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 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']

I want

['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 involved 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

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]) 

Answer 2

I approached the same problem yesterday and found a module called natsort, 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.

Answer 3

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 4

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

It returns a new sorted list.

Answer 5

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, ...].

Answer 6

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.

Answer 7

Seamus Campbell's answer doesn't work on Python 2.x.

list1 = sorted(list1, key=lambda e: int(e)) using lambda function works well.

Answer 8

Try this. It’ll sort the list in-place in descending order (there isn’t any need to specify a key in this case):

Process

listB = [24, 13, -15, -36, 8, 22, 48, 25, 46, -9]
listC = sorted(listB, reverse=True) # listB remains untouched
print listC

output:

 [48, 46, 25, 24, 22, 13, 8, -9, -15, -36]

Answer 9

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 10

The real problem is that 'sort' sorts things alphanumerically.

So if you have a list, ['1', '2', '10', '19'], and run 'sort', you get ['1', '10'. '19', '2']. That is, 10 comes before 2, because it looks at the first character and sorts starting from that.

It seems most methods in Python return things in that order. For example, if you have a directory named 'abc' with the files labelled as 1.jpg, 2.jpg, etc., say, up to 15.jpg and you do file_list=os.listdir(abc), the file_list is not ordered as you expect, but rather as file_list=['1.jpg', '11.jpg'---'15.jpg', '2.jpg].

If the order in which files are processed is important (presumably that's why you named them numerically) the order is not what you think it will be. You can avoid this by using "zeros" padding. For example if you have a list, alist=['01', '03', '05', '10', '02','04', '06] and you run 'sort' on it, you get the order you wanted, alist=['01', '02', etc.], because the first character is 0 which comes before 1. The amount of zeros padding you need is determined by the largest value in the list.

For example, if the largest is, say, between 100 and 1000, you need to pad single digits as 001, 002 ---010, 011--100, 101, etc.

Answer 11

A simple way to sort a numerical list:

numlists = ["5","50","7","51","87","97","53"]
results = list(map(int, numlists))
results.sort(reverse=False)
print(results)

Answer 12

If you want to use strings 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)
# [1, 2, 3, 4, 10, 22, 23, 200]

Answer 13

It may be not the best Python code, 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): example: 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 14

Use:

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.