How to sort a list numerically?

问题:

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.

回答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()

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)

回答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]

回答3:

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

It returns a new sorted list.

回答4:

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.

回答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 for other stuff that you can find on the internet but also works for alphanumericals like [abc0.1, abc0.2..]

回答6:

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

回答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]

回答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) 

回答9:

I approached the same problem yesterday and found a module called natsort which solves the problems. Use:

from natsort import natsorted

# 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\']

回答10:

Simple way to sort a numerical list

    numlists = [5,50,7,51,87,97,53]
    numlists.sort(reverse=False)
    print(numlists)

回答11:

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)

回答12:

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.