Given the following data frame:
import pandas as pd import numpy as np df = pd.DataFrame({'A':['1a',np.nan,'10a','100b','0b'], }) df A 0 1a 1 NaN 2 10a 3 100b 4 0b I'd like to extract the numbers from each cell (where they exist). The desired result is:
A 0 1 1 NaN 2 10 3 100 4 0 I know it can be done with str.extract, but I'm not sure how.
Give it a regex capture group:
df.A.str.extract('(\d+)') Gives you:
0 1 1 NaN 2 10 3 100 4 0 Name: A, dtype: object (\d+) is a regex capturing group, and \d+ specifies a regex pattern that matches only digits. Note that this will only work for whole numbers and not floats.
6,000 a
expand=False to the extract
0.7 mg To answer @Steven G 's question in the comment above, this should work:
df.A.str.extract('(^\d*)') U can replace your column with your result using "assign" function:
df = df.assign(A = lambda x: x['A'].str.extract('(\d+)')) If you have cases where you have multiple disjoint sets of digits, as in 1a2b3c, in which you would like to extract 123, you can do it with Series.str.replace:
>>> df A 0 1a 1 b2 2 a1b2 3 1a2b3c >>> df['A'] = df['A'].str.replace('\D+', '') 0 1 1 2 2 12 3 123 You could also work this around with Series.str.extractall and groupby but I think that this one is easier.
Hope this helps!
