如何无序搜索关键字和无序搜索嵌入特殊字符的关键字

Question

更新 2

我编辑你的建议做 5 个案例，几乎可以做所有案例， 但它有额外的结果，多余的结果，如何改进或有其他更好的解决方案？

case 1
row1 = "search key   $ @ $  words today"
re.findall(r'[(?=k&e&y&w&o&r&d&s)]*[ \$\@]*', row1, re.DOTALL)
['se', '', 'r', '', '', ' ', 'key   $ @ $  ', 'words ', '', 'od', '', 'y', '']

case 2
row1 = "search key      words today"
re.findall(r'[(?=k&e&y&w&o&r&d&s)]*[ \$\@]*', row1, re.DOTALL)
['se', '', 'r', '', '', ' ', 'key      ', 'words ', '', 'od', '', 'y', '']

case 3 need help
row1 = "search key      wrods today"
re.findall(r'[(?=k&e&y&w&o&r&d&s)]*[ \$\@]*', row1, re.DOTALL)
['se', '', 'r', '', '', ' ', 'key      ', 'wrods ', '', 'od', '', 'y', '']

case 4 need help
row1 = "search key  $ @ $     wrods today"
re.findall(r'[(?=k&e&y&w&o&r&d&s)]*[ \$\@]*', row1, re.DOTALL)
['se', '', 'r', '', '', ' ', 'key  $ @ $     ', 'wrods ', '', 'od', '', 'y', '']

case 5 
row1 = "search key      wrds today"
re.findall(r'[(?=k&e&y&w&o&r&d&s)]*[ \$\@]*', row1, re.DOTALL)
['se', '', 'r', '', '', ' ', 'key      ', 'wrds ', '', 'od', '', 'y', '']

更新 1

case 1 is ok
row1 = "search key   $ @ $  words today"
re.sub(r' ',r'[ \$\@]*', r'key words')
re.findall(re.sub(r' ',r'[ \$\@]*', r'key words'), row1, re.DOTALL)

case 2 is ok
row1 = "search key      words today"
re.sub(r' ',r'[ \$\@]*', r'key words')
re.findall(re.sub(r' ',r'[ \$\@]*', r'key words'), row1, re.DOTALL)

case 3 need help
row1 = "search key      wrods today"
re.sub(r' ',r'[ \$\@]*', r'key words')
re.findall(r'(?=k|e|y|[ \$\@]*|\ |w|o|r|d|s)', row1, re.DOTALL)
['', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '']

case 4 need help
row1 = "search key  $ @ $     wrods today"
re.sub(r' ',r'[ \$\@]*', r'key words')
re.findall(r'(?=k|e|y|[ \$\@]*|\ |w|o|r|d|s)', row1, re.DOTALL)
['', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '']

A.

因为

row1 = "search key words today"
searchresult = re.findall(re.sub(r' ', ' *', r'key wrods', flags=re.IGNORECASE), row1, re.DOTALL)

r 和 o 互换，

然后我使用 (?=

searchresult = re.findall(re.sub(r' ', ' *', r'(?=k)(?=e)(?=y)(?=\ )(?=w)(?=r)(?=o)(?=d)(?=s)', flags=re.IGNORECASE), row1, re.DOTALL)
searchresult
[]

返回空值

如果重复字符，关键字是“kkey wods”怎么办？

B.

从嵌入任意顺序的特殊字符的内容中搜索关键字时包含任意顺序的特殊字符时出现错误

row1 = "search key $ @ $ words today"
re.sub(r' ',r'(?=$*)(?= *)*', r'key words')
re.findall(re.sub(r' ',r'(?=$*)(?=@*)(?= *)*', r'key words'), row1, re.DOTALL)

>>> re.findall(re.sub(r' ',r'(?=$*)(?=@*)(?= *)*', r'key words'), row1, re.DOTALL)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python27\lib\re.py", line 181, in findall
return _compile(pattern, flags).findall(string)
File "C:\Python27\lib\re.py", line 251, in _compile
raise error, v # invalid expression

sre_constants.error: 无需重复

只为

row1 = "search key     words today"
def easysearch(content, keywords):
    row1 = content
    searchresult = re.findall(re.sub(r' ', ' *', keywords, flags=re.IGNORECASE), row1, re.DOTALL)
    if count_chars(searchresult[0])/count_chars(keywords) > 0.5:
        return searchresult

easysearch(row1, r'key words')
easysearch(row1, r'key wrods')

row1 = "search key   $ @ $  words today"
def easysearch(content, keywords):
    row1 = content
    searchresult = re.findall(re.sub(r' ',r'(?=$*)(?=@*)(?= *)*', keywords), row1, re.DOTALL)
    if count_chars(searchresult[0])/count_chars(keywords) > 0.5:
        return searchresult

easysearch(row1, r'key words')
easysearch(row1, r'key wrods')

Answer 1

你想要这样的东西吗？

row1 = "search key words today"
searchresult = re.findall(re.sub(r' ', ' *', r'key words', flags=re.IGNORECASE), row1, re.DOTALL)
print(searchresult)

row1 = "search key $ @ $ words today"
searchresult = re.findall(re.sub(r' ',r' .*?', r'key words'), row1, re.DOTALL)
print(searchresult)

输出是

['key words']
['key $ @ $ words']

我只是在re.sub() 函数中更改了您的replacement strings 并将拼写错误wrods 修复为words

对于案例 3、4（更新）

你可以试试这个脚本

row1 = "search eky        sky key                 wrods today"
print(re.findall(r'(?:(?=k|e|y|[ \$\@]+|w|o|r|d|s).)+', row1, re.DOTALL))

row1 = "search yek key $ @   sky                    $     wrods today"
print(re.findall(r'(?:(?=k|e|y|[ \$\@]+|w|o|r|d|s).)+', row1, re.DOTALL))

输出是

['se', 'r', ' eky        sky key                 wrods ', 'od', 'y']
['se', 'r', ' yek key $ @   sky                    $     wrods ', 'od', 'y']

[附注]

你想做的可能是这样的

row1 = "search eky                       wrods today"
print(re.findall(r'(?:k|e|y)+[ \$\@]+(?:w|o|r|d|s)+', row1, re.DOTALL))

row1 = "search yek $ @                      $     wrods today"
print(re.findall(r'(?:k|e|y)+[ \$\@]+(?:w|o|r|d|s)+', row1, re.DOTALL))

输出是

['eky                       wrods']
['yek $ @                      $     wrods']