pandas dataframe 中的正则表达式

Question

I have a dataframe like this a column like this

COL1      
RED[10%(INC)]
RED[12%(INC)]

and I want create col2 as this

COL2
10
12

Could cou help me to find the good regex? I tried this:

RED\[(\d+\.\d+) %INC\]

but it doesn't walk.

Answer 1

If you want to use your regex and only extract numbers in the specified context, you can use

df['COL2'] = df['COL1'].str.extract(r'RED\[(\d+(?:\.\d+)?)%\[INC]]', expand=False)

See the regex demo .

Details

• RED\[ - a RED[ string
• (\d+(?:\.\d+)?) - Capturing group 1: one or more digits followed with an optional sequence of a dot and one or more digits
• %\[INC]] - a %[INC]] literal string.

You could also explore other options:

• Extracting the number followed with a percentage sign: df['COL1'].str.extract(r'(\d+(?:\.\d+)?)%', expand=False)
• Splitting with [ , getting the second item and removing % from it: df['COL1'].str.split("[").str[1].str.replace("%", "")

Answer 2

This solution uses re.findall :

Modules and data:

import pandas as pd
df = pd.DataFrame({'COL1':['RED[10%(INC)','RED[12%(INC)']})

Solution:

df['COL2'] = df['COL1'].apply(lambda x: re.findall('[0-9]+', x))
df['COL2'] = pd.DataFrame(df['COL2'].tolist())

Answer 3

df['COL2']= np.where(df['COL1'].str.extract(r'RED\[(d+)')