在python中从列表中删除项，同时对其进行迭代

Delete item from list in Python while iterating over it

本问题已经有最佳答案，请猛点这里访问。

我正在为锦标赛应用程序编写循环算法。

当玩家数量为奇数时，我将'DELETE'添加到玩家列表中，但稍后，当我想从包含'DELETE'的日程表列表中删除所有项目时，我不能——总是留下一个。请看一下代码——问题很简单，我想是关于列表的；我就是看不到。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45

"""
Round-robin tournament:
1, 2, 3, 4, | 5, 6, 7, 8 => 1, 2, 3, 4 => rotate all but 1 => 1, 5, 2, 3 => repeat => 1, 6, 5, 2 ...
5, 6, 7, 8 6, 7, 8, 4 7, 8, 4, 3
in every round pick l1[0] and l2[0] as first couple, after that l1[1] and l2[1]...
"""

import math

lst = []
schedule = []
delLater = False

for i in range(3): #make list of numbers
lst.append(i+1)

if len(lst) % 2 != 0: #if num of items is odd, add 'DELETE'
lst.append('DELETE')
delLater = True

while len(schedule) < math.factorial(len(lst))/(2*math.factorial(len(lst) - 2)): #!(n)/!(n-k)

mid = len(lst)/2

l1 = lst[:mid]
l2 = lst[mid:]

for i in range(len(l1)):
schedule.append((l1[i], l2[i])) #add lst items in schedule

l1.insert(1, l2[0]) #rotate lst
l2.append(l1[-1])
lst = l1[:-1] + l2[1:]

if delLater == True: #PROBLEM!!! One DELETE always left in list
for x in schedule:
if 'DELETE' in x:
schedule.remove(x)

i = 1
for x in schedule:
print i, x
i+=1

1	schedule[:] = [x for x in schedule if 'DELETE' not in x]

请参阅有关在迭代列表时从列表中删除的其他问题。

相关讨论

要在遍历列表时从列表中删除元素，需要向后：

1
2
3
4

if delLater == True:
for x in schedule[-1::-1]):
if 'DELETE' in x:
schedule.remove(x)

更好的选择是使用列表理解：

1	schedule[:] = [item for item in schedule if item != 'DELETE']

现在，你可以用schedule =代替schedule[:] =，有什么区别？一个例子是：

1
2
3
4
5
6

schedule = [some list stuff here] # first time creating
modify_schedule(schedule, new_players='...') # does validation, etc.

def modify_schedule(sched, new_players):
# validate, validate, hallucinate, illustrate...
sched = [changes here]

此时，modify_schedule所做的所有更改都已丢失。为什么？因为它没有就地修改列表对象，而是将名称sched重新绑定到一个新列表，使原始列表仍然绑定到调用方的名称schedule。

因此，您是否使用list_object[:] =取决于您是否希望将任何其他名称绑定到您的列表以查看更改。