Regex to find if a string contains all numbers 0-9 be it in any order

Question

Regex to find if a string contains all numbers 0-9 be it in any order. Can anyone help?

Suppose I have a string- "01230912747238507156" This has all characters from 0-9. So I should get True.

The question is not clear at all, what do you want to achieve, what are the examples of your strings? — Sudheesh Singanamalla, Jan 26 '18 at 15:09
It'd be much easier to do this without a regex than with. Just write some Python that iterates each char in the string and remembers if it's seen every digit 0-9. — Taylor Wood, Jan 26 '18 at 15:10
Suppose I have a string- "0123091274723507156" This has all characters from 0-9. So I should get True. — Abhisek Roy, Jan 26 '18 at 15:10
`{n for n in s if n in '1234567890'} == {'%i' % i for i in range(10)}` — karlson, Jan 26 '18 at 15:14
Ok, then take this `set(filter(lambda x: x.isdigit(), s)) == set(map(str, range(10)))` :) — karlson, Jan 26 '18 at 15:18
Very demanding for someone asking us to do his homework for him. — pault, Jan 26 '18 at 15:30
`^(?=.*1.*)(?=.*2.*)(?=.*3.*)(?=.*4.*)(?=.*5.*)(?=.*6.*)(?=.*7.*)(?=.*8.*)(?=.*9.*)(?=.*0.*)\d+$` Positive look aheads are definitely for you here. https://regex101.com/r/fl2xd8/1 — KyleFairns, Jan 26 '18 at 16:05

score 4 · Answer 1 · answered Jan 26 '18 at 15:27

4

You can use Python's built-in all() function

s = "0123091274723507156"
n = [0,1,2,3,4,5,6,7,8,9]
print(all(str(i) in s for i in n))

You could also replace [0,1,2,3,4,5,6,7,8,9] with list(range(0,10))

answered Jan 26 '18 at 15:27

ctwheels

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Jan · Answer 2 · 2018-01-26T15:18:04.597

1

You could use a counter like this

string = "01230912747235071568"

cnt = 0
for i in range(0,10):
    if str(i) in string:
        print(i)
        cnt += 1

print(cnt)

If you are really looking for a regex solution (learning purposes?), you could use multiple lookaheads (not advisable, really redundant code):

import re
rx = re.compile(r'^(?=^0*0)(?=^1*1)(?=^2*2)(and so on...).+$')

That is:

(?=not_one zero or more times, then one)...

edited Jan 26 '18 at 15:18

answered Jan 26 '18 at 15:15

Jan

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No loops. I mentioned before – Abhisek Roy Jan 26 '18 at 15:16
@AbhisekRoy: See the updated answer then. – Jan Jan 26 '18 at 15:17
I just want to check if a string has all numbers 0-9, anywhere in the string. It has to be written in regex. And I am coding in Python. – Abhisek Roy Jan 26 '18 at 15:17
TypeError: unsupported operand type(s) for &: 'str' and 'int' This is the error that your code is giving @Jan – Abhisek Roy Jan 26 '18 at 15:20
3

@AbhisekRoy There is no "No loops" in your question. – Aran-Fey Jan 26 '18 at 15:24

Darkonaut · Accepted Answer · 2018-01-26T16:36:46.103

This solution can handle all sorts of strings, not only numeric characters.

s1 = 'ABC0123091274723507156XYZ' # without 8
s2 = 'ABC0123091274723507156XYZ8'

len(set("".join(re.findall(r'([\d])', s1)))) == 10 # False
len(set("".join(re.findall(r'([\d])', s2)))) == 10 # True

How it works:

Find all digits in the string with regex findall. Join all matches to one string and get unique characters by putting it in a set. Then count the length of the set. If all digits are represented the length will be 10.

score 0 · Answer 4 · edited Jun 01 '20 at 20:20

0

You could just use .isdigit() to filter for non numbers it returns a boolean.

This has the limitation of only working for contiguous numbers/integers.

edited Jun 01 '20 at 20:20

notacorn

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answered Jan 26 '18 at 15:18

Mike Tung

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only if the string doesn't contain any non-numeric characters – karlson Jan 26 '18 at 15:19
The string has all numbers only. I just need to find if it has all the numbers 0,1,2,3,4,5,6,7,8,9 somewhere or the other in the string. – Abhisek Roy Jan 26 '18 at 15:24

Regex to find if a string contains all numbers 0-9 be it in any order

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