Python 使用用户 function 进行列表排序。 （错误）

Question

i have a list of strings which are split by a space. i want to sort the list by the second split value of the strings.

it works, if i use lambda:

Names = ['Ccc Eee', 'Bbb Aaa', 'Aaa Bbb', 'Zzz Zzz', 'Ddd Ddd']
# Names.sort(key=lambda name: name.split(' ')[1])
# print(Names) # ['Bbb Aaa', 'Aaa Bbb', 'Ddd Ddd', 'Ccc Eee', 'Zzz Zzz']

but if i want to sort it without using lambda, it shows error:

def mysort(name):
    for i in range(0, len(name)):
        return name[i].split(' ')[1]

Names.sort(key=mysort(Names))
print(mysort(Names))

# output:
# Names.sort(key=mysort(Names))
# TypeError: 'str' object is not callable

what am i doing wrong?

Update:

def mysort(name):
    return name.split(' ')[-1]

Names.sort(key=mysort)
print(Names) # ['Bbb Aaa', 'Aaa Bbb', 'Ddd Ddd', 'Ccc Eee', 'Zzz Zzz']

Answer 1

Two problems.

• You should not call the function.

  Just like you are doing

  Names.sort(key=lambda name: name.split(' ')[1])

  and not

  Names.sort(key=lambda name: name.split(' ')[1](Names))

  , you should do

  Names.sort(key=mysort)

  and not

  Names.sort(key=mysort(Names))

• You don't need the loop in the function:

   def mysort(name): return name.split(' ')[1]

Answer 2

def mysort(name):
    return name.split(' ')[1]

Names = ['Ccc Eee', 'Bbb Aaa', 'Aaa Bbb', 'Zzz Zzz', 'Ddd Ddd']
Names.sort(key=mysort)
print(Names)

This will give you the exact answer.
And the problem with your approach is,

1. inside your mysort function you are using loop which you shouldn't
2. you are passing argument to the function while mentioning key which you shouldn't because the value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes.