Python Regex: Proper way to extract separated numbers (AxBxC -> [A, B, C])

Question

I am now trying to extract sizes from the string, which is a very common pattern i guess: AxBxC where A, B, C separated with x (may be x with spaces also), are the sizes (int or float):

import re

s = 'zzz 3062 0.2 aaa 15.8x20.2x12.2875 mm'

I am expecting to obtain onli three numbers: [15.8, 20.2, 12.2875] The only working approach i have now is ugly:

r1 = re.findall('(\d+\.?\d*)\ *x\ *', s)
r2 = re.findall('\ *x\ *(\d+\.?\d*)', s)
r1.extend(r2)
print(set(r1))

{'15.8', '20.2', '12.2875'}

Is there any way to use single robust regexp for extraction these numbers? Thanks.

wouldn't `\d+\.*\d+` be enough? This should get any numbers separated by anything except a decimal point. For you example it finds [3062, 0.2, 15.8, 20.2, 12.2875] — Zinki, Jul 27 '17 at 12:48
Maybe `re.findall(r'(\d[\d.]*)x(\d[\d.]*)x(\d[\d.]*)', s)`? These numbers always go in triples, don't they? See [the regex demo](https://regex101.com/r/0K5v5P/1). — Wiktor Stribiżew, Jul 27 '17 at 12:50
Yes, this is good, but what if we have AxB (only two dimensions)? — Alexey Trofimov, Jul 27 '17 at 12:51
Make the 3rd one optional - [`r'(\d[\d.]*)x(\d[\d.]*)(?:x(\d[\d.]*))?'`](https://regex101.com/r/0K5v5P/2), it will return `('15.8', '20.2', '')` — Wiktor Stribiżew, Jul 27 '17 at 12:52

score 1 · Answer 1 · answered Jul 27 '17 at 12:59

1

This is a solution if you don't know how many numbers you will have to get:

((?:\d+\.\d+)(?=x)|(?<=x)(?:\d+\.\d+))

It's based on the fact that or the x is before your number or after it.

answered Jul 27 '17 at 12:59

Marco Luzzara

5,540
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score 1 · Accepted Answer · answered Jul 27 '17 at 12:59

It seems you need to match 2 or 3 x separated float values. You may use

r'(\d[\d.]*)x(\d[\d.]*)(?:x(\d[\d.]*))?'

See the regex demo

Details

(\d[\d.]*) - Group 1: a digit and then 0+ digits or/and .
x - a literal x
(\d[\d.]*) - Group 2: a digit and then 0+ digits or/and .
(?:x(\d[\d.]*))? - an optional sequence of x(\d[\d.]*), an x followed with Group 3 capturing a digit and then 0+ digits or/and ..

In Python, use

re.findall(r'(\d[\d.]*)x(\d[\d.]*)(?:x(\d[\d.]*))?', s)

score 1 · Answer 3 · answered Jul 27 '17 at 13:00

Instead, for r1, you can use the following:

r1 = re.split( "x", re.findall('\d*\.\d*x\d*\.\d*x\d*\.\d*', s )[0] )

Unfortunately, it still exists of two commands, and by nesting this, it becomes a bit hard to read. And if you want to use them as numbers, they should still be converted from string to a number, e.g. for each number float(r1[#]) or use the numpy version to convert the whole array, which can be found here.

score 0 · Answer 4 · answered Jul 27 '17 at 13:22

0

I hope this will help you

>>> s.split()[-2].split("x")
['15.8', '20.2', '12.2875']

answered Jul 27 '17 at 13:22

Hugo Xia

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Python Regex: Proper way to extract separated numbers (AxBxC -> [A, B, C])

4 Answers4