Robust regex to match all characters until a floating point number

Question

I need a robust regex that will match all characters until a float.

I have a dict of strings with strings looking like the following mock example:

    'some string 1 some more 2.1 even more 9.2 caracala,domitian2.3'
...

I need a robust regex to substring each string only on the floats, so the end result will look like this:

{
  'some string 1 some more': '2.1'
  'even more': '9.2'
  'caracala,domitian': '2.3'
}

I'll use a for loop with python re to get the end result but I need a robust regex that will match all characters until a float.

I have tried: [-+]?\d*\.\d+|\d+ but it selects numbers as well

I wasn't the one who downvoted, but what have you tried so far? — Haroldo_OK, Sep 25 '19 at 10:41
Possible duplicate of [How to extract a floating number from a string](https://stackoverflow.com/questions/4703390/how-to-extract-a-floating-number-from-a-string) — Tomerikoo, Sep 25 '19 at 10:42

score 2 · Accepted Answer · answered Sep 25 '19 at 10:42

Using re.findall might get you the result you want:

inp = "some string 1 some more 2.1 even more 9.2 caracala,domitian2.3"
matches = re.findall(r'(.*?)\s*(\d+\.\d+)\s*', inp)
print(matches)

[('some string 1 some more', '2.1'), ('even more', '9.2'), ('caracala,domitian', '2.3')]

Explanation of regex:

(.*?)       match all content up the first
\s*         optional space, which is followed by
(\d+\.\d+)  a floating point number

Note that we capture the leading content and float in separate capture groups, which then appear separately in the resulting list.

Robust regex to match all characters until a floating point number

1 Answers1