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']

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

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 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

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

It returns a new sorted list.

Answer 4

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 5

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 6

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

Answer 7

Try this, it’ll sort the list in-place in descending order (there’s no 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 8

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 9

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']. ie 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 10

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 11

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

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)