没有得到而不是空列表

发布于 2025-02-13 11:36:08 字数 473 浏览 2 评论 0 原文

我想回来[]，但是我没有得到。例如，当我将现有的str作为关键字时，例如：“赌场”，我会得到适当的结果：[0]。

def word_search(doc_list, keyword):
    for i in doc_list:
            list = []
            words = i.split()
            for n in words:
                if keyword == n.rstrip('.,').lower():
                    list.append(doc_list.index(i))
                    return list
word_search(['The Learn Python Challenge Casino', 'They bought a car, and a horse', 'Casinoville?'], 'ear')

原文

I want to get back [], but instead i get None.
When i put existing str as a keyword, for example: 'casino', i get a proper result: [0].

def word_search(doc_list, keyword):
    for i in doc_list:
            list = []
            words = i.split()
            for n in words:
                if keyword == n.rstrip('.,').lower():
                    list.append(doc_list.index(i))
                    return list
word_search(['The Learn Python Challenge Casino', 'They bought a car, and a horse', 'Casinoville?'], 'ear')

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归属感 2025-02-20 11:36:09

您需要更加谨慎。当控件在函数末尾掉落而又不遇到返回列表时，您将获得一个隐式返回none 。

def word_search(doc_list, keyword):
    list = []
    for i in doc_list:
        words = i.split()
        for n in words:
            if keyword == n.rstrip('.,').lower():
                list.append(doc_list.index(i))
    return list

可能是一个不错的第一开始，但是我们可以用枚举来做得更好，因此您不需要 index 呼叫（而且您也不需要真正使用 list 作为变量名称，因为它会为列表的内置类型遮蔽）：

def word_search(doc_list, keyword):
    results = []
    for i, doc in enumerate(doc_list):
        words = doc.split()
        for n in words:
            if keyword == n.rstrip('.,').lower():
                results.append(i)
    return results

如果您只喜欢每个文档索引一次出现在结果中：

def word_search(doc_list, keyword):
    results = []
    for i, doc in enumerate(doc_list):
        if any(keyword == word.rstrip('.,').lower() for word in doc.split()):
            results.append(i)
    return results

You'll need to be a bit more careful with your indents. When control falls off the end of the function without encountering that return list, you get an implicit return None.

def word_search(doc_list, keyword):
    list = []
    for i in doc_list:
        words = i.split()
        for n in words:
            if keyword == n.rstrip('.,').lower():
                list.append(doc_list.index(i))
    return list

might be a good first start, but we can do better with enumerate so you don't need the index call (and you also don't really want to use list as a variable name, since it shadows the built-in type for lists):

def word_search(doc_list, keyword):
    results = []
    for i, doc in enumerate(doc_list):
        words = doc.split()
        for n in words:
            if keyword == n.rstrip('.,').lower():
                results.append(i)
    return results

We can still improve things a bit, if you'd only like each document index to appear once in the results:

def word_search(doc_list, keyword):
    results = []
    for i, doc in enumerate(doc_list):
        if any(keyword == word.rstrip('.,').lower() for word in doc.split()):
            results.append(i)
    return results

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