Issues with "if-else" statement-CodePudding

I have the following dataframe:

df = pd.DataFrame({'y': [4,3,6,1], 'x': [0,0,2,1]})

I would like to compute the ratio of the two columns. However, since there are some 0 in the denominator, I would like to fix if with a if else statement. The condition should be: if in the denominator there is 0, replace it with 1 and compute the ratio, otherwise as usual.

I did this (and other variations) but it doesn't work (see below error):

if df['x'] == 0:
   df['x'] = 1
   df['ratio'] = df['y']/df['x']
else:
   df['ratio'] = df['y']/df['x']

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

Can anyone help me?

Thanks!

CodePudding user response：

The issue here is that df['x'] is actually a pandas series, it's not being evaluated for each individual value, but as a single element (pd Series) itself, and since it contains multiple values, the evaluation is returning an error because it's ambiguous.

For a fast and efficient solution without the need of apply, you can use np.where(). So that:

df['ratio'] = np.where(df['x'] == 0,df['y'],df['y']/df['x'])

Outputs:

   y  x  ratio
0  4  0    4.0
1  3  0    3.0
2  6  2    3.0
3  1  1    1.0

CodePudding user response：

If you wanna make it simple and use if/else

import pandas as pd
df = pd.DataFrame({'y': [4,3,6,1], 'x': [0,0,2,1]})


if 0 in df['x']:
    df['x'].replace(0, 1, inplace=True)
df['ratio'] = df['y']/df['x']
print(df)

Here the output

   y  x  ratio
0  4  1    4.0
1  3  1    3.0
2  6  2    3.0
3  1  1    1.0

CodePudding user response：

You can do this:

def x(a,b):
  if b == 0:
    b = 1
  return a / b

df['ratio']  = df.apply(lambda f: x(f['y'],f['x']), axis=1)