

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