Extracting Numbers from a String Without Regular Expressions - python-2.7

Question

I am trying to extract all the numbers from a string composed of digits, symbols and letters.
If the numbers are multi-digit, I have to extract them as multidigit (e.g. from "shsgd89shs2011%%5swts"), I have to pull the numbers out as they appear (89, 2011 and 5).
So far what I have done just loops through and returns all the numbers incrementally, which I like but I cannot figure out how to make it stop
after finishing with one set of digits:
def StringThings(strng):
nums = []
number = ""
for each in range(len(strng)):
if strng[each].isdigit():
number += strng[each]
else:
continue
nums.append(number)
return nums
Running this value: "6wtwyw66hgsgs" returns ['6', '66', '666']
w
hat simple way is there of breaking out of the loop once I have gotten what I needed?

Answer 1

Using your function, just use a temp variable to concat each sequence of digits, yielding the groups each time you encounter a non-digit if the temp variable is not an empty string:
def string_things(strng):
temp = ""
for ele in strng:
if ele.isdigit():
temp += ele
elif temp: # if we have a sequence
yield temp
temp = "" # reset temp
if temp: # catch ending sequence
yield temp
Output
In [9]: s = "shsgd89shs2011%%5swts"
In [10]: list(string_things(s))
Out[10]: ['89', '2011', '5']
In [11]: s ="67gobbledegook95"
In [12]: list(string_things(s))
Out[12]: ['67', '95']
Or you could translate the string replacing letters and punctuation with spaces then split:
from string import ascii_letters, punctuation, maketrans
s = "shsgd89shs2011%%5swts"
replace = ascii_letters+punctuation
tbl = maketrans(replace," " * len(replace))
print(s.translate(tbl).split())
['89', '2011', '5']

Answer 2

L2 = []
file_Name1 = 'shsgd89shs2011%%5swts'
from itertools import groupby
for k,g in groupby(file_Name1, str.isdigit):
a = list(g)
if k == 1:
L2.append("".join(a))
print(L2)
Result ['89', '2011', '5']

Answer 3

Updated to account for trailing numbers:
def StringThings(strng):
nums = []
number = ""
for each in range(len(strng)):
if strng[each].isdigit():
number += strng[each]
if each == len(strng)-1:
if number != '':
nums.append(number)
if each != 0:
if strng[each].isdigit() == False:
if strng[each-1].isdigit():
nums.append(number)
number = ""
continue;
return nums
print StringThings("shsgd89shs2011%%5swts34");
// returns ['89', '2011', '5', '34']
So, when we reach a character which is not a number, and if the previously observed character was a number, append the contents of number to nums and then simply empty our temporary container number, to avoid it containing all the old stuff.
Note, I don't know Python so the solution may not be very pythonic.
Alternatively, save yourself all the work and just do:
import re
print re.findall(r'\d+', 'shsgd89shs2011%%5swts');