I have columns with values like this:
Col1
1/1/100 'BA1
1/1/102Packe
1/1/102 'to_
And need to extract just 1/1/100 (from the first row) and so on (1/1/102...)
I am using:
df['col1'] = df['col1'].str.extract('(\d+)/(\d+)/(\d+)', expand=True)
But I'm getting only 1.
Not sure why this is not working, is there a problem with regex or I need some kind of mapping?
You need to only use a single capturing group:
df['col1'] = df['col1'].str.extract('(\d+/\d+/\d+)', expand=True)
^ ^
The str.extract method returns the value captured with the first capturing group, and your regex captures the first 1 into that group.
Test:
>>> import pandas as pd
>>> df = pd.DataFrame({"col1":["1/1/100 'BA1", "1/1/102Packe", "1/1/102 'to_"]})
>>> df['col1'].str.extract('(\d+/\d+/\d+)', expand=True)
0
0 1/1/100
1 1/1/102
2 1/1/102
you can try this also,
df['Col1']=df['Col1'].str.replace('\d+|/','')
Note: Regex is more powerful than .str.replace.
I suggest this Regex:
df['col1'].str.extract('\b(\d/?)+', expand=True)
