循环“忘记"删除一些物品 [英] Loop "Forgets" to Remove Some Items

问题描述

在这段代码中，我尝试创建一个函数anti_vowel，该函数将从字符串中删除所有元音(aeiouAEIOU).我认为它应该可以正常运行，但是当我运行它时，示例文本嘿，看字！"返回为"Hy lk Words！".它忘记"删除最后一个"o".怎么会这样?

in this code I am trying to create a function anti_vowel that will remove all vowels (aeiouAEIOU) from a string. I think it should work ok, but when I run it, the sample text "Hey look Words!" is returned as "Hy lk Words!". It "forgets" to remove the last 'o'. How can this be?

text = "Hey look Words!"

def anti_vowel(text):

    textlist = list(text)

    for char in textlist:
        if char.lower() in 'aeiou':
            textlist.remove(char)

    return "".join(textlist)

print anti_vowel(text)

推荐答案

您正在修改要遍历的列表，这肯定会导致某些不直观的行为.相反，请复制列表，这样就不会从迭代中删除元素.

You're modifying the list you're iterating over, which is bound to result in some unintuitive behavior. Instead, make a copy of the list so you don't remove elements from what you're iterating through.

for char in textlist[:]: #shallow copy of the list
    # etc

---

要弄清您所看到的行为，请检查一下.将print char, textlist放在(原始)循环的开头.您可能希望，这会在列表的旁边垂直打印出您的字符串，但实际上您会得到以下信息:

---

To clarify the behavior you're seeing, check this out. Put print char, textlist at the beginning of your (original) loop. You'd expect, perhaps, that this would print out your string vertically, alongside the list, but what you'll actually get is this:

H ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
e ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
  ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # !
l ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
k ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # Problem!!
  ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
W ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] 
d ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
s ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
! ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
Hy lk Words!

那是怎么回事? Python中不错的for x in y循环实际上只是语法糖:它仍然按索引访问列表元素.因此，当您在列表上进行迭代时从列表中删除元素时，您将开始跳过值(如上所示).结果，您再也看不到"look"中的第二个o了.您跳过了它，因为删除上一个元素时索引具有高级的粘贴"它.然后，当您到达"Words"中的o时，您将删除第一次出现的'o'，这是您之前跳过的那个.

So what's going on? The nice for x in y loop in Python is really just syntactic sugar: it still accesses list elements by index. So when you remove elements from the list while iterating over it, you start skipping values (as you can see above). As a result, you never see the second o in "look"; you skip over it because the index has advanced "past" it when you deleted the previous element. Then, when you get to the o in "Words", you go to remove the first occurrence of 'o', which is the one you skipped before.

正如其他人提到的那样，列表理解可能是一种更好(更清晰)的方法.利用Python字符串可迭代的事实:

As others have mentioned, list comprehensions are probably an even better (cleaner, clearer) way to do this. Make use of the fact that Python strings are iterable:

def remove_vowels(text): # function names should start with verbs! :)
    return ''.join(ch for ch in text if ch.lower() not in 'aeiou')

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