当您尝试在遍历它时删除列表元素会发生什么情况 [英] What happens when you try to delete a list element while iterating over it
问题描述
some_list = [1,2,3,4]
another_list = [1,2,3,4]
为idx,枚举项目(some_list):
del some_list [idx]
another_list:
another_list.remove(item)
当我列出列表的内容
>>> some_list
[2,4]
>>> another_list
[2,4]
t支持修改一个 list
而迭代它,正确的方法是迭代列表的副本,而不是(所以请不要downvote)。但是我想知道幕后究竟发生了什么,也就是说为什么上面的代码片段的输出 [2,4]
?
class CustomIterator(object):
def __init __(self,seq):
self.seq = seq
self.idx = 0
def __iter __(self):
return self
$ b $ def __next __(self):
print( 'self.idx)
for idx,enumerate(self.seq)中的项:
if id == self.idx:
print(idx,' - - >',item)
else:
print(idx,'',item)
try:
nxtitem = self.seq [self.idx]
除了IndexError:
提高StopIteration
self.idx + = 1
返回nxtitem
下一个=然后在你想检查的列表中使用它: $ b
$ p
$ b $
$ b
some_list = [1,2,3,4]
为idx,枚举中的项(CustomIterator(some_list)):
del some_list [idx]
这个例子说明了这种情况:
给下一个元素:0
0 ---> 1
1 2
2 3
3 4
给下一个元素:1
0 2
1 ---> 3
2 4
给下一个元素:2
0 2
1 4
它只适用于序列。这对于映射或集合来说更为复杂。
I'm iterating a list as follows:
some_list = [1, 2, 3, 4]
another_list = [1, 2, 3, 4]
for idx, item in enumerate(some_list):
del some_list[idx]
for item in another_list:
another_list.remove(item)
When I print out the contents of the lists
>>> some_list
[2, 4]
>>> another_list
[2, 4]
I'm aware that Python doesn't support modifying a list
while iterating over it and the right way is to iterate over copy of list instead (so please don't downvote). But I want to know what exactly happens behind the scenes i.e. Why is the output of the above snippet [2, 4]
?
You can use a self-made iterator that shows (in this case print
s) the state of the iterator:
class CustomIterator(object):
def __init__(self, seq):
self.seq = seq
self.idx = 0
def __iter__(self):
return self
def __next__(self):
print('give next element:', self.idx)
for idx, item in enumerate(self.seq):
if idx == self.idx:
print(idx, '--->', item)
else:
print(idx, ' ', item)
try:
nxtitem = self.seq[self.idx]
except IndexError:
raise StopIteration
self.idx += 1
return nxtitem
next = __next__ # py2 compat
Then use it around the list you want to check:
some_list = [1, 2, 3, 4]
for idx, item in enumerate(CustomIterator(some_list)):
del some_list[idx]
This should illustrate what happens in that case:
give next element: 0
0 ---> 1
1 2
2 3
3 4
give next element: 1
0 2
1 ---> 3
2 4
give next element: 2
0 2
1 4
It only works for sequences though. It's more complicated for mappings or sets.
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