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从enumerate()交换索引和值?

[英]swap index and value from enumerate()?

Python's enumerate() returns tuples of index and value: Python的enumerate()返回索引和值的元组:

enumerate('abc')
((0,'a'),(1,'b'),(2,'c'))

I'd like to get those tuples in item,index order ( ('a', 0) ) instead. 我想让这些元组以item,index顺序( ('a', 0) )代替。

How can I do that? 我怎样才能做到这一点?

I'd like to use the reversed tuples to create a dictionary like: 我想使用反向元组来创建像这样的字典:

{'a':0,'b':1,'c':2}

Use dict comprehension to reverse it: 使用dict理解来逆转它:

result = {v: i for i, v in enumerate('abc')}

Addressing @karakfa 's point - this will overwrite potentially repeated elements. 解决@karakfa的问题-这将覆盖可能重复的元素。 If your string was abca , the index value assigned for a will contain 3 , not 0 . 如果您的字符串是abca ,则为a分配的索引值将包含3而不是0

One solution is to use the itertools library: 一种解决方案是使用itertools库:

import itertools
dict(zip('abc', itertools.count()))

itertools.count() is a generator object which generates 0, 1, 2... and the zip function just ... well zip the two together. itertools.count()是一个生成器对象,它生成0、1、2 ...,而zip函数只是...将两者压缩在一起。

There's always good, old-fashioned anonymous functions to do the work for you. 总是有很好的老式匿名函数可以为您完成工作。 It's not the cleanest solution, but it gets the job done. 这不是最干净的解决方案,但可以完成工作。

dict(map(lambda x: (x[1], x[0]), enumerate('abc')))

Another way is to use zip: 另一种方法是使用zip:

>>> s = 'abc'
>>> dict(zip(s, range(len(s))))
{'a': 0, 'b': 1, 'c': 2}

Using a generator: 使用生成器:

>>> dict((v, i) for i, v in enumerate('abc'))
{'a': 0, 'b': 1, 'c': 2}

You can also try with list comprehension : 您也可以尝试使用列表理解:

new_data=((0,'a'),(1,'b'),(2,'c'))

print(tuple([(i[1],i[0])for i in new_data]))

output: 输出:

(('a', 0), ('b', 1), ('c', 2))

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