字符串拆分问题

cod*_*nob 6 python string split

问题:通过作为列表传入的分隔符将字符串拆分为单词列表.

串: "After the flood ... all the colors came out."

期望的输出: ['After', 'the', 'flood', 'all', 'the', 'colors', 'came', 'out']

我写了以下函数 - 注意我知道有更好的方法来使用函数内置的一些pythons来分割字符串但是为了学习我想我会这样做:

def split_string(source,splitlist):
    result = []
    for e in source:
           if e in splitlist:
                end = source.find(e)
                result.append(source[0:end])
                tmp = source[end+1:]
                for f in tmp:
                    if f not in splitlist:
                        start = tmp.find(f)
                        break
                source = tmp[start:]
    return result

out = split_string("After  the flood   ...  all the colors came out.", " .")

print out

['After', 'the', 'flood', 'all', 'the', 'colors', 'came out', '', '', '', '', '', '', '', '', '']
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我无法弄清楚为什么"出来"不会分成"来"和"出"两个单独的词.就好像两个单词之间的空白字符被忽略一样.我认为输出的其余部分是垃圾,源于与"出来"问题相关的问题.

编辑:

我按照@ Ivc的建议,提出了以下代码:

def split_string(source,splitlist):
    result = []
    lasti = -1
    for i, e in enumerate(source):
        if e in splitlist:
            tmp = source[lasti+1:i]
            if tmp not in splitlist:
                result.append(tmp)
            lasti = i
        if e not in splitlist and i == len(source) - 1:
            tmp = source[lasti+1:i+1]
            result.append(tmp)
    return result

out = split_string("This is a test-of the,string separation-code!"," ,!-")
print out
#>>> ['This', 'is', 'a', 'test', 'of', 'the', 'string', 'separation', 'code']

out = split_string("After  the flood   ...  all the colors came out.", " .")
print out
#>>> ['After', 'the', 'flood', 'all', 'the', 'colors', 'came', 'out']

out = split_string("First Name,Last Name,Street Address,City,State,Zip Code",",")
print out
#>>>['First Name', 'Last Name', 'Street Address', 'City', 'State', 'Zip Code']

out = split_string(" After  the flood   ...  all the colors came out...............", " ."
print out
#>>>['After', 'the', 'flood', 'all', 'the', 'colors', 'came', 'out']
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lvc*_*lvc 2

你似乎在期待:

source = tmp[start:]
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修改source外部 for 循环正在迭代的 。它不会 - 该循环将继续遍历您给它的字符串,而不是现在使用该名称的任何对象。这可能意味着你所扮演的角色可能不在剩下的内容中source

不要尝试这样做,而是以这种方式跟踪字符串中的当前索引:

for i, e in enumerate(source):
   ...
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并且您要附加的内容将始终是source[lasti+1:i],并且您只需要跟踪即可lasti