Is it possible to use python dict comprehension when updating values?

Question

counts = defaultdict(int)
for elem in sets:         #list of sets
    for _ in elem:        #each set contains elements that may be duplicates among the sets
        counts[_] += 1

Is there a way to use dict comprehension for code like this?

Answer 1

Don't use a list comprehension for side effects (i.e. updating counts).

Instead, if you replace defaultdict(int) with Counter, you can write the solution in one line. (I assume your end goal is to make your code shorter/cleaner.)

sets = [
    {1, 2, 3},
    {1, 2},
    {3},
    {3}]
# ---
from collections import Counter
counts = Counter(x for s in sets for x in s)
print(counts)  # -> Counter({3: 3, 1: 2, 2: 2})

• Note: Instead of the nested generator expression, you might prefer to use itertools.chain.from_iterable(sets).

Answer 2

If you are adamant at using list comprehension, you can do this way:

>>> counts = {1:4, 2:5, 3:8, 4:2, 5:9}
>>> sets = [{2,3}, {3,1}, {2}]
>>> [[counts.update({e: counts[e]+1}) for e in si] for si in sets]
[[None, None], [None, None], [None]]
>>> counts
{1: 5, 2: 7, 3: 10, 4: 2, 5: 9}

But yes, as others suggested, you can do it using Counter:

#1: use Counter to get new counts from sets

#2: merge both new & default counters

#3: get Counter back into your dictionary

With sample input:

>>> counts = {1:4, 2:5, 3:8, 4:2, 5:9}
>>> sets = [{2,3}, {3,1}, {2}]
>>> from collections import Counter
>>> cntr_new = Counter(x for s in sets for x in s)
>>> cntr_new
Counter({2: 2, 3: 2, 1: 1})
>>> cntr_orig = Counter(counts)
>>> cntr_final = cntr_orig + cntr_new

Final result:

>>> dict(cntr_final)
{1: 5, 2: 7, 3: 10, 4: 2, 5: 9}