Question

我正在编写以下代码并面临一个令人沮丧的问题，而且在被困两天之后我无法解决它。

这是简化的代码：

def crawl_web(url, depth):
    toCrawl = [url]
    crawled = ['https://index.html']
    i = 0
    while i <= depth:
        interim = []
        for x in toCrawl:
            if x not in toCrawl and x not in crawled and x not in interim:
                print("NOT IN")
            crawled.append(x)
        toCrawl = interim
        i += 1
    return crawled

print(crawl_web("https://index.html", 1))

我期望的结果应该只是：

['https://index.html']

但不知何故，“if not in”不起作用并继续将此作为输出：

['https://index.html','https://index.html']

Answer 1

无论if语句是什么，都会调用crawled.append，因为它与if语句在同一缩进级别上。你需要把它移到里面。

def crawl_web(url, depth):
    toCrawl = [url]
    crawled = ['https://index.html']
    i = 0
    while i <= depth:
        interim = []
        for x in toCrawl:
            if x not in toCrawl and x not in crawled and x not in interim:
                print("NOT IN")
                crawled.append(x)
        toCrawl = interim
        i += 1
    return crawled

print(crawl_web("https://index.html", 1))

Python的问题“如果不在”

1 个答案: