划分对齐的 DataFrame 列时获取 NaN

Question

I have a dataframe of the form:

            A             B               C
Cat-1    798.26        456.65          187.56
Cat-2 165165.53      45450.00         4897.57
Cat-3 488565.65      15198.56        15654.65
Cat-4      0.00      54256.35        49878.65
Cat-5   1156.61        789.05        89789.54
Cat-6      0.00       1644.78         6876.15

I am attempting to get a percentage by dividing B by A. To achieve this I used the following:

if_condition = df['A'] != 0
then = (1 - df['B'].div(df['A']))
else_= 0
df['New Col'] = np.where(if_condition, then, else_)

I expected the following result:

            A             B               C       New Col
Cat-1    798.26        456.65          187.56        .5720
Cat-2 165165.53      45450.00         4897.57        .2751 
Cat-3 488565.65      15198.56        15654.65        .0311
Cat-4      0.00      54256.35        49878.65        0
Cat-5   1156.61        789.05        89789.54        .6822
Cat-6      0.00       1644.78         6876.15        0

However, I got the following result:

            A             B               C        New Col
Cat-1    798.26        456.65          187.56        NaN
Cat-2 165165.53      45450.00         4897.57        0.2751 
Cat-3 488565.65      15198.56        15654.65        0.0311
Cat-4      0.00      54256.35        49878.65        0
Cat-5   1156.61        789.05        89789.54        NaN
Cat-6      0.00       1644.78         6876.15        0

I have tried some other solutions which involved the alignment of the two columns, however that did not alter the end result. What could potentially generate these NaN values?

Answer 1

import pandas as pd
import numpy as np
import io

df = pd.read_csv(io.StringIO("""            A             B               C
Cat-1    798.26        456.65          187.56
Cat-2     165165.53      45450.00         4897.57
Cat-3     488565.65      15198.56        15654.65
Cat-4      0.00      54256.35        49878.65
Cat-5   1156.61        789.05        89789.54
Cat-6      0.00       1644.78         6876.15"""), sep="\s\s+", engine="python")

df

# output
               A         B         C
Cat-1     798.26    456.65    187.56
Cat-2  165165.53  45450.00   4897.57
Cat-3  488565.65  15198.56  15654.65
Cat-4       0.00  54256.35  49878.65
Cat-5    1156.61    789.05  89789.54
Cat-6       0.00   1644.78   6876.15

if_condition = df['A'] != 0
then = (1 - df['B'].div(df['A']))
else_= 0
df['New Col'] = np.where(if_condition, then, else_)

# output
               A         B         C   New Col
Cat-1     798.26    456.65    187.56  0.427943
Cat-2  165165.53  45450.00   4897.57  0.724822
Cat-3  488565.65  15198.56  15654.65  0.968891
Cat-4       0.00  54256.35  49878.65  0.000000
Cat-5    1156.61    789.05  89789.54  0.317791
Cat-6       0.00   1644.78   6876.15  0.000000

Seems to be correct. I use pandas version '1.2.5'

Also you could do this "if else" condition a bit easier:

df["New col"] = df.apply(lambda x: 1 - x["B"] / x["A"] if x["A"] != 0 else 0, axis=1)

Answer 2

You don't need a condition, replace -np.inf by 0:

# df['New Col'] = (1 - df['B'] / df['A']).replace(-np.inf, 0)
df['New Col'] = ((1 - df['B'] / df['A']) * 100).round(2).replace(-np.inf, 0)
print(df)

# Output:
               A         B         C  New Col
Cat-1     798.26    456.65    187.56    42.79
Cat-2  165165.53  45450.00   4897.57    72.48
Cat-3  488565.65  15198.56  15654.65    96.89
Cat-4       0.00  54256.35  49878.65     0.00
Cat-5    1156.61    789.05  89789.54    31.78
Cat-6       0.00   1644.78   6876.15     0.00

Answer 3

I was able to resolve this issue, by simply not diving by 0 and then replacing the NaN values with 0. It produced the anticipated result:

df['New Col'] = (1 - df['B']/df['A'][df['A'] != 0]).fillna(0)

I basically was able to divide everything but 0, and the remaining NaN values are a result of not dividing 0 and can thus be replaced by 0.