这三种从列表中删除元素的方法有什么区别吗?
>>> a = [1, 2, 3]
>>> a.remove(2)
>>> a
[1, 3]
>>> a = [1, 2, 3]
>>> del a[1]
>>> a
[1, 3]
>>> a = [1, 2, 3]
>>> a.pop(1)
2
>>> a
[1, 3]
这三种从列表中删除元素的方法有什么区别吗?
>>> a = [1, 2, 3]
>>> a.remove(2)
>>> a
[1, 3]
>>> a = [1, 2, 3]
>>> del a[1]
>>> a
[1, 3]
>>> a = [1, 2, 3]
>>> a.pop(1)
2
>>> a
[1, 3]
当前回答
Remove主要作用于值。 删除和弹出工作在索引上
Remove基本上删除了第一个匹配的值。 Delete从指定索引中删除项 Pop基本上接受一个索引并返回该索引的值。下次打印列表时,该值不会出现。
其他回答
Remove主要作用于值。 删除和弹出工作在索引上
Remove基本上删除了第一个匹配的值。 Delete从指定索引中删除项 Pop基本上接受一个索引并返回该索引的值。下次打印列表时,该值不会出现。
也可以使用remove按索引删除值。
n = [1, 3, 5]
n.remove(n[1])
N则表示[1,5]
列表上的删除操作给定一个要删除的值。它搜索列表以查找具有该值的项,并删除找到的第一个匹配项。如果没有匹配项,则是一个错误,引发ValueError。
>>> x = [1, 0, 0, 0, 3, 4, 5]
>>> x.remove(4)
>>> x
[1, 0, 0, 0, 3, 5]
>>> del x[7]
Traceback (most recent call last):
File "<pyshell#1>", line 1, in <module>
del x[7]
IndexError: list assignment index out of range
del语句可用于删除整个列表。如果你有一个特定的列表项作为del的参数(例如listname[7]专门引用列表中的第8项),它会删除该项。甚至可以从列表中删除“slice”。如果索引超出范围,则会引发IndexError。
>>> x = [1, 2, 3, 4]
>>> del x[3]
>>> x
[1, 2, 3]
>>> del x[4]
Traceback (most recent call last):
File "<pyshell#1>", line 1, in <module>
del x[4]
IndexError: list assignment index out of range
The usual use of pop is to delete the last item from a list as you use the list as a stack. Unlike del, pop returns the value that it popped off the list. You can optionally give an index value to pop and pop from other than the end of the list (e.g listname.pop(0) will delete the first item from the list and return that first item as its result). You can use this to make the list behave like a queue, but there are library routines available that can provide queue operations with better performance than pop(0) does. It is an error if there index out of range, raises a IndexError.
>>> x = [1, 2, 3]
>>> x.pop(2)
3
>>> x
[1, 2]
>>> x.pop(4)
Traceback (most recent call last):
File "<pyshell#1>", line 1, in <module>
x.pop(4)
IndexError: pop index out of range
有关更多细节,请参阅collections.deque。
因为没有人提到它,请注意del(不像pop)允许删除一系列索引,因为列表切片:
>>> lst = [3, 2, 2, 1]
>>> del lst[1:]
>>> lst
[3]
这也允许避免IndexError如果索引不在列表中:
>>> lst = [3, 2, 2, 1]
>>> del lst[10:]
>>> lst
[3, 2, 2, 1]
pop
获取index(如果给定,则获取last),删除该索引处的值,并返回值
删除
获取值,删除第一次出现的内容,并不返回任何内容
删除
获取索引,删除该索引处的值,并不返回任何值