python throws_Python 2.7 throws ValueError: list.remove(x): x not in list

最新推荐文章于 2023-05-15 14:43:58 发布

原创最新推荐文章于 2023-05-15 14:43:58 发布 · 883 阅读

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#python throws

本文介绍了在Python中处理列表时的一些常见陷阱，特别是在循环中删除列表元素时的问题。为了避免这些问题，文章推荐使用列表副本进行操作，并提供了一段示例代码说明如何在遍历列表的同时安全地移除元素。此外，还展示了一个更复杂的例子，涉及两个列表间的碰撞检测和健康值更新。

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You should not remove items from a list you are looping over. Create a copy instead: for a in aliens[:]:

and for b in bolts[:]:

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]

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

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]

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']