如何从列表中删除第一个项目?

问题:如何从列表中删除第一个项目?

我有表[0, 1, 2, 3, 4],我想将它做成[1, 2, 3, 4]。我该怎么办?

I have the list [0, 1, 2, 3, 4] I’d like to make it into [1, 2, 3, 4]. How do I go about this?


回答 0

Python清单

list.pop(索引)

>>> l = ['a', 'b', 'c', 'd']
>>> l.pop(0)
'a'
>>> l
['b', 'c', 'd']
>>> 

删除列表[索引]

>>> l = ['a', 'b', 'c', 'd']
>>> del l[0]
>>> l
['b', 'c', 'd']
>>> 

这些都将修改您的原始列表。

其他人建议使用切片:

  • 复制清单
  • 可以返回一个子集

另外,如果要执行许多pop(0),则应查看collections.deque

from collections import deque
>>> l = deque(['a', 'b', 'c', 'd'])
>>> l.popleft()
'a'
>>> l
deque(['b', 'c', 'd'])
  • 从列表的左端提供更高的性能

Python List

list.pop(index)

>>> l = ['a', 'b', 'c', 'd']
>>> l.pop(0)
'a'
>>> l
['b', 'c', 'd']
>>> 

del list[index]

>>> l = ['a', 'b', 'c', 'd']
>>> del l[0]
>>> l
['b', 'c', 'd']
>>> 

These both modify your original list.

Others have suggested using slicing:

  • Copies the list
  • Can return a subset

Also, if you are performing many pop(0), you should look at collections.deque

from collections import deque
>>> l = deque(['a', 'b', 'c', 'd'])
>>> l.popleft()
'a'
>>> l
deque(['b', 'c', 'd'])
  • Provides higher performance popping from left end of the list

回答 1

切片:

x = [0,1,2,3,4]
x = x[1:]

这实际上将返回原始的子集,但不会对其进行修改。

Slicing:

x = [0,1,2,3,4]
x = x[1:]

Which would actually return a subset of the original but not modify it.


回答 2

>>> x = [0, 1, 2, 3, 4]
>>> x.pop(0)
0

更多关于此这里

>>> x = [0, 1, 2, 3, 4]
>>> x.pop(0)
0

More on this here.


回答 3

使用列表切片,请参阅有关列表的Python教程以获取更多详细信息:

>>> l = [0, 1, 2, 3, 4]
>>> l[1:]
[1, 2, 3, 4]

With list slicing, see the Python tutorial about lists for more details:

>>> l = [0, 1, 2, 3, 4]
>>> l[1:]
[1, 2, 3, 4]

回答 4

你会这样做

l = [0, 1, 2, 3, 4]
l.pop(0)

要么 l = l[1:]

利弊

使用pop可以获取值

x = l.pop(0) x0

you would just do this

l = [0, 1, 2, 3, 4]
l.pop(0)

or l = l[1:]

Pros and Cons

Using pop you can retrieve the value

say x = l.pop(0) x would be 0


回答 5

然后将其删除:

x = [0, 1, 2, 3, 4]
del x[0]
print x
# [1, 2, 3, 4]

Then just delete it:

x = [0, 1, 2, 3, 4]
del x[0]
print x
# [1, 2, 3, 4]

回答 6

您可以list.reverse()用来反转列表,然后list.pop()删除最后一个元素,例如:

l = [0, 1, 2, 3, 4]
l.reverse()
print l
[4, 3, 2, 1, 0]


l.pop()
0
l.pop()
1
l.pop()
2
l.pop()
3
l.pop()
4

You can use list.reverse() to reverse the list, then list.pop() to remove the last element, for example:

l = [0, 1, 2, 3, 4]
l.reverse()
print l
[4, 3, 2, 1, 0]


l.pop()
0
l.pop()
1
l.pop()
2
l.pop()
3
l.pop()
4

回答 7

您也可以使用list.remove(a[0])pop删除列表中的第一个元素。

>>>> a=[1,2,3,4,5]
>>>> a.remove(a[0])
>>>> print a
>>>> [2,3,4,5]

You can also use list.remove(a[0]) to pop out the first element in the list.

>>>> a=[1,2,3,4,5]
>>>> a.remove(a[0])
>>>> print a
>>>> [2,3,4,5]

回答 8

如果使用numpy,则需要使用delete方法:

import numpy as np

a = np.array([1, 2, 3, 4, 5])

a = np.delete(a, 0)

print(a) # [2 3 4 5]

If you are working with numpy you need to use the delete method:

import numpy as np

a = np.array([1, 2, 3, 4, 5])

a = np.delete(a, 0)

print(a) # [2 3 4 5]

回答 9

有一个称为“双端队列”或双头队列的数据结构,它比列表更快,更高效。您可以使用列表并将其转换为双端队列,并在其中进行所需的转换。您也可以将双端队列转换回列表。

import collections
mylist = [0, 1, 2, 3, 4]

#make a deque from your list
de = collections.deque(mylist)

#you can remove from a deque from either left side or right side
de.popleft()
print(de)

#you can covert the deque back to list
mylist = list(de)
print(mylist)

Deque还提供了非常有用的功能,例如将元素插入列表的任一侧或任何特定的索引。您也可以旋转或反转双端队列。试试看!!

There is a datastructure called “deque” or double ended queue which is faster and efficient than a list. You can use your list and convert it to deque and do the required transformations in it. You can also convert the deque back to list.

import collections
mylist = [0, 1, 2, 3, 4]

#make a deque from your list
de = collections.deque(mylist)

#you can remove from a deque from either left side or right side
de.popleft()
print(de)

#you can covert the deque back to list
mylist = list(de)
print(mylist)

Deque also provides very useful functions like inserting elements to either side of the list or to any specific index. You can also rotate or reverse a deque. Give it a try!!