如何检查一个列表的一部分是否在另一个列表中

Question

Suppose I have two lists like:

list_of_urls = ['https://en.wikipedia.org/wiki/Barack_Obama', 'https://en.wikipedia.org/wiki/President_of_the_United_States', 'google.com']

list_of_blacklisted_urls = ['wikipedia']

How to return True if any part of the blacklisted url is in the list_of_urls? I have tried:

for url in list_of_urls:
        if any(URL in URLs for URL in list_of_blacklisted_urls):
                return True

But I'm quite sure this doesn't work.

Answer 1

You're pretty close... But the any function doesn't work the way you seem to think it does. You have to use a nested loop instead.

Here's an example:

list_of_urls = ['https://en.wikipedia.org/wiki/Barack_Obama', 'https://en.wikipedia.org/wiki/President_of_the_United_States', 'google.com']

list_of_blacklisted_urls = ['wikipedia']

for url in list_of_urls:
    for keyword in list_of_blacklisted_urls:
        if keyword in url:
            print("FOUND", keyword, "in", url)

Answer 2

data = pd.DataFrame(list_of_urls)
data  = data[data[0].str.contains(*list_of_blacklisted_urls)]

then you can see the result checking data.

Answer 3

How about this:

def in_black_urls():
    for black_url in list_of_blacklisted_urls :
        if black_url in list_of_urls:
            return True
    return False

Answer 4

只需一行，保持简单：

len([x for x in list_of_urls if any(y in x for y in list_of_blacklisted_urls)]) > 0

Answer 5

You can use nested loop and 'in':

list_of_urls = ['https://en.wikipedia.org/wiki/Barack_Obama', 'https://en.wikipedia.org/wiki/President_of_the_United_States', 'google.com']
list_of_blacklisted_urls = ['wikipedia']

def checker(urls,blacklist):
    for url in urls:
        for URL in blacklist:
            if URL in url:
                print(True, url, URL)
            else:
                return False
checker(list_of_urls,list_of_blacklisted_urls)

Answer 6

Using nested list comprehension:

def blacklisted(all_urls, blacklist):
  if len([word for url in all_urls for word in blacklist if word in url]) > 0:
    return True