jQuery自动完成标记插件，无需重复检查(jQuery autocomplete tagging plug-in without duplicate check)

 我不能在一行中作为正则表达式，但这是我尝试使用itertools和任何： 
 
import pandas as pd
from itertools import product

url = pd.DataFrame({'urls' : ['www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA', 'www.ulta.com/beautyservices/benefitbrowbar/']})
string_list = ['Benefit Cosmetics', 'Anastasia Beverly Hills']

"""
For each of Cartesian product (the different combinations) of 
string_list and urls.
"""
for x in list(product(string_list, url['urls'])):
    """
    If any of the words in the string (x[0]) are present in 
    the URL (x[1]) disregarding case.
    """
    if any (word.lower() in x[1].lower() for word in x[0].split()):
        """
        Show the match.
        """
        print ("Match String: %s URL: %s" % (x[0], x[1])) 
 
 输出： 
 
Match String: Benefit Cosmetics URL: www.ulta.com/beautyservices/benefitbrowbar/
Match String: Anastasia Beverly Hills URL: www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA
 
 更新： 
 
 您可以选择使用以下方式： 
 
import pandas as pd
import warnings
pd.set_option('display.width', 100)
"""
Supress the warning it will give on a match.
"""
warnings.filterwarnings("ignore", 'This pattern has match groups')
string_list = ['Benefit Cosmetics', 'Anastasia Beverly Hills']
"""
Create a pandas DataFrame.
"""
url = pd.DataFrame({'urls' : ['www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA', 'www.ulta.com/beautyservices/benefitbrowbar/']})
"""
Using one string at a time.
"""
for string in string_list:
    """
    Get the individual words in the string and concatenate them 
    using a pipe to create a regex pattern. 
    """
    s = "|".join(string.split())
    """
    Update the DataFrame with True or False where the regex 
    matches the URL.
    """
    url[string] = url['urls'].str.contains('('+s+')', case = False)
"""
Show the result
"""
print (url)
 
 哪个会输出： 
 
                                                urls Benefit Cosmetics Anastasia Beverly Hills
0  www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00...             False                    True
1        www.ulta.com/beautyservices/benefitbrowbar/              True                   False
 
 我想，如果你想在DataFrame中使用它，可能会更好，但我更喜欢第一种方式。 

I can't do it as a regex in one line but here is my attempt using itertools and any:
 
import pandas as pd
from itertools import product

url = pd.DataFrame({'urls' : ['www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA', 'www.ulta.com/beautyservices/benefitbrowbar/']})
string_list = ['Benefit Cosmetics', 'Anastasia Beverly Hills']

"""
For each of Cartesian product (the different combinations) of 
string_list and urls.
"""
for x in list(product(string_list, url['urls'])):
    """
    If any of the words in the string (x[0]) are present in 
    the URL (x[1]) disregarding case.
    """
    if any (word.lower() in x[1].lower() for word in x[0].split()):
        """
        Show the match.
        """
        print ("Match String: %s URL: %s" % (x[0], x[1])) 
 
Outputs:
 
Match String: Benefit Cosmetics URL: www.ulta.com/beautyservices/benefitbrowbar/
Match String: Anastasia Beverly Hills URL: www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA
 
Updated:
 
The way you were looking at it you could alternatively use:
 
import pandas as pd
import warnings
pd.set_option('display.width', 100)
"""
Supress the warning it will give on a match.
"""
warnings.filterwarnings("ignore", 'This pattern has match groups')
string_list = ['Benefit Cosmetics', 'Anastasia Beverly Hills']
"""
Create a pandas DataFrame.
"""
url = pd.DataFrame({'urls' : ['www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00GI21NZA', 'www.ulta.com/beautyservices/benefitbrowbar/']})
"""
Using one string at a time.
"""
for string in string_list:
    """
    Get the individual words in the string and concatenate them 
    using a pipe to create a regex pattern. 
    """
    s = "|".join(string.split())
    """
    Update the DataFrame with True or False where the regex 
    matches the URL.
    """
    url[string] = url['urls'].str.contains('('+s+')', case = False)
"""
Show the result
"""
print (url)
 
which would output:
 
                                                urls Benefit Cosmetics Anastasia Beverly Hills
0  www.amazon.com/ANASTASIA-Beverly...Brow/dp/B00...             False                    True
1        www.ulta.com/beautyservices/benefitbrowbar/              True                   False
 
Which I guess, if you want it in a DataFrame, may be better but I prefer the first way.