Python for循环保存包含某个值的键和值

Question

假设我有一个 python 列表和字典结构，如下所示：

[ {'href': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'},
  {'href': 'https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'},
  {'href': 'https://www.simplyrecipes.com/recipes/type/condiment/'},
  {'href': 'https://www.simplyrecipes.com/recipes/ingredient/adobado/'}]

我正在寻找最有效的方法

(i) 仅循环遍历 = 'href' 的键和值包含 ''https://www.simplyrecipes.com/recipes/' 的 'href' 键并识别值 ('http...')包含 'recipes/cuisine'、'recipes/season' 和 'recipes/ingredient'
(ii) 将每个完整的 url 值保存到单独的列表中（取决于它们满足哪些 'recipe/...' 条件）并命名为适当的。

预期结果：

cuisine = ['https://www.simplyrecipes.com/recipes/cuisine/portuguese/']
season = ['https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/']
type = ['https://www.simplyrecipes.com/recipes/type/condiment/']
ingredient = ['https://www.simplyrecipes.com/recipes/ingredient/adobado/']

任何不符合上述条件的键和/或值都将被跳过。

任何指针将不胜感激。

Answer 1

假设 url 在所附问题中遵循相同的格式。更好的方法是创建不同食谱的字典。

In [50]: from collections import defaultdict

In [51]: sep_data = defaultdict(list)

In [52]: lst = [ {'href': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'},
    ...:   {'href': 'https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'},
    ...:   {'href': 'https://www.simplyrecipes.com/recipes/type/condiment/'},
    ...:   {'href': 'https://www.simplyrecipes.com/recipes/ingredient/adobado/'}]

In [59]: for i in lst: sep_data[i["href"].split("/")[-3]].append(i["href"])

In [60]: sep_data
Out[60]:
defaultdict(list,
            {'cuisine': ['https://www.simplyrecipes.com/recipes/cuisine/portuguese/'],
             'season': ['https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'],
             'type': ['https://www.simplyrecipes.com/recipes/type/condiment/'],
             'ingredient': ['https://www.simplyrecipes.com/recipes/ingredient/adobado/']})

Answer 2

这是一个简单的方法，希望对您有所帮助

import re

trash = [ {'href': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'},
          {'href': 'https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'},
          {'href': 'https://www.simplyrecipes.com/recipes/type/condiment/'},
          {'href': 'https://www.simplyrecipes.com/recipes/ingredient/adobado/'}]

for x in trash:
    for y in x.values():
        txt = ''
        for i in re.findall("recipes/.*", y):
            txt += i
            title = txt.split('/')[1]
            print({title: y})

输出

{'cuisine': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'}
{'season': 'https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'}
{'type': 'https://www.simplyrecipes.com/recipes/type/condiment/'}
{'ingredient': 'https://www.simplyrecipes.com/recipes/ingredient/adobado/'}

Answer 3

所以，大致

from itertools import groupby
import re

lst = [ {'href': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'},
  {'href': 'https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/'},
  {'href': 'https://www.simplyrecipes.com/recipes/type/condiment/'},
  {'href': 'https://www.simplyrecipes.com/recipes/ingredient/adobado/'}]

def f(i):
    x = re.findall("https://www.simplyrecipes.com/recipes/([^/ ]+)/(?:[^/ ]+/?)+", i["href"]) 
    return x and x[0] or None

r = filter(lambda i: i[0] in ('cuisine', 'season', 'ingredient'), groupby(lst, f))
for i in r:
    print(f"{i[0]} = {list(map(lambda j: j['href'], i[1]))}")

# result:
# cuisine = ['https://www.simplyrecipes.com/recipes/cuisine/portuguese/']
# season = ['https://www.simplyrecipes.com/recipes/season/seasonal_favorites_spring/']
# ingredient = ['https://www.simplyrecipes.com/recipes/ingredient/adobado/']

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