使用 List Comprehension 获取前 n 项

Question

我试过了：

x = [1,2,3,4,5,6,7,100,-1]
y = [a for a in x.sort()[0:5]]
print(y)
y = y[0:4]
print(y)

但这行不通，因为 sort() 不会 return 排序列表

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-21-d13e11c1e8ab> in <module>
      1 x = [1,2,3,4,5,6,7,100,-1]
----> 2 y = [a for a in x.sort()[0:5]]
      3 print(y)
      4 y = y[0:4]
      5 print(y)

TypeError: 'NoneType' object is not subscriptable

Answer 1

.sort() 就地排序列表

list.sort() sorts the list in-place, mutating the list indices, and returns None (like all in-place operations)

所以你必须这样做：

x = [1,2,3,4,5,6,7,100,-1]
x.sort()
y = [a for a in x[0:5]]
print(y)
y = y[0:4]
print(y)

输出：

[-1, 1, 2, 3, 4]
[-1, 1, 2, 3]

The difference between .sort() and sorted() in Python.

使用 List Comprehension 获取前 n 项

Use List Comprehension to get the top n items

python

list-comprehension