如何将元组转换为字典？

Question

I have a list of tuple(each tuple is of 3 elements) and I want to convert it into dictionary, how to do it in most efficient way? Here is the example:

[(980898, 9977, 1),
 (899979879, 23, 1),
 (1, 1, 2),
 (980898, 98789797, 1),
 (98789797, 980898, 1),
 (1, 756735, 1),
 (1, 3344, 1),
 (23, 4, 1),
 (534, 23, 1),
 (756735, 1, 1),
 (9977, 980898, 1),
 (23, 899979879, 1),
 (4, 23, 1),
 (756735, 980898, 1),
 (3344, 1, 1),
 (980898, 756735, 1),
 (23, 534, 1)]

I want the following dict:

{1: {1: 2, 3344: 1, 756735: 1},
 4: {23: 1},
 23: {4: 1, 534: 1, 899979879: 1},
 534: {23: 1},
 3344: {1: 1},
 9977: {980898: 1},
 756735: {1: 1, 980898: 1},
 980898: {9977: 1, 756735: 1, 98789797: 1},
 98789797: {980898: 1},
 899979879: {23: 1}}

Here first element of the tuple is the key and next element becomes the key in the dict of first element and final element becomes the value for the second key.

I have tried the following but it gives incomplete dictionary :

finalDict = {a:{b:c} for a,b,c in e}
print finalDict

{1: {3344: 1}, 980898: {756735: 1}, 4: {23: 1}, 98789797: {980898: 1}, 899979879: {23: 1}, 3344: {1: 1}, 534: {23: 1}, 23: {534: 1}, 9977: {980898: 1}, 756735: {980898: 1}}

Answer 1

d = [(980898, 9977, 1),
 (899979879, 23, 1),
 (1, 1, 2),
 (980898, 98789797, 1),
 (98789797, 980898, 1),
 (1, 756735, 1),
 (1, 3344, 1),
 (23, 4, 1),
 (534, 23, 1),
 (756735, 1, 1),
 (9977, 980898, 1),
 (23, 899979879, 1),
 (4, 23, 1),
 (756735, 980898, 1),
 (3344, 1, 1),
 (980898, 756735, 1),
 (23, 534, 1)]

from collections import defaultdict

D = defaultdict(dict)

for a, b, c in d:
    D[a][b] = c

for k in sorted(D):
    print k, D[k]

---

1 {3344: 1, 1: 2, 756735: 1}
4 {23: 1}
23 {4: 1, 534: 1, 899979879: 1}
534 {23: 1}
3344 {1: 1}
9977 {980898: 1}
756735 {1: 1, 980898: 1}
980898 {9977: 1, 98789797: 1, 756735: 1}
98789797 {980898: 1}
899979879 {23: 1}

Answer 2

Your

{a:{b:c} for a,b,c in e}

overwrites the values of the main dictionary. You can use setdefault to access the value of the main dictionary if it already exists:

d = {}
for a, b, c in e:
    d.setdefault(a, {})[b] = c

Answer 3

Your code only keeps the first encountered value for a given key. You can use defaultdict from the collections module to initialize result with a fresh dict and then set the key to the value as encountered in your list of tuples.

#all_tuples = [(...), (...), ...]
from collections import defaultdict

result = defaultdict(dict)
for uid, k, v in all_tuples :
    result[uid][k] = v

dict(result)

Output:

{1: {1: 2, 3344: 1, 756735: 1},
 4: {23: 1},
 23: {4: 1, 534: 1, 899979879: 1},
 534: {23: 1},
 3344: {1: 1},
 9977: {980898: 1},
 756735: {1: 1, 980898: 1},
 980898: {9977: 1, 756735: 1, 98789797: 1},
 98789797: {980898: 1},
 899979879: {23: 1}}