python - 比较三个文本列表以查看匹配的单词

Question

您好，我编写了一个函数来读取和比较三个句子列表之间的单词，如果任何单词匹配，则该函数将返回文本，否则False，基本上采用在来自 selenium 的网络元素列表中并检查如果文本与任何关键字列表匹配，我想要做的是修改它，如果1个或3个或更多，则在检查后返回链接，即如果只有两个单词匹配则返回False.(如果任何单词匹配且其中一个关键词匹配链接，此函数将返回链接)我想要这个函数将返回链接，如果(1,3,4,5...)的单词匹配并且其中一个关键词匹配链接(只有0,2返回False)链接和文本长度相等。

from selenium import webdriver
d = webdriver.Chrome(executable_path=r"C:\Users\test\PycharmProjects\chromedriver")
sentence = "hello world from python"
url_keywords = [".com",".edu"]
d.get("https://google.com/search?q={}".format(sentence))
y=d.find_elements_by_xpath("//a[@href]")
a=check(y,url_keywords)
li=[]
if a:
    check(y)
else:
    pass

def check(y,url_keywords):
    links = [i.get('href') for i in y]
    texts = [i.text_content() for i in y]
    for i, link in enumerate(links):
        for keyword in url_keywords:
            if keyword in link:
                for word in sentence.lower().split():
                    if word in texts[i].lower():
                        return link

    return False

如果有更简单的方法，请指教

Answer 1

from selenium import webdriver

# Use descriptive names for variables, not single letters.
driver = webdriver.Chrome(executable_path=r"C:\Users\test\PycharmProjects\chromedriver")

# Use UPPERCASE for constants
SENTENCE = "hello world from python"
URL_KEYWORDS = [".com",".edu"]

driver.get("https://google.com/search?q={}".format(sentence))
elements  = driver.find_elements_by_xpath("//a[@href]")
result = check(elements, url_keywords)


def check(elements, url_keywords):
    links = [i.get('href') for i in elements]
    texts = [i.text_content() for i in elements]

    # Use zip to avoid so much nesting! Also means you can drop the index variable "i"
    search_space = zip(links, texts)

    for link, text in search_space:
        #Let's keep track
        number_of_matches = 0  
        for keyword in url_keywords:
            # Create a separate function, again to avoid so much nesting! (see "Zen of Python")
            match = is_match(keyword, link, text)
            #If match is true int(match) will be 1, otherwise 0
            number_of_matches += int(match)
        if has_correct_number_of_matches(number_of_matches):
            return link
        else:
            return False

def normalise(string):
    """
    There is often quite a bit that we want to do to normalise strings. And you might want to extend this later. For this reason, I again make a new function, and also add in the "strip" method for good measure and as an example of extending the normalisation behaviour.
    """
    return string.lower().strip()

def is_match(keyword, link, text):
    if keyword in link:
        for word in normalise(sentence).split():
            if word in normalise(text):
                return True
        else:
            return False
     else:
        return False

def has_correct_number_of_matches(number_of_matches):
    """Now that this function is isolated, you can define it however you want!
    """
    return number_of_matches not in (0, 2)