需要从某列中提取数据，如果存在特定字符，提取该字符前的substring

Question

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

Wherever the pattern is of xy 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.

Answer 1

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)

Answer 2

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