Find intersection of dictionary values which are lists

Question

I have dictionaries in a list with same keys, while the values are variant:

[{1:[1,2,3,4,5], 2:[6,7,8], 3:[1,3,5,7,9]},
 {1:[2,3,4], 2:[6,7], 3:[1,3,5]},
 ...]

I would like to get intersection as dictionary under same keys like this:

{1:[2,3,4], 2:[6,7], 3:[1,3,5]}

Do you have just 2 items that you want to intersect or could your list of dict have more than just 2 items? — mgilson, Jul 15 '16 at 07:49
I've tried: set(mylist[0][key]).intersection(*mylist[1:][key]) but got error message "list index out of range". I have in total 12 dictionaries in the list. — Fei-man Hsu, Jul 15 '16 at 07:52
You could gather all lists for a specific key, e.g. `lists` and then do `list(set.intersection(*map(set, lists)))`. If you do this for all keys you basically have what you desire. Do you need a fixed order in your "values"? If not, you could consider making them `set`s in the first place. Then, it would come down to `set.intersection(*sets)` after gathering all sets for a specific key in the variable `sets`. — Michael Hoff, Jul 15 '16 at 07:53

score 0 · Answer 1 · answered Jul 15 '16 at 07:55

Give this a try?

dicts = [{1:[1,2,3,4,5], 2:[6,7,8], 3:[1,3,5,7,9]},{1:[2,3,4], 2:[6,7], 3:[1,3,5]}]

result = { k: set(dicts[0][k]).intersection(*(d[k] for d in dicts[1:])) for k in dicts[0].keys() }

print(result)

# Output:
# {1: {2, 3, 4}, 2: {6, 7}, 3: {1, 3, 5}}

If you want lists instead of sets as the output value type, just throw a list(...) around the set intersection.

score 0 · Answer 2 · answered Jul 15 '16 at 07:55

For a list of dictionaries, reduce the whole list as this:

>>> from functools import reduce
>>> d = [{1:[1,2,3,4,5], 2:[6,7,8], 3:[1,3,5,7,9]},{1:[2,3,4], 2:[6,7], 3:[1,3,5]}]
>>> reduce(lambda x, y: {k: sorted(list(set(x[k])&set(y[k]))) for k in x.keys()}, d)
{1: [2, 3, 4], 2: [6, 7], 3: [1, 3, 5]}

score 0 · Answer 3 · answered Jul 15 '16 at 08:05

I would probably do something along these lines:

# Take the first dict and convert the values to `set`.
output = {k: set(v) for k, v in dictionaries[0].items()}

# For the rest of the dicts, update the set at a given key by intersecting it with each of the other lists that have the same key.
for d in dictionaries[1:]:
    for k, v in output.items():
        output[k] = v.intersection(d[k])

There are different variations on this same theme, but I find this one to be about as simple to read as it gets (and since code is read more often than it is written, I consider that a win :-)

Thank you mgilson for the detailed code interpretation!! It works! — Fei-man Hsu, Jul 15 '16 at 08:50

score 0 · Answer 4 · edited May 23 '17 at 12:25

use dict.viewkeys and dict.viewitems

In [103]: dict.viewkeys?
Docstring: D.viewkeys() -> a set-like object providing a view on D's keys

dict.viewitems?
Docstring: D.viewitems() -> a set-like object providing a view on D's items

a = [{1: [1, 2, 3, 4, 5], 2: [6, 7, 8], 3: [1, 3, 5, 7, 9]},
 {1: [2, 3, 4], 2: [6, 7], 3: [1, 3, 5]}]

In [100]: dict(zip(a[0].viewkeys() and a[1].viewkeys(), a[0].viewvalues() and a[1].viewvalues()))
Out[100]: {1: [2, 3, 4], 2: [6, 7], 3: [1, 3, 5]}

Find intersection of dictionary values which are lists

4 Answers4