从单一列表构建字典的最Pythonic方法

Question

I have a list of day names (typically Monday-Saturday, though special cases apply) that I want to create a dictionary out of. I want to initialize the value of each day to zero.

If I had a list of zeroes the same length of the list of days, this would be a simple use case of zip() . However, a list of zeroes is a waste of space, and if that were the only solution I'd just as soon do something like:

for day in weekList:
    dayDict[day] = 0

Is there a more pythonic way?

Answer 1

Use the .fromkeys() class method :

dayDict = dict.fromkeys(weekList, 0)

It builds a dictionary using the elements from the first argument (a sequence) as the keys, and the second argument (which defaults to None ) as the value for all entries.

By it's very nature, this method will reuse the value for all keys; don't pass it a mutable value such as a list or dict and expect it to create separate copies of that mutable value for each key. In that case, use a dict comprehension instead:

dayDict = {d: [] for d in weekList}

Answer 2

Apart from dict.fromkeys you can also use dict-comprehension , but fromkeys() is faster than dict comprehensions:

In [27]: lis = ['a', 'b', 'c', 'd']

In [28]: dic = {x: 0 for x in lis}

In [29]: dic
Out[29]: {'a': 0, 'b': 0, 'c': 0, 'd': 0}

For 2.6 and earlier:

In [30]: dic = dict((x, 0) for x in lis)

In [31]: dic
Out[31]: {'a': 0, 'b': 0, 'c': 0, 'd': 0}

timeit comparisons:

In [38]: %timeit dict.fromkeys(xrange(10000), 0)         # winner
1000 loops, best of 3: 1.4 ms per loop

In [39]: %timeit {x: 0 for x in xrange(10000)}
100 loops, best of 3: 2.08 ms per loop

In [40]: %timeit dict((x, 0) for x in xrange(10000))
100 loops, best of 3: 4.63 ms per loop

As mentioned in comments by @Eumiro and @mgilson it is important to note that fromkeys() and dict-comprehensions may return different objects if the values used are mutable objects:

In [42]: dic = dict.fromkeys(lis, [])

In [43]: [id(x) for x in dic.values()]
Out[43]: [165420716, 165420716, 165420716, 165420716] # all point to a same object

In [44]: dic = {x: [] for x in lis}

In [45]: [id(x) for x in dic.values()]
Out[45]: [165420780, 165420940, 163062700, 163948812]  # unique objects