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社区首页 >问答首页 >按关键字和分组结果对行进行排序

问按关键字和分组结果对行进行排序
EN

Stack Overflow用户
提问于 2021-04-25 03:55:39
回答 2查看 54关注 0票数 2

我想像在期望的输出中那样对链接进行排序，但我不知道如何更改我的代码

那么，如何更改代码以获得所需的输出呢？

提前感谢您的帮助

代码：

代码语言：javascript
复制
def my_sort(line):
    social_folders = {'engine': 1,
                    'wormix_mm': 2,
                    'wormix_ok': 3}
    line_fields = line.strip().split("/")
    social = line_fields[3]
    print(line_fields[3])
    return social_folders[social]

numbers = 'First', 'Second', 'Third', 'Fourth'
with open('./testsort.txt') as testsortf, \
     open('./test_out999.txt', "w") as test_out:
    contents = testsortf.readlines()
    contents[-1] = f'{contents[-1]}\n'
    contents.sort(key=my_sort)
    for i, line in enumerate(contents):
        test_out.write(f'{numbers[i]}:\n{line}')
        if i+1 < len(contents):
            test_out.write('\n')

我来自.txt文件的输入：

代码语言：javascript
复制
https://markus.rmart.ru/engine/preloader/somefold
https://markus.rmart.ru/wormix_ok/preloader/somefold3
https://markus.rmart.ru/engine/preloader/somefold3
https://markus.rmart.ru/engine/preloader/somefold1
https://markus.rmart.ru/wormix_ok/preloader/somefold4
https://markus.rmart.ru/wormix_mm/preloader/somefold2
https://markus.rmart.ru/wormix_mm/preloader/somefold1
https://markus.rmart.ru/engine/preloader/somefold2
https://markus.rmart.ru/engine/preloader/somefold5
https://markus.rmart.ru/wormix_mm/preloader/somefold5
https://markus.rmart.ru/wormix_ok/preloader/somefold1

因此，没有任何排序的输入

所需输出：

代码语言：javascript
复制
First:
https://markus.rmart.ru/engine/preloader/somefold
https://markus.rmart.ru/engine/preloader/somefold3
https://markus.rmart.ru/engine/preloader/somefold1
https://markus.rmart.ru/engine/preloader/somefold2
https://markus.rmart.ru/engine/preloader/somefold5

Second:
https://markus.rmart.ru/wormix_mm/preloader/somefold2
https://markus.rmart.ru/wormix_mm/preloader/somefold1
https://markus.rmart.ru/wormix_mm/preloader/somefold5

Third:
https://markus.rmart.ru/wormix_ok/preloader/somefold1
https://markus.rmart.ru/wormix_ok/preloader/somefold4
https://markus.rmart.ru/wormix_ok/preloader/somefold3

现在输出：

代码语言：javascript
复制
First:
https://markus.rmart.ru/engine/preloader/somefold

Second:
https://markus.rmart.ru/engine/preloader/somefold1

Third:
https://markus.rmart.ru/engine/preloader/somefold3

Fourth:
https://markus.rmart.ru/engine/preloader/somefoldtest
python
python-3.x
sorting
EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2021-04-25 05:48:36

尝试这段代码(我在代码中注释了我所做的事情)。

代码语言：javascript
复制
def my_sort(conts):
    social_folders = {'engine': 1, 'wormix_mm': 2, 'wormix_ok': 3}
    line_fields = conts.strip().split("/")
    social = line_fields[3]
    return social_folders[social]


# I didn't know what is the differences between First and second section.
# So I put them together. You can handle that yourself.
numbers = 'First', 'Second', 'Third'#, 'Fourth'
folds = ['engine', 'wormix_mm', 'wormix_ok']
with open('./testsort.txt') as testsortf, open('./test_out999.txt', "w") as test_out:
    contents = testsortf.readlines()
    contents[-1] = f'{contents[-1]}\n'
    contents.sort(key=my_sort)
    # It needs 2 for loops
    for k, fold in enumerate(numbers):
        # Put enter before every category, except the first one
        if k != 0:
            test_out.write(f'\n')
        # Put the label of each category
        test_out.write(f'{numbers[k]}:\n')
        for i, line in enumerate(contents):
            # Put the right label in each category
            if line.strip().split("/")[3] == folds[k]:
                test_out.write(f'{line}')
票数 0
EN

Stack Overflow用户

发布于 2021-04-25 06:17:27

标准库中的itertools.groupby()将按照您想要的方式对链接列表进行集群，但它需要一些设置。具体地说，除了链接的排序迭代之外，它还需要知道分组依据的是链接字符串的哪一部分。为此，需要一些类似于正则表达式的东西来隔离链接的关键部分。

示例：

代码语言：javascript
复制
import re
from itertools import groupby

sorted_links = sorted([
    "https://markus.rmart.ru/engine/preloader/somefold",
    "https://markus.rmart.ru/wormix_ok/preloader/somefold3",
    "https://markus.rmart.ru/engine/preloader/somefold3",
    "https://markus.rmart.ru/engine/preloader/somefold1",
    "https://markus.rmart.ru/wormix_ok/preloader/somefold4",
    "https://markus.rmart.ru/wormix_mm/preloader/somefold2",
    "https://markus.rmart.ru/wormix_mm/preloader/somefold1",
    "https://markus.rmart.ru/engine/preloader/somefold2",
    "https://markus.rmart.ru/engine/preloader/somefold5",
    "https://markus.rmart.ru/wormix_mm/preloader/somefold5",
    "https://markus.rmart.ru/wormix_ok/preloader/somefold1",
])

# Finds the category part of the path that follows the domain name (e.g., "engine")
category = re.compile(r"https.*\.[a-z]{2,3}\/([^\/]*)")

for _, group in groupby(sorted_links, lambda url: category.search(url).group(1)):
    for url in group:
        print(url)
    print()

输出：

代码语言：javascript
复制
https://markus.rmart.ru/engine/preloader/somefold
https://markus.rmart.ru/engine/preloader/somefold1
https://markus.rmart.ru/engine/preloader/somefold2
https://markus.rmart.ru/engine/preloader/somefold3
https://markus.rmart.ru/engine/preloader/somefold5

https://markus.rmart.ru/wormix_mm/preloader/somefold1
https://markus.rmart.ru/wormix_mm/preloader/somefold2
https://markus.rmart.ru/wormix_mm/preloader/somefold5

https://markus.rmart.ru/wormix_ok/preloader/somefold1
https://markus.rmart.ru/wormix_ok/preloader/somefold3
https://markus.rmart.ru/wormix_ok/preloader/somefold4
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：

https://stackoverflow.com/questions/67246902

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