Question

我有这段代码：

test = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O"]

for i in test:
    if i not in ["C","D"]:
        test.remove(i)

print(test)

由于运行上面的代码，我希望得到['C','D']，但是我得到了['B', 'C', 'D', 'F', 'H', 'J', 'L', 'N']

如何成功遍历字符串列表并删除使用Python 3不需要的元素？

注意：我不想使用理解列表

感谢

Answer 1

从其他语言的列表中删除时，我曾经反向走过列表：

test = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O"]

for i in reversed(test):
    if i not in ["C","D"]:
        test.remove(i)

print(test)

请注意reversed会创建一个新列表，因此这可能不是大型列表的最佳解决方案。现在，既然你已经走了一份列表副本，如果你需要按照正确的顺序解析，你可以使用copy：

import copy
for i in copy.copy(test):
    if i not in ["C","D"]:
        test.remove(i)

并避免导入（来自here）：

for i in test[:]:
    if i not in ["C","D"]:
        test.remove(i)

最后，对我来说最好的解决方案是传统的，就地反向迭代而不复制列表（“借用”并修改自this答案）

for i in range(len(test) - 1, -1, -1):
    if test[i] not in ["C","D"]:
        del test[i]

Answer 2

第一个循环：i ='A'，i不在['C'，'D'] - ＆gt;删除i。现在test的第一项是'B'。所以在下一个循环i将等于'C'。这就是出问题的地方......

如何遍历字符串列表？

2 个答案: