如何找到用于匹配两个字典中的值的字典键？

Question

I have two dictionaries mapping IDs to values. For simplicity, lets say those are the dictionaries:

d_source = {'a': 1, 'b': 2, 'c': 3, '3': 3}
d_target = {'A': 1, 'B': 2, 'C': 3, '1': 1}

As named, the dictionaries are not symmetrical. I would like to get a dictionary of keys from dictionaries d_source and d_target whose values match. The resulting dictionary would have d_source keys as its own keys, and d_target keys as that keys value (in either a list , tuple or set format).

This would be The expected returned value for the above example should be the following list:

{'a': ('1', 'A'),
 'b': ('B',),
 'c': ('C',),
 '3': ('C',)}

There are two somewhat similar questions , but those solutions can't be easily applied to my question.

Some characteristics of the data:

1. Source would usually be smaller than target. Having roughly few thousand sources (tops) and a magnitude more targets.
2. Duplicates in the same dict (both d_source and d_target ) are not too likely on values.
3. matches are expected to be found for (a rough estimate) not more than 50% than d_source items.
4. All keys are integers.

What is the best (performance wise) solution to this problem? Modeling data into other datatypes for improved performance is totally ok, even when using third party libraries (i'm thinking numpy )

Answer 1

All answers have O(n^2) efficiency which isn't very good so I thought of answering myself.

I use 2(source_len) + 2(dict_count)(dict_len) memory and I have O(2n) efficiency which is the best you can get here I believe.

Here you go:

from collections import defaultdict

d_source = {'a': 1, 'b': 2, 'c': 3, '3': 3}
d_target = {'A': 1, 'B': 2, 'C': 3, '1': 1}

def merge_dicts(source_dict, *rest):
    flipped_rest = defaultdict(list)
    for d in rest:
        while d:
            k, v = d.popitem()
            flipped_rest[v].append(k)
    return {k: tuple(flipped_rest.get(v, ())) for k, v in source_dict.items()}

new_dict = merge_dicts(d_source, d_target)

By the way, I'm using a tuple in order not to link the resulting lists together.

---

As you've added specifications for the data, here's a closer matching solution:

d_source = {'a': 1, 'b': 2, 'c': 3, '3': 3}
d_target = {'A': 1, 'B': 2, 'C': 3, '1': 1}

def second_merge_dicts(source_dict, *rest):
    """Optimized for ~50% source match due to if statement addition.

    Also uses less memory.
    """
    unique_values = set(source_dict.values())
    flipped_rest = defaultdict(list)
    for d in rest:
        while d:
            k, v = d.popitem()
            if v in unique_values:
                flipped_rest[v].append(k)
    return {k: tuple(flipped_rest.get(v, ())) for k, v in source_dict.items()}

new_dict = second_merge_dicts(d_source, d_target)

Answer 2

from collections import defaultdict
from pprint import pprint

d_source  = {'a': 1, 'b': 2, 'c': 3, '3': 3}
d_target = {'A': 1, 'B': 2, 'C': 3, '1': 1}

d_result = defaultdict(list)
{d_result[a].append(b) for a in d_source for b in d_target if d_source[a] == d_target[b]}

pprint(d_result)

Output:

{'3': ['C'],
 'a': ['A', '1'],
 'b': ['B'],
 'c': ['C']}

---

Timing results:

from collections import defaultdict
from copy import deepcopy
from random import randint
from timeit import timeit


def Craig_match(source, target):
    result = defaultdict(list)
    {result[a].append(b) for a in source for b in target if source[a] == target[b]}
    return result

def Bharel_match(source_dict, *rest):
    flipped_rest = defaultdict(list)
    for d in rest:
        while d:
            k, v = d.popitem()
            flipped_rest[v].append(k)
    return {k: tuple(flipped_rest.get(v, ())) for k, v in source_dict.items()}

def modified_Bharel_match(source_dict, *rest):
    """Optimized for ~50% source match due to if statement addition.

    Also uses less memory.
    """
    unique_values = set(source_dict.values())
    flipped_rest = defaultdict(list)
    for d in rest:
        while d:
            k, v = d.popitem()
            if v in unique_values:
                flipped_rest[v].append(k)
    return {k: tuple(flipped_rest.get(v, ())) for k, v in source_dict.items()}

# generate source, target such that:
# a) ~10% duplicate values in source and target
# b) 2000 unique source keys, 20000 unique target keys
# c) a little less than 50% matches source value to target value
# d) numeric keys and values
source = {}
for k in range(2000):
    source[k] = randint(0, 1800)
target = {}
for k in range(20000):
    if k < 1000:
        target[k] = randint(0, 2000)
    else:
        target[k] = randint(2000, 19000)

best_time = {}
approaches = ('Craig', 'Bharel', 'modified_Bharel')
for a in approaches:
    best_time[a] = None

for _ in range(3):
    for approach in approaches:
        test_source = deepcopy(source)
        test_target = deepcopy(target)

        statement = 'd=' + approach + '_match(test_source,test_target)'
        setup = 'from __main__ import test_source, test_target, ' + approach + '_match'
        t = timeit(stmt=statement, setup=setup, number=1)
        if not best_time[approach] or (t < best_time[approach]):
            best_time[approach] = t

for approach in approaches:
    print(approach, ':', '%0.5f' % best_time[approach])

Output:

Craig : 7.29259
Bharel : 0.01587
modified_Bharel : 0.00682

Answer 3

Here is another solution. There are a lot of ways to do this

for key1 in d1:
    for key2 in d2:
        if d1[key1] == d2[key2]:
            stuff

Note that you can use any name for key1 and key2.

Answer 4

This maybe "cheating" in some regards, although if you are looking for the matching values of the keys regardless of the case sensitivity then you might be able to do:

import sets

aa = {'a': 1, 'b': 2, 'c':3}
bb = {'A': 1, 'B': 2, 'd': 3}

bbl = {k.lower():v for k,v in bb.items()}

result = {k:k.upper() for k,v in aa.iteritems() & bbl.viewitems()}
print( result )

Output:

{'a': 'A', 'b': 'B'}

The bbl declaration changes the bb keys into lowercase (it could be either aa , or bb ).

_{* I only tested this on my phone, so just throwing this idea out there I suppose... Also, you've changed your question radically since I began composing my answer, so you get what you get.}

Answer 5

It is up to you to determine the best solution. Here is a solution:

def dicts_to_tuples(*dicts):
    result = {}
    for d in dicts:
        for k,v in d.items():
            result.setdefault(v, []).append(k)
    return [tuple(v) for v in result.values() if len(v) > 1]

d1 = {'a': 1, 'b': 2, 'c':3}
d2 = {'A': 1, 'B': 2}
print dicts_to_tuples(d1, d2)