手动删除字符串中的重复空格

Question

我有个关于某个代码的问题。我在python中做了一个关于字符串的练习。我想出了正确的逻辑，但由于某种原因，for循环中的输出没有正确返回。相反，返回全局值。我不太熟悉Python，但是有什么方法可以解决这个问题吗？

def song_decoder(song):
    global Ret

    Ret = ""

    Ret = song.replace("WUB", " ")
    Ret = Ret.strip()
    Ret += "1"

    space = False
    for i in range(0, len(Ret)):
        if Ret[i] == "1":
            Ret = Ret[:i]
            break
        elif Ret[i] == " ":
            if space is False:
                space = True
            else:
                if i+1 == len(Ret):
                    Ret = Ret[:i]
                else:
                    Ret = Ret[:i] + Ret[(i+1):]
        else:
            space = False

    return Ret

测试代码：

def test_song_decoder(self):
    self.assertEquals(song_decoder("AWUBBWUBC"), "A B C","WUB should be replaced by 1 space")
    self.assertEquals(song_decoder("AWUBWUBWUBBWUBWUBWUBC"), "A B C","multiples WUB should be replaced by only 1 space")
    self.assertEquals(song_decoder("WUBAWUBBWUBCWUB"), "A B C","heading or trailing spaces should be removed")

第二个测试失败，然后返回'A B C'。

Answer 1

首先，您没有必要在这里使Ret成为全局的。所以最好去掉那条线。

第二，缺少一个测试，它将给您另一个提示：

>>> song_decoder("AWUBBWUBC")
'A B C'
>>> song_decoder("AWUBWUBBWUBWUBC")
'A B C'
>>> song_decoder("AWUBWUBWUBBWUBWUBWUBC")
'A  B  C'

如您所见，两个WUB被正确地替换为一个空格。当有三个时，问题就出现了。这应该会给您一个提示，即在进行替换之后，空间检测不能正常工作。这样做的原因其实相当简单：

# you iterate over the *initial* length of Ret
for i in range(0, len(Ret)):
    # ...
    elif Ret[i] == " ":
        if space is False:
            space = True
        else:
            # when you hit a space and you have seen a space directly
            # before then you check the next index…
            if i+1 == len(Ret):
                Ret = Ret[:i]
            else:
                # … and remove the next index from the string
                Ret = Ret[:i] + Ret[(i+1):]

    # now at the end of the loop, `i` is incremented to `i + 1`
    # although you have already removed the character at index `i`
    # making the next character you would have to check belong to
    # index `i` too

因此，结果是跳过第二个空格后直接出现的字符(移除该字符)。所以不可能这样检测到三个空格，因为你总是跳过第三个空格。

一般来说，在进行修改时迭代一些内容是一个非常糟糕的主意。在您的例子中，您正在迭代字符串的长度，但是字符串实际上一直在变短。所以你真的应该避免那样做。

与其迭代Ret字符串，不如迭代原始字符串，保持不变：

def song_decoder(song):
    # replace the WUBs and strip spaces
    song = song.replace("WUB", " ").strip() 
    ret = ''

    space = False
    # instead of iterating over the length, just iterate over
    # the characters of the string
    for c in song:
        # since we iterate over the string, we don’t need to check
        # where it ends

        # check for spaces
        if c == " ":
            # space is a boolean, so don’t compare it against booleans
            if not space:
                space = True
            else:
                # if we saw a space before and this character is a space
                # we can just skip it
                continue
        else:
            space = False

        # if we got here, we didn’t skip a later space, so we should
        # include the current character
        ret += c

    return ret

Answer 2

你在试图将多个空格折叠成一个空间时遇到了太多的麻烦：

def song_decoder(song, delimiter="WUB"):
    splits = song.split(delimiter)  # instead of replacing, just split on your delimiter
    cleaned = filter(None, splits)  # remove empty elements caused by consecutive WUBs
    return ' '.join(cleaned)        # join them up with a single space in between