Numeric Sort in Python

Question

I know that this sounds trivial but I did not realize that the sort() function of Python was weird. I have a list of "numbers" that are actually in string form, so I first convert them to ints, then attempt a sort.

list1=["1","10","3","22","23","4","2","200"]
for item in list1:
    item=int(item)

list1.sort()
print list1

Gives me: ['1', '10', '2', '200', '22', '23', '3', '4']

What I want is ['1','2','3','4','10','22','23','200']

I've looked around for some of the algorithms associated with sorting numeric sets, but the ones I found all involve sorting alphanumeric sets. I know this is probably a no brainer problem but google and my textbook don't offer anything more or less useful than the .sort() function. Thanks for your help.

Answer 1

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

list1.sort(key=int)

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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]

Answer 2

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()

However, python makes it even easier for you: sort takes a named parameter, key, which is a function that is called on each element before it is compared (but without modifying the list)

list1 = ["1","10","3","22","23","4","2","200"]
# call int(x) on each element before comparing it
list1.sort(key=int)

Answer 3

Python's sort isn't weird. It's just that this code:

for item in list1:
   item=int(item)

isn't doing what you think it is - item is not replaced back into the list, it is simply thrown away.

Anyway, the correct solution is to use key=int as others have shown you.