如何对 python 中的 URL 进行分层排序？

Question

给定从网站抓取的 URL 的初始列表：

https://somesite.com/
https://somesite.com/advertise
https://somesite.com/articles
https://somesite.com/articles/read
https://somesite.com/articles/read/1154
https://somesite.com/articles/read/1155
https://somesite.com/articles/read/1156
https://somesite.com/articles/read/1157
https://somesite.com/articles/read/1158
https://somesite.com/blogs

我正在尝试将列表转换为按选项卡组织的树状层次结构：

https://somesite.com
    /advertise
    /articles
        /read
            /1154
            /1155
            /1156
            /1157
            /1158
    /blogs

我试过使用列表、元组和字典。到目前为止，我已经想出了两种有缺陷的输出内容的方法。

如果元素在层次结构中具有相同的名称和位置，则方法 1 将丢失这些元素：

Input:
https://somesite.com
https://somesite.com/missions
https://somesite.com/missions/playit
https://somesite.com/missions/playit/extbasic
https://somesite.com/missions/playit/extbasic/0
https://somesite.com/missions/playit/stego
https://somesite.com/missions/playit/stego/0

Output:
https://somesite.com/
    /missions
        /playit
            /extbasic
                /0
            /stego

----------------^ Missing expected output "/0"

方法二不会遗漏任何元素，但会打印多余的内容：

Input:
https://somesite.com
https://somesite.com/missions
https://somesite.com/missions/playit
https://somesite.com/missions/playit/extbasic
https://somesite.com/missions/playit/extbasic/0
https://somesite.com/missions/playit/stego
https://somesite.com/missions/playit/stego/0

Output:
https://somesite.com/
    /missions
        /playit
            /extbasic
                /0
    /missions       <- Redundant content
        /playit     <- Redundant content
            /stego      
                /0

我不确定如何正确执行此操作，我的谷歌搜索只找到了对 urllib 的引用，但似乎不是我需要的。也许有更好的方法，但我一直找不到。

我将内容放入可用列表的代码：

#!/usr/bin/python3

import re

# Read the original list of URLs from file
with open("sitelist.raw", "r") as f:
    raw_site_list = f.readlines()

# Extract the prefix and domain from the first line
first_line = raw_site_list[0]
prefix, domain = re.match("(http[s]://)(.*)[/]" , first_line).group(1, 2)

# Remove instances of prefix and domain, and trailing newlines, drop any lines that are only a slash
clean_site_list = []
for line in raw_site_list:
    clean_line = line.strip(prefix).strip(domain).strip()
    if not clean_line == "/":
        if not clean_line[len(clean_line) - 1] == "/":
            clean_site_list += [clean_line]

# Split the resulting relative paths into their component parts and filter out empty strings
split_site_list = []
for site in clean_site_list:
    split_site_list += [list(filter(None, site.split("/")))]

这给出了一个要操作的列表，但我运行不知道如何在不丢失元素或输出冗余元素的情况下输出它。

谢谢

编辑：这是我根据下面选择的答案组合在一起的最终工作代码：

# Read list of URLs from file
with open("sitelist.raw", "r") as f:
    urls = f.readlines()

# Remove trailing newlines
for url in urls:
    urls[urls.index(url)] = url[:-1]

# Remove any trailing slashes
for url in urls:
    if url[-1:] == "/":
        urls[urls.index(url)] = url[:-1]

# Remove duplicate lines
unique_urls = []
for url in urls:
    if url not in unique_urls:
        unique_urls += [url]

# Do the actual work (modified to use unique_urls and use tabs instead of 4x spaces, and to write to file)
base = unique_urls[0]
tabdepth = 0
tlen = len(base.split('/'))

final_urls = []
for url in unique_urls[1:]:
    t = url.split('/')
    lt = len(t)
    if lt != tlen:
        tabdepth += 1 if lt > tlen else -1
        tlen = lt
    pad = ''.join(['\t' for _ in range(tabdepth)])
    final_urls += [f'{pad}/{t[-1]}']

with open("sitelist.new", "wt") as f:
    f.write(base + "\n")
    for url in final_urls:
        f.write(url + "\n")

Answer 1

这适用于您的示例数据：

urls = ['https://somesite.com',
        'https://somesite.com/missions',
        'https://somesite.com/missions/playit',
        'https://somesite.com/missions/playit/extbasic',
        'https://somesite.com/missions/playit/extbasic/0',
        'https://somesite.com/missions/playit/stego',
        'https://somesite.com/missions/playit/stego/0']


base = urls[0]
print(base)
tabdepth = 0
tlen = len(base.split('/'))

for url in urls[1:]:
    t = url.split('/')
    lt = len(t)
    if lt != tlen:
        tabdepth += 1 if lt > tlen else -1
        tlen = lt
    pad = ''.join(['    ' for _ in range(tabdepth)])
    print(f'{pad}/{t[-1]}')

Answer 2

此代码将帮助您完成任务。我同意这段代码可能有点大，可能包含一些冗余代码和检查，但这将创建一个包含 url 层次结构的字典，您可以随意使用该字典，打印或存储它。

此代码的更多内容还将解析不同的 url 并创建它们的单独树（参见代码和输出）

编辑：这也将处理冗余网址

代码：

    from json import dumps


def process_urls(urls: list):
    tree = {}

    for url in urls:
        url_components = url.split("/")
        # First three components will be the protocol
        # an empty entry
        # and the base domain 
        base_domain = url_components[:3]
        base_domain = base_domain[0] + "//" + "".join(base_domain[1:])
        # Add base domain to tree if not there.
        try:
            tree[base_domain]
        except:
            tree[base_domain] = {}

        structure = url_components[3:]
        
        for i in range(len(structure)):
            # add the first element
            if i == 0 :
                try:
                    tree[base_domain]["/"+structure[i]]
                except:
                    tree[base_domain]["/"+structure[i]] = {}
            else:
                base = tree[base_domain]["/"+structure[0]]
                for j in range(1, i):
                    base = base["/"+structure[j]]

                try:
                    base["/"+structure[i]]
                except:
                    base["/"+structure[i]] = {}

    return tree


def print_tree(tree: dict, depth=0):
    for key in tree.keys():
        print("\t"*depth+key)

        # redundant checks
        if type(tree[key]) == dict:
            
            # if dictionary is empty then do nothing
            # else call this function recuressively
            # increase depth by 1
            if tree[key]:
                print_tree(tree[key], depth+1)


if __name__ == "__main__":
        urls = [
            'https://somesite.com',
            'https://somesite.com/missions',
            'https://somesite.com/missions/playit',
            'https://somesite.com/missions/playit/extbasic',
            'https://somesite.com/missions/playit/extbasic/0',
            'https://somesite.com/missions/playit/extbasic/0',
            'https://somesite.com/missions/playit/extbasic/0',
            'https://somesite.com/missions/playit/extbasic/0',
            'https://somesite.com/missions/playit/stego',
            'https://somesite.com/missions/playit/stego/0',
            'https://somesite2.com/missions/playit',
            'https://somesite2.com/missions/playit/extbasic',
            'https://somesite2.com/missions/playit/extbasic/0',
            'https://somesite2.com/missions/playit/stego',
            'https://somesite2.com/missions/playit/stego/0'
        ]
    tree = process_urls(urls)
    print_tree(tree)

输出：

https://somesite.com
    /missions
            /playit
                    /extbasic
                            /0
                    /stego
                            /0
https://somesite2.com
    /missions
            /playit
                    /extbasic
                            /0
                    /stego
                            /0

如何对 python 中的 URL 进行分层排序？

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python-3.x