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.
我知道这听起来微不足道，但我没有意识到Python的sort（）函数很奇怪。我有一个实际上是字符串形式的“数字”列表，所以我首先将它们转换为int，然后尝试排序。
list1=["1","10","3","22","23","4","2","200"]
for item in list1:
item=int(item)
list1.sort()
print list1
Gives me:
给我：
['1', '10', '2', '200', '22', '23', '3', '4']
What I want is
我想要的是
['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.
我查看了一些与排序数字集相关的算法，但我发现的算法都涉及排序字母数字集。
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.
我知道这可能是一个毫无疑问的问题，但谷歌和我的教科书没有提供比.sort（）函数更多或更少有用的东西。
9 个解决方案
#1
140
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:
您实际上没有将字符串转换为整数。或者说，你做了，但后来你没有对结果做任何事情。你想要的是：
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)
但是，python使你更容易：sort接受一个命名参数key，这是一个在比较每个元素之前调用的函数（但不修改列表）
list1 = ["1","10","3","22","23","4","2","200"]
# call int(x) on each element before comparing it
list1.sort(key=int)
#2
30
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.
您可以将函数传递给.sort方法的key参数。有了这个，系统将按键（x）而不是x排序。
list1.sort(key=int)
BTW, to convert the list to integers permanently, use the map function
顺便说一句，要将列表永久转换为整数，请使用map函数
list1 = list(map(int, list1)) # you don't need to call list() in Python 2.x
or list comprehension
或列表理解
list1 = [int(x) for x in list1]
#3
13
In case you want to use sorted() function: sorted(list1, key=int)
如果你想使用sorted（）函数：sorted（list1，key = int）
It returns a new sorted list.
它返回一个新的排序列表。
#4
10
Python's sort isn't weird. It's just that this code:
Python的排序并不奇怪。只是这段代码：
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.
没有做你认为的事情 - 项目没有被替换回列表，它被简单地扔掉了。
Anyway, the correct solution is to use key=int as others have shown you.
无论如何，正确的解决方案是使用key = int，正如其他人向您展示的那样。
#5
5
You can also use:
您还可以使用：
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..]
这与你可以在互联网上找到的其他东西非常相似，但也适用于像[abc0.1，abc0.2 ..]这样的字母数字
#6
3
Seamus Campbell's answer doesnot work on python2.x.list1 = sorted(list1, key=lambda e: int(e)) using lambda function works well.
Seamus Campbell的答案对python2.x不起作用。 list1 = sorted（list1，key = lambda e：int（e））使用lambda函数效果很好。
#7
0
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.
最新的解决方案是正确的。您正在以字符串形式阅读解决方案，在这种情况下，顺序为1，然后是100，然后是104，然后是2，然后是21，然后是2001001010，3，依此类推。
You have to CAST your input as an int instead:
你必须将你的输入作为int来代替：
sorted strings:
排序字符串：
stringList = (1, 10, 2, 21, 3)
stringList =（1,10,2,21,3）
sorted ints:
排序的整数：
intList = (1, 2, 3, 10, 21)
intList =（1,2,3,10,21）
To cast, just put the stringList inside int ( blahblah ).
要转换，只需将stringList放在int（blahblah）中。
Again:
再次：
stringList = (1, 10, 2, 21, 3)
newList = int (stringList)
print newList
=> returns (1, 2, 3, 10, 21)
#8
0
I approached the same problem yesterday and found a module called natsort which solves the problems. Use:
我昨天遇到了同样的问题，发现了一个名为natsort的模块来解决问题。使用：
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']
#9
-2
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.
这对我来说使用python版本3，虽然它没有在版本2中。
#1
140
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:
您实际上没有将字符串转换为整数。或者说，你做了，但后来你没有对结果做任何事情。你想要的是：
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)
但是，python使你更容易：sort接受一个命名参数key，这是一个在比较每个元素之前调用的函数（但不修改列表）
list1 = ["1","10","3","22","23","4","2","200"]
# call int(x) on each element before comparing it
list1.sort(key=int)
#2
30
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.
您可以将函数传递给.sort方法的key参数。有了这个，系统将按键（x）而不是x排序。
list1.sort(key=int)
BTW, to convert the list to integers permanently, use the map function
顺便说一句，要将列表永久转换为整数，请使用map函数
list1 = list(map(int, list1)) # you don't need to call list() in Python 2.x
or list comprehension
或列表理解
list1 = [int(x) for x in list1]
#3
13
In case you want to use sorted() function: sorted(list1, key=int)
如果你想使用sorted（）函数：sorted（list1，key = int）
It returns a new sorted list.
它返回一个新的排序列表。
#4
10
Python's sort isn't weird. It's just that this code:
Python的排序并不奇怪。只是这段代码：
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.
没有做你认为的事情 - 项目没有被替换回列表，它被简单地扔掉了。
Anyway, the correct solution is to use key=int as others have shown you.
无论如何，正确的解决方案是使用key = int，正如其他人向您展示的那样。
#5
5
You can also use:
您还可以使用：
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..]
这与你可以在互联网上找到的其他东西非常相似，但也适用于像[abc0.1，abc0.2 ..]这样的字母数字
#6
3
Seamus Campbell's answer doesnot work on python2.x.list1 = sorted(list1, key=lambda e: int(e)) using lambda function works well.
Seamus Campbell的答案对python2.x不起作用。 list1 = sorted（list1，key = lambda e：int（e））使用lambda函数效果很好。
#7
0
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.
最新的解决方案是正确的。您正在以字符串形式阅读解决方案，在这种情况下，顺序为1，然后是100，然后是104，然后是2，然后是21，然后是2001001010，3，依此类推。
You have to CAST your input as an int instead:
你必须将你的输入作为int来代替：
sorted strings:
排序字符串：
stringList = (1, 10, 2, 21, 3)
stringList =（1,10,2,21,3）
sorted ints:
排序的整数：
intList = (1, 2, 3, 10, 21)
intList =（1,2,3,10,21）
To cast, just put the stringList inside int ( blahblah ).
要转换，只需将stringList放在int（blahblah）中。
Again:
再次：
stringList = (1, 10, 2, 21, 3)
newList = int (stringList)
print newList
=> returns (1, 2, 3, 10, 21)
#8
0
I approached the same problem yesterday and found a module called natsort which solves the problems. Use:
我昨天遇到了同样的问题，发现了一个名为natsort的模块来解决问题。使用：
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']
#9
-2
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.
这对我来说使用python版本3，虽然它没有在版本2中。