Question

我有一个包含单词和数字的文字。我将给出一个代表性的文本示例：

string = "This is a 1example of the text. But, it only is 2.5 percent of all data"

我想把它转换成类似的东西：

"This is a  1 example of the text But it only is  2.5  percent of all data"

删除标点符号（可以是. ,或string.punctuation中的任何其他内容），并在连接时在数字和单词之间放置一个空格。但在我的例子中保持浮点数像2.5。

我使用了以下代码：

item = "This is a 1example of the text. But, it only is 2.5 percent of all data"
item = ' '.join(re.sub( r"([A-Z])", r" \1", item).split())
# This a start but not there yet !
#item = ' '.join([x.strip(string.punctuation) for x in item.split() if x not in string.digits])
item = ' '.join(re.split(r'(\d+)', item) )
print item

结果是：

 >> "This is a  1 example of the text. But, it only is  2 . 5  percent of all data"

我几乎在那里，但无法弄清楚最后的平安。

Answer 1

你可以使用这样的正则表达式：

(?<!\d)[.,;:](?!\d)

<强> Working demo

这个想法是让一个字符类收集你想要替换的标点符号，并使用lookarounds来匹配没有数字的标点符号

regex = r"(?<!\d)[.,;:](?!\d)"

test_str = "This is a 1example of the text. But, it only is 2.5 percent of all data"

result = re.sub(regex, "", test_str, 0)

结果是：

This is a 1example of the text But it only is 2.5 percent of all data

Answer 2

好的伙计们，这是一个答案（最好的？我不知道，但似乎有效）：

item = "This is a 1example 2Ex of the text.But, it only is 2.5 percent of all data?"
#if there is two strings contatenated with the second starting with capital letter
item = ' '.join(re.sub( r"([A-Z])", r" \1", item).split())
#if a word starts with a digit like "1example"
item = ' '.join(re.split(r'(\d+)([A-Za-z]+)', item) )
#Magical line that removes punctuation apart from floats
item = re.sub('\S+', lambda m: re.match(r'^\W*(.*\w)\W*$', m.group()).group(1), item)
item = item.replace("  "," ")
print item

Answer 3

我与Python脱节，但对regexp有一些了解。我建议使用或？我会使用这个正则表达式："(\d+)([a-zA-Z])|([a-zA-Z])(\d+)"，然后作为替换字符串使用： "\1 \2"
如果某些极端情况困扰你，你可以将反向引用传递给一个过程，然后处理1-by-1，可能是通过检查你的＆＃34; \ 1 \ 2＆＃34;可以翻译为浮动。 TCL有这样的内置功能，Python也应该。

Answer 4

我试过这个并且效果很好。

a = "This is a 1example of the text. But, it only is 2.5 percent of all data" a.replace(". ", " ").replace(", "," ")

请注意，在替换函数中，标点符号后面有空格。我只用空格替换了标点符号和空格。

Answer 5

<强>代码：

from itertools import groupby

s1 = "This is a 1example of the text. But, it only is 2.5 percent of all data"
s2 = [''.join(g) for _, g in groupby(s1, str.isalpha)]
s3 = ' '.join(s2).replace("   ", "  ").replace("  ", " ")

#you can keep adding a replace for each ponctuation
s4 = s3.replace(". ", " ").replace(", "," ").replace("; "," ").replace(", "," ").replace("- "," ").replace("? "," ").replace("! "," ").replace(" ("," ").replace(") "," ").replace('" '," ").replace(' "'," ").replace('... '," ").replace('/ '," ").replace(' “'," ").replace('” '," ").replace('] '," ").replace(' ['," ")

s5 = s4.replace("  ", " ")
print(s5)

<强>输出：

'This is a 1 example of the text But it only is 2.5 percent of all data'

P.s。：您可以查看Punctuation Marks并继续将其添加到.replace()函数中。

Answer 6

这是一种正则表达式方法

([^ ]?)(?:[^\P{punct}.]|(?<!\d)\.(?!\d))([^ ]?)

替换回调：

如果$ 1长度＆gt; 0和$ 2长度＆gt; 0
替换为$ 1 +空格+ $ 2
其他替换为$ 1 $ 2

扩展

 ( [^ ]? )                     # (1)
 (?:
      [^\P{punct}.] 
   |  
      (?<! \d )
      \.
      (?! \d )
 )
 ( [^ ]? )                     # (2)

如果您不想使用旁路旁边的字符逻辑使用(?:[^\P{punct}.]|(?<!\d)\.(?!\d))并替换为空。

从字符串中删除所有标点符号，除非它在数字之间

6 个答案: