如何将列表转换为键和值都相同的字典？

Question

This is based on this question: python convert list to dictionary

The asker provides the following:

 l = ["a", "b", "c", "d", "e"]

I want to convert this list to a dictionary like:

 d = {"a": "b", "c": "d", "e": ""}

While the grouper recipe is quite interesting and I have been "playing" around with it, however, what I would like to do is, unlike the output dictionary wanted by that asker, I would like to convert a list with strings like the one above into a dictionary where all values and keys are the same. Example:

d = {"a": "a", "c": "c", "d": "d", "e": "e"}

How would I do this?

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Edit :

Implementation code:

def ylimChoice():

    #returns all permutations of letter capitalisation in a certain word.
    def permLet(s):
        return(''.join(t) for t in product(*zip(s.lower(), s.upper())))

    inputList = []
    yesNo = input('Would you like to set custom ylim() arguments? ')
    inputList.append(yesNo)
    yes = list(permLet("yes"))
    no = list(permLet("no"))

    if any(yesNo in str({y: y for y in yes}.values()) for yesNo in inputList[0]):
        yLimits()
    elif any(yesNo in str({n: n for n in no}.values()) for yesNo in inputList[0]):
        labelLocation = number.arange(len(count))
        plot.bar(labelLocation, list(count.values()), align='center', width=0.5)
        plot.xticks(labelLocation, list(count.keys()))
        plot.xlabel('Characters')
        plot.ylabel('Frequency')
        plot.autoscale(enable=True, axis='both', tight=False)
        plot.show()

Answer 1

You can just zip the same list together:

dict(zip(l, l))

or use a dict comprehension:

{i: i for i in l}

The latter is faster:

>>> from timeit import timeit
>>> timeit('dict(zip(l, l))', 'l = ["a", "b", "c", "d", "e"]')
0.850489666001522
>>> timeit('{i: i for i in l}', 'l = ["a", "b", "c", "d", "e"]')
0.38318819299456663

This holds even for large sequences:

>>> timeit('dict(zip(l, l))', 'l = range(1000000)', number=10)
1.28369528199255
>>> timeit('{i: i for i in l}', 'l = range(1000000)', number=10)
0.9533485669962829

Answer 2

You can use a dict-comprehension like this:

l = ["a", "b", "c", "d", "e"]
d = {s: s for s in l}