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python - 演变一个 for 循环以跳过不等于的 dict 键和值

问题描述

背景
 我的结构如下：

trash = [ {'href': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'},
          {'href': 'https://www.simplyrecipes.com/recipes/cuisine/german/'},
          {'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/'}]
          {'href': 'https://www.simplyrecipes.com/', 
          'title': 'Simply Recipes Food and Cooking Blog', 'rel': ['home']},]

如您所见，大多数键是'href'，大多数值包含'https://www.simplyrecipes.com/recipes/'. 问题在于那些不符合此命名约定的键和值...

代码：
此代码遍历结构并使用re.findall之间的字符串值'recipes/'并继续/为其对应的值创建一个新的键名。

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：
如果我删除了不符合命名约定的keys和包含代码的字符串值的命名约定，则可以正常工作，如下所示：values'href''https://www.simplyrecipes.com/recipes/'

{'cuisine': 'https://www.simplyrecipes.com/recipes/cuisine/portuguese/'}
{'cuisine': 'https://www.simplyrecipes.com/recipes/cuisine/german/'}
{'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/'}

问题：代码的问题是，如果结构的键和值不符合代码中的命名约定
，我会得到一个。问题：我将如何改进此代码，以便它跳过任何未命名的键，如果它们被命名，如果它们的值不包含则将跳过？TypeError: expected string or bytes-like object


'href''href''https://www.simplyrecipes.com/recipes/'

标签： pythonlistdictionaryfor-loopfindall

解决方案

for d in trash:
    for key, value in d.items():
        if key != "href" or "https://www.simplyrecipes.com/recipes/" not in value:
            continue  # Move onto next iteration
        txt = ''
        for i in re.findall("recipes/.*", value):
            txt += i
            title = txt.split('/')[1]
            print({title: value})

您可以使用 遍历字典的键和值dict.items()。然后您可以使用if语句来检查您的条件以及continue是否不满足。

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