Python Regex to get Float number from string

Question

I am using regex to parse float number from the string.

re.findall("[^a-zA-Z:][-+]?\d+[\.]?\d*", t)

is the code that I used. There is a problem with this code. It is not parse the number if there is no space between number and any character. For Example, the expect output from "0|1|2|3|4|5|6|7|8|9" is [0,1,2,3,4,5,6,7,8,9], but it returns "[|1,|2,|3,...].

Is there any way to solve this kind of problem?

Try re.findall(r"[^a-zA-Z:]([-+]?\d*\.?\d+)", t) or Try re.findall(r"(?<![a-zA-Z:])[-+]?\d*\.?\d+", t) — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 9, 2017 at 16:53
@WiktorStribiżew It is working, but somehow it loses the first digit. if t is 120, it returns 20. — Bryan K.
– Bryan K., Commented Feb 9, 2017 at 16:57
@WiktorStribiżew In the string, there is a case like M1, M2, and I would like to avoid the number from that case. — Bryan K.
– Bryan K., Commented Feb 9, 2017 at 17:04

Wiktor Stribiżew · Accepted Answer · 2017-02-09 19:10:14Z

4

Use

re.findall(r"(?<![a-zA-Z:])[-+]?\d*\.?\d+", t)

See the regex demo

It will match integer and float numbers not preceded with letters or colon.

Details:

(?<![a-zA-Z:]) - a negative lookbehind that makes sure there is no ASCII letter or colon immediately before the current location
[-+]? - an optional + or -
\d* - zero or more digits
\.? - an optional dot
\d+ - 1+ digits

edited Feb 9, 2017 at 19:10

answered Feb 9, 2017 at 18:45

Wiktor Stribiżew

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2 Comments

Bryan K. Over a year ago

Thank you, and one last question. In the follow case "3monthSummary : month1: 60.5 month2: 60.24 month3: 60.25", it parses 3 to the result. Is there any way to prevent that?

Wiktor Stribiżew Over a year ago

Try (?<![a-zA-Z:])[-+]?\d*\.?\d+(?![a-zA-Z0-9]). Or (?<![a-zA-Z:])[-+]?\d*\.?\d+\b.

Shawn Tabrizi · Accepted Answer · 2017-02-09 16:57:26Z

1

The easiest thing you should be able to do here is just wrap the "number" part of your regular expression into a capture group, and then look at those capture groups.

re.findall("[^a-zA-Z:]([-+]?\d+[\.]?\d*)", t)

I just added parentheses around the "number" part of your search.

answered Feb 9, 2017 at 16:57

Shawn Tabrizi

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2 Comments

Bryan K. Over a year ago

Thank you. It is working, but somehow it loses the first digit. From the example, it returns [1,2,3,4,5,6,7,8,9].

Shawn Tabrizi Over a year ago

Yeah, the issue here is that you are forcing the first character in the match to be some non-letter non-colon, whereas the first number in the string has nothing in front of it. You can add a ? to that block, but then you might get some other weird behaviors. I think you will need to provide more overall samples of what you want to happen in order to get an accurate RegEx to do what you want. Here is the regex with the change: [^a-zA-Z:]?([-+]?\d+[\.]?\d*)

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