Python dataframe replace strings in specific condition

Question

import numpy as np
import pandas as pd

I have a df looks like this

My goal is to change them (string) to integer and divide the number by 10 inside the string if it contains 'A'

I have tried to use for loop for the whole column

for i in df.X:
    if 'A' in i:
        y = i.replace('A','') 
        y = int(y)/10
        print(y)
    else:
        k = i.replace('B','')
        k = int(k)
        print(k)

But I can only print them out and I dont know how to replace the value directly and store them in the column. Also, this method seems to be slow? Is there a better methods in pandas? Thanks!

Daniil Ryzhkov · Accepted Answer · 2021-08-05 15:56:42Z

1

Try this:

a_index = df[df[0].str.endswith('A')]
df = df[0].str.slice(stop=-1).astype(int)
df[a_index.index] = df[a_index.index] / 10

First, I save indexes of every row that ends with A. Then, I convert everything to integer. Finally, we divide previously indexed items (a_index) by 10

Demo

answered Aug 5, 2021 at 15:56

Daniil Ryzhkov

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Daniel Aben-Athar Bemerguy · Accepted Answer · 2021-08-05 16:20:09Z

0

You can solve this using regular expression:

import re
import pandas as pd

df = pd.DataFrame({'A':['100A','100B','200A']})

for row in range(0,len(df)):
    if df.iloc[row].str.contains('A').any():
       new_value = int(re.sub('\D', '', df.iloc[row].values[0]))/10
       df.iloc[row] = new_value
    else:
        new_value = int(re.sub('\D', '', df.iloc[row].values[0]))
        df.iloc[row] = new_value

OUTPUT

edited Aug 5, 2021 at 16:20

answered Aug 5, 2021 at 16:14

Daniel Aben-Athar Bemerguy

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