[python] How to get integer values from a string in Python?

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?

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

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

returns:

'123453120'

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))

Iterator version

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

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

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

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.

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