将元组列表（每个 3 个元素）转换为 dict {a: {b: set(c)}}

Question

I have a list of 3-element tuples, such as [(a1, b1, c1), (a2, b2, c2), (a2, b2, c3)] .

I want to convert the list of tuples to a dict like

{a1: {b1: set(c1)}, a2: {b2: set (c2, c3)}}

Essentially, the c needs to be put into a set .

How to write a beautiful piece of code for this?

---

Here is my current code:

s = {}
for a,b,c in abc:
    s.setdefault(a, {}).setdefault(b, set()).add(c)

Answer 1

You can use collections.defaultdict , which will automatically create default values for missing keys, to do this:

>>> from collections import defaultdict
>>> structure = defaultdict(lambda: defaultdict(set))
>>> for a, b, c in ['foo', 'bar', 'baz']:
    structure[a][b].add(c)


>>> structure
defaultdict(<function <lambda> at 0x02D16F30>, 
            {'b': defaultdict(<type 'set'>, {'a': set(['r', 'z'])}), 
             'f': defaultdict(<type 'set'>, {'o': set(['o'])})})

Note the use of lambda ; the argument to each defaultdict must be a callable returning the default value to use for missing keys. However, a slightly simpler structure would make your life easier. With {(a, b): set([c1, c2, ..]), ...} you could use defaultdict(set) or justdict.setdefault :

>>> structure = {}
>>> for a, b, c in ['foo', 'bar', 'baz']:
    structure.setdefault((a, b), set()).add(c)


>>> structure
{('b', 'a'): set(['r', 'z']), ('f', 'o'): set(['o'])}

Note that setdefault could also be used for the first approach, but:

structure.setdefault(a, dict()).setdefault(b, set()).add(c)

is rather less readable than:

structure[a][b].add(c)

Answer 2

You can do this using .setdefault . It creates new keys and assigns default value, so that we can create any level of dictionary in one statement. Code:

tuple_list = [(a1, b1, c1), (a2, b2, c2), (a2, b2, c3)]
new_dict = {}
for grp in tuple_list:
    new_dict.setdefault(grp[0], {}).setdefault(grp[1], set()).add(grp[2])