How to sort a list of strings 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 for

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)

---

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

may be not the best python, but for string lists like ['1','1.0','2.0','2', '1.1', '1.10', '1.11', '1.2','7','3','5']with the expected target ['1', '1.0', '1.1', '1.2', '1.10', '1.11', '2', '2.0', '3', '5', '7'] helped me...

unsortedList = ['1','1.0','2.0','2', '1.1', '1.10', '1.11', '1.2','7','3','5']
sortedList = []
sortDict = {}
sortVal = []
#set zero correct (integer): examp: 1.000 will be 1 and breaks the order
zero = "000"
for i in sorted(unsortedList):
  x = i.split(".")
  if x[0] in sortDict:
    if len(x) > 1:
        sortVal.append(x[1])
    else:
        sortVal.append(zero)
    sortDict[x[0]] = sorted(sortVal, key = int)
  else:
    sortVal = []
    if len(x) > 1:
        sortVal.append(x[1])
    else:
        sortVal.append(zero)
    sortDict[x[0]] = sortVal
for key in sortDict:
  for val in sortDict[key]:
    if val == zero:
       sortedList.append(str(key))
    else:
       sortedList.append(str(key) + "." + str(val))
print(sortedList)

Answer 3

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)

Answer 4

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 5

I approached the same problem yesterday and found a module called [natsort][1], which solves your problem. Use:

from natsort import natsorted # pip install natsort

# Example list of strings
a = ['1', '10', '2', '3', '11']

[In]  sorted(a)
[Out] ['1', '10', '11', '2', '3']

[In]  natsorted(a)
[Out] ['1', '2', '3', '10', '11']

# Your array may contain strings
[In]  natsorted(['string11', 'string3', 'string1', 'string10', 'string100'])
[Out] ['string1', 'string3', 'string10', 'string11', 'string100']

It also works for dictionaries as an equivalent of sorted. [1]: https://pypi.org/project/natsort/

Answer 6

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

It returns a new sorted list.