how to split value in dataframe and replace with another value-CodePudding

dataframe

import pandas as pd
d = {'A': ['B_502_ZZZ_01', 'B_400__01']}
df = pd.DataFrame(data=d)

I tried

def f(x):
 x = x.split('_')[2]
 if x=='':
   x = x.replace('', 'ZZZ') 
 return x
else: 
 return x
df['A'].apply(f)

output required full value like not only 2nd position. I can split 3rd position in another column but I want position changes at that location directly.

['B_502_ZZZ_01', 'B_400_ZZZ_01']

CodePudding user response：

You can use a regex:

df['A'] = df['A'].str.replace(r'((?:[^_] _){2})(?=_)', r'\1ZZZ', regex=True)

NB. The 2 in the regex indicates the number of _ after which to perform the replacement

output (as new column A2 for clarity):

              A            A2
0  B_502_ZZZ_01  B_502_ZZZ_01
1     B_400__01  B_400_ZZZ_01

regex demo

CodePudding user response：

def code(x = 'x are values in the columns'):
  '''It will look for 3rd key in the code and fit 'ZZZ' in that place if there is no value in that'''
  
  y = x.split('_')[2]
  z = x.split('_')[3]
  a = x.split('_')[:2]
  if y =='':
    y = y.replace('', 'ZZZ')
    return '_'.join(a   [y]   [z])
  else:
    return x

Df['A'].apply(code)