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Python 向后枚举列表的大小,但保持列表向前_Python_List_Readability_Enumerate - Fatal编程技术网

Python 向后枚举列表的大小,但保持列表向前

Python 向后枚举列表的大小,但保持列表向前,python,list,readability,enumerate,Python,List,Readability,Enumerate,所以我有一些代码如下: amount_of_episodes = len(episodes_from_user_show) for x in episodes_from_user_show: print("[{0}] Episode {1}: {2}".format(amount_of_episodes, x.episode_number, x.name)) # Prints the available

所以我有一些代码如下:

amount_of_episodes = len(episodes_from_user_show)
for x in episodes_from_user_show:
    print("[{0}] Episode {1}: {2}".format(amount_of_episodes, x.episode_number,
                                          x.name))  # Prints the available list of episodes.
    amount_of_episodes -= 1
它所做的是,它需要一个列表的长度,它向后迭代,直到它达到0,并且我正在使用的api中的插曲名称和编号也正在打印


所以我想知道的是,是否有一种更具可读性的方法。类似于使用enumerate,它可以向后显示列表的大小,但可以向前显示列表的内容(如果有意义的话)

一种简单的方法是只打印
len(values)-index

shows = [{'id': 1, 'name': 'show_1'},{'id': 2, 'name': 'show_2'}]
for idx, show in enumerate(shows):
    print("[{}] Episode {}: {}".format(len(shows) - idx, show['id'], show['name']))
输出:

[2] Episode 1: show_1
[1] Episode 2: show_2

您可以在enumerate中对i,x执行
(来自用户秀的剧集)
,然后
格式化(剧集数量-i…)