How to sort a list of strings numerically?

You haven't actually converted your strings to ints. Or rather, you did, but then you didn't do anything with the results. What you want is:

list1 = ["1","10","3","22","23","4","2","200"]
list1 = [int(x) for x in list1]
list1.sort()

If for some reason you need to keep strings instead of ints (usually a bad idea, but maybe you need to preserve leading zeros or something), you can use a key function. sort takes a named parameter, key, which is a function that is called on each element before it is compared. The key function's return values are compared instead of comparing the list elements directly:

list1 = ["1","10","3","22","23","4","2","200"]
# call int(x) on each element before comparing it
list1.sort(key=int)
# or if you want to do it all in the same line
list1 = sorted([int(x) for x in list1]) 

You could pass a function to the key parameter to the .sort method. With this, the system will sort by int(x) instead of x.

list1.sort(key=int)

BTW, to convert the list to integers permanently, use the map function

list1 = list(map(int, list1))   # you don't need to call list() in Python 2.x

or list comprehension

list1 = [int(x) for x in list1]

In case you want to use sorted() function: sorted(list1, key=int)

It returns a new sorted list.

You can also use:

import re

def sort_human(l):
    convert = lambda text: float(text) if text.isdigit() else text
    alphanum = lambda key: [convert(c) for c in re.split('([-+]?[0-9]*\.?[0-9]*)', key)]
    l.sort(key=alphanum)
    return l

This is very similar to other stuff that you can find on the internet but also works for alphanumericals like [abc0.1, abc0.2, ...].