Need to extract data from a column, if a particular character exists, extracting the substring befor-CodePudding

I've got a column which I am trying to clean, the data is like this:

Wherever the pattern is of x-y year, I want to extract only the 'x' value and leave it in the string. For any other value, I want to keep it as is.

Using str.extract('(.{,2}(-))') is returning a NaN value for all the other rows.

CodePudding user response：

The solution first compiles the regex then the compiled regex will be used on each row. The lambda also relies on the walrus operator :=. Assumes that your 2nd column is named col2.

import re

pattern = re.compile("([\d] )-[\d]  year")
df["col2"] = df["col2"].map(lambda x: m[1] if (m:=pattern.match(x)) else x)

CodePudding user response：

You want series.str.replace(), I believe.

Does this give you the desired output?

df = pd.DataFrame.from_records([[1778, '3-5 year'], [961, np.nan], [2141, 'h 3  year']], columns=['a','b'])

repl = lambda m: m.group(1)
df.b = df.b.str.replace(r'(\d )-\d \syear', repl, regex=True)
df

which takes the original df:

      a          b
0  1778   3-5 year
1   961        NaN
2  2141  h 3  year

and gives the output:

      a          b
0  1778          3
1   961        NaN
2  2141  h 3  year