使列表的元组包括字母顺序数字,字母,计数

Question

我有一个包含小写小写的列表:

l = ['a','z','y','y','z','y','b','b']

如何制作如下所示的元组列表:

[(1, 'a', 1), (2, 'b', 2), (25, 'y', 3), (26, 'z', 2)]

我试图调用Counter我的列表中的字母数,但我不知道如何做.

from collections import Counter

l = ['a','z','y','y','z','y','b','b']

c = Counter(l)
# [('a', 1), ('b', 2), ('y', 3), ('z', 2)]

Answer 1

获取字母序号的一种方法是调用ord函数,并将其与调用结果进行比较ord('a').例如:

>>> ord('s')
115
>>> ord('a')
97
>>> ord('s') - ord('a') + 1
19

另一种方法是构建映射并仅引用它:

>>> import string
>>> ordinals = {letter: ordinal 
                for ordinal, letter in enumerate(string.ascii_lowercase, 1)}
>>> ordinals['s']
19

无论哪种方式,一旦你得到了序数,将它们与Counter计数合并很容易:

>>> l = ['a','z','y','y','z','y','b','b']
>>> c = Counter(l)
>>> [(ordinals[letter], letter, count) for letter, count in c.items()]
[(25, 'y', 3), (1, 'a', 1), (26, 'z', 2), (2, 'b', 2)]

并且,如果您希望按顺序排序,则元组的默认排序顺序会按字典顺序查看元素 - 换句话说,(1, <anything>)就是之前(2, <anything>),依此类推.所以:

>>> sorted((ordinals[letter], letter, count) for letter, count in c.items())
[(1, 'a', 1), (2, 'b', 2), (25, 'y', 3), (26, 'z', 2)]