设置为 dict Python
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Set to dict Python
提问by Sven Bamberger
is there any pythonic way to convert a set into a dict?
有没有任何pythonic方法可以将集合转换为字典？
I got the following set
我得到了以下套装
s = {1,2,4,5,6}
and want the following dict
并想要以下字典
c = {1:0, 2:0, 3:0, 4:0, 5:0, 6:0}
with a list you would do
有一个你会做的清单
a = [1,2,3,4,5,6]
b = []
while len(b) < len(a):
b.append(0)
c = dict(itertools.izip(a,b))
采纳答案by Martijn Pieters
Use dict.fromkeys():
c = dict.fromkeys(s, 0)
Demo:
演示：
>>> s = {1,2,4,5,6}
>>> dict.fromkeys(s, 0)
{1: 0, 2: 0, 4: 0, 5: 0, 6: 0}
This works for lists as well; it is the most efficient method to create a dictionary from a sequence. Note all values are references to that one default you passed into dict.fromkeys(), so be careful when that default value is a mutable object.
这也适用于列表；这是从序列创建字典的最有效方法。请注意，所有值都是对您传入的默认值的引用dict.fromkeys()，因此当该默认值是可变对象时要小心。
回答by papirrin
Besides the method given by @Martijn Pieters, you can also use a dictionary comprehension like this:
除了@Martijn Pieters给出的方法，您还可以使用这样的字典理解：
s = {1,2,4,5,6}
d = {e:0 for e in s}
This method is slower than dict.fromkeys(), but it allows you to set the values in the dict to whatever you need, in case you don't always want it to be zero.
此方法比 dict.fromkeys() 慢，但它允许您将 dict 中的值设置为您需要的任何值，以防您不总是希望它为零。
You can also use it to create lists, lists comprehensions are faster and more pythonic that the loop that you have in your question. You can learn more about comprehensions here: http://docs.python.org/2/tutorial/datastructures.html#list-comprehensions
您还可以使用它来创建列表，与您在问题中的循环相比，列表推导式更快、更具有 Python 风格。您可以在此处了解有关理解的更多信息：http: //docs.python.org/2/tutorial/datastructures.html#list-comprehensions
回答by asit_dhal
This is also another way to do
这也是另一种方式
s = {1,2,3,4,5}
dict([ (elem, 0) for elem in s ])
