Python 2.7给出了ValueError list.remove（x）x不在列表中 [英] Python 2.7 gives ValueError list.remove(x) x not in list

问题描述

每次运行此程序时，我都会收到以下错误：

Every time I run this program, I get this error:

ValueError list.remove（x）x不在列表中

我试图降低一个外星人的健康，当它被一个螺栓击中。如果外星人的健康状况为= 0，外星人也应该被摧毁。同样地，螺栓也会被破坏。这是我的代码：

I am trying to lower the health of an alien whenever it is hit by a bolt. The alien should also be destroyed if its health is <= 0. Similarly, the bolt would also be destroyed. Here is my code:

def manage_collide(bolts, aliens):
    # Check if a bolt collides with any alien(s)
    for b in bolts:
        for a in aliens:
            if b['rect'].colliderect(a['rect']):
                for a in aliens:
                    a['health'] -= 1
                    bolts.remove(b)
                    if a['health'] == 0:
                        aliens.remove(a)
    # Return bolts, aliens dictionaries
    return bolts, aliens

ValueError出现在行 aliens.remove（a）。只是为了澄清，外星人和螺栓都是字典列表。

The ValueError appears on the line aliens.remove(a). Just to clarify, both the aliens and bolts are lists of dictionaries.

推荐答案

您不应该从正在循环的列表中删除项目。创建副本：

You should not remove items from a list you are looping over. Create a copy instead:

for a in aliens[:]:

和

for b in bolts[:]:

在循环时修改列表会影响循环：

Modifying a list while looping over it, affects the loop:

>>> lst = [1, 2, 3]
>>> for i in lst:
...     print i
...     lst.remove(i)
... 
1
3
>>> lst
[2]

从循环两次的列表中删除项目，使事情成为一个稍微复杂一点，导致ValueError：

Removing items from a list you are looping over twice makes things a little more complicated still, resulting in a ValueError:

>>> lst = [1, 2, 3]
>>> for i in lst:
...     for a in lst:
...         print i, a, lst
...         lst.remove(i)
... 
1 1 [1, 2, 3]
1 3 [2, 3]
Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
ValueError: list.remove(x): x not in list

创建副本时您正在每个级别的循环中修改的列表，您可以避免此问题：

When creating a copy of the lists you are modifying at each level of your loops, you avoid the problem:

>>> lst = [1, 2, 3]
>>> for i in lst[:]:
...     for i in lst[:]:
...         print i, lst
...         lst.remove(i)
... 
1 [1, 2, 3]
2 [2, 3]
3 [3]

当您发生碰撞时，您只需删除 b bolt 一次，而不是在你伤害外星人的循环中。稍后清理外星人：

When you have a collision, you only need to remove the b bolt once, not in the loop where you hurt the aliens. Clean out the aliens separately later:

def manage_collide(bolts, aliens):
    for b in bolts[:]:
        for a in aliens:
            if b['rect'].colliderect(a['rect']) and a['health'] > 0:
                bolts.remove(b)
                for a in aliens:
                    a['health'] -= 1
    for a in aliens[:]:
        if a['health'] <= 0:
            aliens.remove(a)
    return bolts, aliens

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