regex to get value and it's proper unit

Question

i use the following regex to extract values that appear before certain units:

([.\d]+)\s*(?:kg|gr|g)

What i want, is to include the unit of that specific value for example from this string :

"some text 5kg another text 3 g more text 11.5gr end"

i should be getting :

["5kg", "3 g", "11.5gr"]

can't wrap my head on how to modify the above expression to get the wanted result. Thank you.

You already have the match, see https://regex101.com/r/1IPomV/1 But re.findall returns only the capture group values. See [re.findall behaves weird](https://stackoverflow.com/questions/31915018/re-findall-behaves-weird) — The fourth bird, Nov 17 '22 at 00:02

score 2 · Accepted Answer · answered Nov 16 '22 at 23:54

2

import re

p = re.compile('(?<!\d|\.)\d+(?:\.\d+)?\s*?(?:gr|kg|g)(?!\w)')
print(p.findall("some text 5kg another text 3 g more text 11.5gr end"))

answered Nov 16 '22 at 23:54

Ricardo

score 1 · Answer 2 · answered Nov 17 '22 at 00:00

Other solution (regex demo):

(?i)\b\d+\.?\d*\s*(?:kg|gr?)\b

(?i) - case insensitive
\b - word boundary
- \d+\.?\d* - match the amount
- \s* - any number of spaces
- (?:kg|gr?) - match kg, g or gr
\b - word boundary

import re

p = re.compile(r"(?i)\b\d+\.?\d*\s*(?:kg|gr?)\b")
print(p.findall("some text 5kg another text 3 g more text 11.5gr end"))

Prints:

['5kg', '3 g', '11.5gr']

2 Answers2