Converting a list of "pairs" into a dictionary of dictionaries?

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This question was previously asked here with an egregious typo: Counting "unique pairs" of numbers into a python dictionary?

This is an algorithmic problem, and I don't know of the most efficient solution. My idea would be to somehow cache values in a list and enumerate pairs...but that would be so slow. I'm guessing there's something useful from itertools.

Let's say I have a list of integers whereby are never repeats:

list1 = [2, 3]

In this case, there is a unique pair 2-3 and 3-2, so the dictionary should be:

{2:{3: 1}, 3:{2: 1}}

That is, there is 1 pair of 2-3 and 1 pair of 3-2.

For larger lists, the pairing is the same, e.g.

list2 = [2, 3, 4]

has the dicitonary

{2:{3:1, 4:1}, 3:{2:1, 4:1}, 4:{3:1, 2:1}}

(1) Once the size of the lists become far larger, how would one algorithmically find the "unique pairs" in this format using python data structures?

(2) I mentioned that the lists cannot have repeat integers, e.g. 

[2, 2, 3]

is impossible, as there are two 2s.

However, one may have a list of lists:

list3 = [[2, 3], [2, 3, 4]]

whereby the dictionary must be 

{2:{3:2, 4:1}, 3:{2:2, 4:1}, 4:{2:1, 3:1}}

as there are two pairs of 2-3 and 3-2. How would one "update" the dictionary given multiple lists within a list?

EDIT: My ultimate use case is, I want to iterate through hundreds of lists of integers, and create a single dictionary with the "counts" of pairs. Does this make sense? There might be another data structure which is more useful.

itertools python-3.x dictionary python
3 Answers

For the nested list example, you can do the following, making use of itertools.permutations and dict.setdefault:

from itertools import permutations

list3 = [[2, 3], [2, 3, 4]]

d = {}
for l in list3:
    for a, b in permutations(l, 2):
        d[a][b] = d.setdefault(a, {}).setdefault(b, 0) + 1

# {2: {3: 2, 4: 1}, 3: {2: 2, 4: 1}, 4: {2: 1, 3: 1}}

For flat lists l, use only the inner loop and omit the outer one

user_23901820

For this example I'll just use a list with straight numbers and no nested list:

values = [3, 2, 4]
result = dict.from_keys(values)
for key, value in result.items():
    value = {}
    for num in values:
        if num != key:
            value[num] = 1

This creates a dict with each number as a key. Now in each key, make the value a nested dict who's contents are num: 1 for each number in the original values list if it isn't the name of the key that we're in

user_99176940

use defaultdict with permutations

from collections import defaultdict
from itertools import permutations

d = defaultdict(dict)
for i in [x for x in permutations([4,2,3])]:
    d[i[0]] = {k: 1 for k in i[1:]}

output is 

In [22]: d
Out[22]: defaultdict(dict, {2: {3: 1, 4: 1}, 4: {2: 1, 3: 1}, 3: {2: 1, 4: 1}})

for inherit list of lists https://stackoverflow.com/a/52206554/8060120

user_80601200
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