python排序错误：TypeError：'list'对象不可调用

Question

I am trying to sort a list of list based on the first item of each element:

def getKey(item):
    return item[0]    

def myFun(x):
    return sorted(x, key= getKey(x))

my_x = [[1,100], [5,200], [3,30]]

myFun(my_x)

I want the sorting based on the first item of each element, ie 1, 5 and 3. The expected result should be: [[1,100], [3,30], [5,200]]

However, I got the error below:

TypeErrorTraceback (most recent call last)
<ipython-input-7-4c88bbc1a944> in <module>()
      3 
      4 my_x = [[1,100], [5,200], [3,30]]
----> 5 myFun(my_x)

<ipython-input-7-4c88bbc1a944> in myFun(x)
      1 def myFun(x):
----> 2     return sorted(x, key= getKey(x))
      3 
      4 my_x = [[1,100], [5,200], [3,30]]
      5 myFun(my_x)

TypeError: 'list' object is not callable

Any idea what I did wrong here? Thanks!

Answer 1

The problem is that you're assigning the result of getKey(x) to key=getKey(x) . You want to assign the function pointer.

def getKey(item):
    return item[0]    

def myFun(x):
    return sorted(x, key=getKey)

my_x = [[1,100], [5,200], [3,30]]

myFun(my_x)

Having said that, if you're just sorting based on the first element, that's the default behavior of sorted .

You can just do sorted(x) .

Answer 2

You don't need to call the function getKey . The signature of sorted requires you pass the key argument as a callable:

sorted(x, key=getKey)

Answer 3

getKey is a function, (a type of "callable" object). Its output, getKey(x) , on the other hand, is a list object which is not callable. Your mistake is that you're setting key=getKey(x) and hence assigning a list object to the argument key , whereas sorted expects something callable attached to that name. So that explains why, when the internal sorting code tries to call your key , it fails with the error "'list' object is not callable". Really, you should have just said key=getKey .

Answer 4

UPDATE

sorted 's key attribute should be callable, so instead of assigning key=getKey() you need to key=getKey

INITIAL ANSWER (could be useful if you want to make code look better):

There is no such method as getKey , butthere is itemgetter from operator package

from operator import itemgetter


def myFun(x):
    return sorted(x, key=itemgetter(0))

my_x = [[1,100], [5,200], [3,30]]

myFun(my_x)

To the downvoters :

It's clearly said that

I want the sorting based on the first item of each element

And if you think that you would sort presented list only with sorted() , you are wrong.

>>> my_x = [[1,100], [5,200], [3,30], [1,30]]
>>> myFun(my_x)
[[1, 100], [1, 30], [3, 30], [5, 200]]
>>> sorted(my_x)
[[1, 30], [1, 100], [3, 30], [5, 200]]

So sorted is not based only on first argument, as OP wanted

INITIAL ANSWER'S UPDATE

Also, when i answered there was no getKey method, you can look at OP post edit history

And so OP post now have getKey method you can look at the appropriate @MosesKoledoye 's answer

Answer 5

If you are coming here because you got the exact same error as mentioned in the question

TypeError: 'list' object is not callable

chances are you simply did something like the following:

sorted = sorted(someList) 

The error will go away if the variable name is changed from sorted to something else, like

sortedList = sorted(someList)