从字符串中删除所有标点符号，除非它位于数字之间

Question

我有一篇包含单词和数字的文本。我将给出一个有代表性的文本示例：

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

我想将其转换为：

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

因此删除 标点符号（可以是

.

,

或

string.punctuation

中的任何其他标点符号），并在连接时在数字和单词之间放置一个空格。但在我的示例中保持浮点数为 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"

我快到了，但无法弄清楚最后的平静。

Answer 1

您可以像这样使用正则表达式查找：

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

工作演示

这个想法是让一个字符类收集您想要替换的标点符号，并使用环视来匹配周围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

好吧，伙计们，这是一个答案（最好的？我不知道，但它似乎有效）：

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

我对Python不太了解，但对正则表达式有一些了解。 我建议使用 or? 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您，您可以将反向引用传递给过程，然后进行1×1处理，可能通过检查你的“”是否可以转换为浮动。 TCL 有这样的内置功能，Python 也应该如此。

Answer 4

我尝试过这个，效果非常好。

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。我只是用空格替换了标点符号和空格。

Answer 5

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)

输出：

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

P.s.：您可以查看标点符号并继续在

.replace()

函数中添加它们。

Answer 6

这是一种正则表达式方法

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

在回调中替换：

如果 $1 长度 > 0 并且 $2 长度 > 0
替换为 $1 + 空格 + $2
别的 替换为 $1$2

扩大

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

如果您不想对与 punct 相邻的字符使用逻辑
使用

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

并替换为任何内容。

Answer 7

要完成接受的答案，您可以通过在正则表达式中使用 OR 来强制仅采用数字之间的标点符号：

import re
test_str = "This is a .12 23. 1example of the text. But, it only is 2.5 percent of all data"
punctuation = ".,()/"

regex_expr = '(?<!\d)?[{}](?!\d)|(?<!\d)[{}](?!\d)?'.format(punctuation, punctuation)
result = re.sub(regex, "", test_str, 0)

print(result)

输出：

'This is a 12 23 1example of the text But it only is 2.5 percent of all data'