How to get integer values from a string in Python?

Question

Suppose I had a string

string1 = "498results should get"

Now I need to get only integer values from the string like 498. Here I don't want to use list slicing because the integer values may increase like these examples:

string2 = "49867results should get" 
string3 = "497543results should get"

So I want to get only integer values out from the string exactly in the same order. I mean like 498,49867,497543 from string1,string2,string3 respectively.

Can anyone let me know how to do this in a one or two lines?

jamylak · Accepted Answer · 2015-03-19T14:41:05.680

148

>>> import re
>>> string1 = "498results should get"
>>> int(re.search(r'\d+', string1).group())
498

If there are multiple integers in the string:

>>> map(int, re.findall(r'\d+', string1))
[498]

edited Mar 19 '15 at 14:41

answered Jul 05 '12 at 06:57

jamylak

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3

this is not exactly correct, yes it will do for string **starting** with a integer, but if the integer is in the middle of the string, it won't do. Maybe it should be better to use `int(re.search(r'\d+', string1).group())` – eLRuLL Mar 20 '14 at 17:38
1

If `string1` is **(020) 3493**, it fails. – Hussain Mar 19 '15 at 12:53
Right it was the wrong code, should ahve used `re.search` – jamylak Mar 19 '15 at 14:40
1

@Hussain well, the example only showed numbers at the fron though but anyway now it's safe either way – jamylak Mar 19 '15 at 14:41
2

This will work fine. But, one should check if result of `re.search` is not **None**. Otherwise, it'll throw an **AttributeError** (`'NoneType' object has no attribute 'group'`) if there is no digit present in the string after accessing `.group()`. – Mohsin Aljiwala Jun 15 '17 at 10:14
@gofmtyourself Easier to ask for forgiveness than permission. The error can be handled in `try` `except AttributeError` – jamylak Jun 19 '17 at 22:05

score 61 · Answer 2 · edited May 23 '17 at 12:10

61

An answer taken from ChristopheD here: https://stackoverflow.com/a/2500023/1225603

r = "456results string789"
s = ''.join(x for x in r if x.isdigit())
print int(s)
456789

edited May 23 '17 at 12:10

Community

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answered Jul 26 '12 at 16:08

Sepero

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score 28 · Answer 3 · answered Nov 17 '16 at 18:54

28

Here's your one-liner, without using any regular expressions, which can get expensive at times:

>>> ''.join(filter(str.isdigit, "1234GAgade5312djdl0"))

returns:

'123453120'

answered Nov 17 '16 at 18:54

Bobort

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score 19 · Answer 4 · answered May 24 '16 at 11:07

19

if you have multiple sets of numbers then this is another option

>>> import re
>>> print(re.findall('\d+', 'xyz123abc456def789'))
['123', '456', '789']

its no good for floating point number strings though.

answered May 24 '16 at 11:07

jacanterbury

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score 10 · Answer 5 · answered Jul 05 '12 at 07:02

10

Iterator version

>>> import re
>>> string1 = "498results should get"
>>> [int(x.group()) for x in re.finditer(r'\d+', string1)]
[498]

answered Jul 05 '12 at 07:02

John La Rooy

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score 7 · Answer 6 · answered Jul 05 '12 at 07:06

7

>>> import itertools
>>> int(''.join(itertools.takewhile(lambda s: s.isdigit(), string1)))

answered Jul 05 '12 at 07:06

Craig Citro

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This will only work if the number is at the start of the string. Also, why not use `str.isdigit` instead of the lambda? – John La Rooy Jul 05 '12 at 07:15
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This could also be written as `int(''.join(itertools.takewhile(str.isdigit, string1)))`. I would never actually use either method since this is overcomplicating it. – jamylak Jul 05 '12 at 07:17
https://groups.google.com/forum/?hl=en&fromgroups#!msg/alt.religion.emacs/DR057Srw5-c/Co-2L2BKn7UJ – Craig Citro Jul 05 '12 at 07:23
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I'm planning ahead for the case where the strings are thousands of characters long. – Craig Citro Jul 05 '12 at 07:46

score 4 · Answer 7 · answered May 16 '18 at 15:50

4

With python 3.6, these two lines return a list (may be empty)

>>[int(x) for x in re.findall('\d+', your_string)]

Similar to

>>list(map(int, re.findall('\d+', your_string))

answered May 16 '18 at 15:50

huseyin39

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score 1 · Answer 8 · answered Jul 05 '21 at 05:34

this approach uses list comprehension, just pass the string as argument to the function and it will return a list of integers in that string.

def getIntegers(string):
        numbers = [int(x) for x in string.split() if x.isnumeric()]
        return numbers

Like this

print(getIntegers('this text contains some numbers like 3 5 and 7'))

Output

[3, 5, 7]

hey alex · Answer 9 · 2022-10-29T14:09:37.597

1

  integerstring=""
  string1 = "498results should get"
  for i in string1:
      if i.isdigit()==True 
      integerstring=integerstring+i
   print(integerstring)

edited Oct 29 '22 at 14:09

answered Oct 29 '22 at 14:05

hey alex

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score 0 · Answer 10 · edited Oct 30 '17 at 15:46

0

def function(string):  
    final = ''  
    for i in string:  
        try:   
            final += str(int(i))   
        except ValueError:  
            return int(final)  
print(function("4983results should get"))

edited Oct 30 '17 at 15:46

JochenJung

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answered Oct 30 '17 at 15:20

Kishor Subedi

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not sure why you had to do `return int(final)` just returning `return final` also works. – AhmFM Nov 20 '20 at 00:40
also can you explain what is making it bring only digits? – AhmFM Nov 20 '20 at 00:41

score 0 · Answer 11 · answered Feb 01 '22 at 23:40

Another option is to remove the trailing the letters using rstrip and string.ascii_lowercase (to get the letters):

import string
out = [int(s.replace(' ','').rstrip(string.ascii_lowercase)) for s in strings]

Output:

[498, 49867, 497543]

How to get integer values from a string in Python?

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