Python相当于Haskell的[1 ..]（用于索引列表）

问题描述

我有一个Python中的元素列表。我不知道列表中元素的数量。我希望将索引添加到列表中。

在Haskell中，我可以执行以下操作：

pre >

只要执行 list（enumerate（s））即可。它遍历枚举对象并将其转换为 list 。

I have a list of elements in python. I don't know the number of elements in the list. I would like to add indexes to the list.

In Haskell, I could do the following

zip [1..] "abcdefghijklmnop"
[(1,'a'),(2,'b'),(3,'c'),(4,'d'),(5,'e'),(6,'f'),(7,'g'),(8,'h'),(9,'i'),(10,'j'),(11,'k'),(12,'l'),(13,'m'),(14,'n'),(15,'o'),(16,'p')]

Now imagine that the string was of unknown size. This would still work in Haskell, and the integer list gives as many integers as necessary until the string runs out.

How would one do the equivalent in Python?

I have tried this:

s = "abcdefghijklmnop"
indexedlist = []
for i,c in enumerate(s):
    indexedlist.append((i,c))

>>> indexedlist
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g'), (7, 'h'), (8, 'i'), (9, 'j'), (10, 'k'), (11, 'l'), (12, 'm'), (13, 'n'), (14, 'o'), (15, 'p')]

And it works, but I'm wondering if there is a shorter/cleaner way, since it is 4 lines of code and feels much.

解决方案

Just do list(enumerate(s)). This iterates over the enumerate object and converts it to a list.

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