how to replace part of string using regular expression?

Question

I have a dataframe like this:

fict={'well':['10B23','10B23','10B23','10B23','10B23','10B23'],
      'tag':['15B22|TestSep_OutletFlow','15B22|TestSep_GasOutletFlow','15B22|TestSep_WellNum','15B22|TestSep_GasPresValve','15B22|TestSep_Temp','WHT']}
df=pd.DataFrame(dict)
df

    well    tag
0   10B23   15B22|TestSep_OutletFlow
1   10B23   15B22|TestSep_GasOutletFlow
2   10B23   15B22|TestSep_WellNum
3   10B23   15B22|TestSep_GasPresValve
4   10B23   15B22|TestSep_Temp
5   10B23   WHT

Now I'd like to replace anything before | in column of tag to a string like 11A22, so the dataframe after replace should look like this:

well    tag
0   10B23   11A22|TestSep_OutletFlow
1   10B23   11A22|TestSep_GasOutletFlow
2   10B23   11A22|TestSep_WellNum
3   10B23   11A22|TestSep_GasPresValve
4   10B23   11A22|TestSep_Temp
5   10B23   WHT

I am thinking to use regular expression with group to replace group by a string, something in my mind look like this

df['tag2']=df['tag'].str.replace(r'([a-z0-9]*)|TestSep_[a-z0-9]*','11A22',regex=True)

then i got result of

well    tag tag2
0   10B23   15B22|TestSep_OutletFlow    11A2211A22B11A2211A22|11A2211A2211A22O11A2211A...
1   10B23   15B22|TestSep_GasOutletFlow 11A2211A22B11A2211A22|11A2211A2211A22G11A2211A...
2   10B23   15B22|TestSep_WellNum   11A2211A22B11A2211A22|11A2211A2211A22W11A2211A...
3   10B23   15B22|TestSep_GasPresValve  11A2211A22B11A2211A22|11A2211A2211A22G11A2211A...
4   10B23   15B22|TestSep_Temp  11A2211A22B11A2211A22|11A2211A2211A22T11A2211A22
5   10B23   WHT 11A22W11A22H11A22T11A22

Thanks for your help

Timeless · Accepted Answer · 2023-01-25 05:12:15Z

4

(|) is a special character in regex, you need to escape it.

df["tag2"] = df["tag"].str.replace(r"^\w*\|", "11A22|", regex=True)

Output :

print(df)

    well                          tag                         tag2
0  10B23     15B22|TestSep_OutletFlow     11A22|TestSep_OutletFlow
1  10B23  15B22|TestSep_GasOutletFlow  11A22|TestSep_GasOutletFlow
2  10B23        15B22|TestSep_WellNum        11A22|TestSep_WellNum
3  10B23   15B22|TestSep_GasPresValve   11A22|TestSep_GasPresValve
4  10B23           15B22|TestSep_Temp           11A22|TestSep_Temp
5  10B23                          WHT                          WHT

answered Jan 25, 2023 at 5:12

Timeless

38.3k6 gold badges33 silver badges54 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

roudan Over a year ago

oh yeah, it works beautifully! Thanks Timeless, I appreciate your help. so we don't need to use group?

Timeless Over a year ago

You're welcome. It is not necessary in this case to capture any group since you just need to replace a word that occurs at (the beginning of the string/column + before a delimiter).

Timeless Over a year ago

You might use df["tag"].str.replace(r"^\w*\|(.*)", r"11A22|\1", regex=True). Here, we capture any character found after the delimiter | with (.*). Then, we use "\1" to place this captured group right after the 11A22|.

roudan Over a year ago

I see, that is nice. that is really helpful! I definitely learn this trick tonight. Thank you Timeless, I appreciate it.

Timeless Over a year ago

You're welcome roudan.

|

Collectives™ on Stack Overflow

how to replace part of string using regular expression?

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related