Regex for 10,000 through 100,000 to 150,000,000 and beyond

Question

I'm trying to dollar amounts between 10,000 to 150,000,000.

I got this from a stack user previously but only catches from 1,000,000 through 150,000,000

(?<!\d)(\d{1,3}(?:,\d{3}){2,})(?!\d)

I tried reworking it for the last hour but can't and regex is a notorious head wreck :D anyone can update it to start catching from 10,000? Thanks!

Why would you even need a regex for that? where does the input come from? how does it look like? why not use `int` and numeric comparisons? — DeepSpace, Jun 18 '20 at 10:36
@WiktorStribiżew That regex pattern would also admit 160,000,000, [see the demo](https://regex101.com/r/uwcV8E/1). — Tim Biegeleisen, Jun 18 '20 at 10:38
Thanks @WiktorStribiżew, you provided the last ones I asked for too. I'll sit down for some regex tutorials after this. Sick of wasting my time and others with regex questions! — Digital Moniker, Jun 18 '20 at 10:38
@TimBiegeleisen thats fines its monetary values that can exceed 150,000,00 sorry my question title was not clear enough — Digital Moniker, Jun 18 '20 at 10:39
[Posted the answer](https://stackoverflow.com/a/66532366/3832970). — Wiktor Stribiżew, Mar 08 '21 at 15:17

score 3 · Answer 1 · answered Jun 18 '20 at 10:52

3

One approach might be to use the generic regex for thousands, and then add a lookahead to restrict the lengths to the range you want:

^(?=.{6,11}$)\d{1,3}(?:,\d{3})*$

Demo

answered Jun 18 '20 at 10:52

Tim Biegeleisen

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score 1 · Accepted Answer · answered Mar 08 '21 at 15:17

1

You can use

(?<!\d)(\d{1,3}(?:,\d{3})+)(?!\d)

See the regex demo.

Details:

(?<!\d) - no digit allowed immediately to the left of the current location
(\d{1,3}(?:,\d{3})+) - Group 1: one to three digits followed with one or more (due to + quantifier) occurrences of a comma and thee digits
(?!\d) - no digit allowed immediately to the right of the current location.

answered Mar 08 '21 at 15:17

Wiktor Stribiżew

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Thanks, @Wiktor Stribiżew – Digital Moniker Mar 08 '21 at 16:56

Jan · Answer 3 · 2020-06-18T11:56:06.023

Just cast it and compare it programmatically:

import pandas as pd

dct = {"numbers": ["10", "100", "200", "5,000", "10,000", "15000", "some weird stuff", "160,000,000"]}


def tester(number):
    try:
        number = float(number.replace(",", ""))
        if 10 * 10 ** 3 <= number <= 150 * 10 ** 6:
            return True
    except:
        pass
    return False

df = pd.DataFrame(dct)
df["in_range"] = df["numbers"].apply(tester)
print(df)

This yields

            numbers  in_range
0                10     False
1               100     False
2               200     False
3             5,000     False
4            10,000      True
5             15000      True
6  some weird stuff     False
7       160,000,000     False

score 0 · Answer 4 · answered Jun 20 '20 at 12:59

You current pattern could also possibly match 999,999,999,999 due to the repeating of 2 or more times for this part (?:,\d{3}){2,}

The pattern also uses only the \d which can match 0-9 and is not limited to 5 anywhere in the pattern.

Matching 3 digits after the comma, you could use use an alternation | to match the separate range parts:

(?<!\S)(?:[1-9]\d\d?,\d{3}|(?:[1-9]\d?|1[0-4]\d),\d{3},\d{3}|150,000,000)(?!\S)

(?<!\S) Assert whitespace boundary to the left
(?: Non capture group
- [1-9]\d\d?,\d{3} Match range 10,000 - 999,999
- | Or
- (?: Non capture group
  - [1-9]\d? Match range 1 - 99
  - | Or
  - 1[0-4]\d Match digits 100 - 149
- ) Close non capture group
- ,\d{3},\d{3} Match range ,000,000 - ,999,999
- | Or
- 150,000,000 Match the max value
) Close non capture group
(?!\S) Assert whitespace boundary to the right

Regex demo

Regex for 10,000 through 100,000 to 150,000,000 and beyond

4 Answers4

Demo