How to sort a list numerically?

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.

Answer 1

scores = ['91','89','87','86','85']
scores.sort()
print (scores)

This worked for me using python version 3, though it didn't in version 2.

Answer 2

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)

---

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 3

The most recent solution is right. You are reading solutions as a string, in which case the order is 1, then 100, then 104 followed by 2 then 21, then 2001001010, 3 and so forth.

You have to CAST your input as an int instead:

sorted strings:

stringList = (1, 10, 2, 21, 3)

sorted ints:

intList = (1, 2, 3, 10, 21)

To cast, just put the stringList inside int ( blahblah ).

Again:

stringList = (1, 10, 2, 21, 3)

newList = int (stringList)

print newList

=> returns (1, 2, 3, 10, 21) 

Answer 4

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 5

if you want to use string of the numbers better take another list as shown in my code it will work fine.

list1=["1","10","3","22","23","4","2","200"]

k=[]

for item in list1:

         k.append(int(item))

k.sort()

print(k)

Answer 6

Seamus Campbell's answer doesnot work on python2.x.
list1 = sorted(list1, key=lambda e: int(e)) using lambda function works well.