不使用正则表达式拆分标点符号答案

【问题标题】：split punctuations without using regex不使用正则表达式拆分标点符号
【发布时间】：2021-06-19 13:33:38
【问题描述】：

给定输入：

Democr _acy , is overrat _ed .

期望的输出：

Democracy, is overrated.

这是我的代码：

sentence=input()
punctuation = "!\"#$%&'()*+,-./:;<=>?@[\]^`{|}~"
suffixes = ["acy", "ance", "ence", "dom", "er", "or", "ism", "ist",
             "ty", "ment", "ness", "ship", "sion", "tion", "ate",
            "en", "fy", "ize", "able", "ible", "al",
            "esque", "ful", "ic", "ous", "ish", "ive",
            "less", "ed", "ing", "ly", "ward", "wise"]
sentence_list = sentence.split('_')
c=""
if c not in punctuation:

 print("".join(sentence_list))
elif c in punctuation:
        for c in sentence:
         print("".join(sentence_list).split(c))

如您所见，我的输出有 29 个不同的列表，但我只想要其中一个。我想从单词中删除 '' 并加入我从中删除 '' 的标点符号和单词。当我写这样的代码时：

sentence_list = sentence.split('_')
print("".join(sentence_list))

'_'和标点符号消失。我在哪里做错了？

【问题讨论】：

标签： python-3.x join split punctuation

【解决方案1】：

这就是我要解决这个问题的方法。

def combineSentence(si):
    punctuation = "!\"#$%&'()*+,-./:;<=>?@[\]^`{|}~"
    rslt = ''
    ptr = 0
    while ptr < len(si):
        if si[ptr] not in punctuation and si[ptr] != '_':
            rslt += si[ptr]
        else:
            rslt = rslt[:-1]
            if si[ptr] in punctuation:
                rslt += si[ptr]
        ptr += 1   
    return rslt

执行combineSentence('Democr _acy , is overrat _ed .') 将产生：

'Democracy, is overrated.'

【讨论】：