In order to quickly compare the keys of 2 dictionaries, I'm creating sets of the keys using this method:

dict_1 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_2 = {"file_1":10, "file_2":20, "file_3":30}
set_1 = {file for file in dict_1}
set_2 = {file for file in dict_2}

Than I use diff_set = set_1 - set_2 to see which keys are missing from set_2.

Is there a faster way? I see that using set(dict.keys()) is less of a workarou, so I'll switch to it - but is it more efficient?

CodePudding user response：

Let's measure more properly (not just measuring a single execution and also not including the setup) and include faster solutions:

300 ns  300 ns  300 ns  {*dict_1} - {*dict_2}
388 ns  389 ns  389 ns  {file for file in dict_1 if file not in dict_2}
389 ns  390 ns  390 ns  dict_1.keys() - dict_2
458 ns  458 ns  458 ns  set(dict_1) - set(dict_2)
472 ns  472 ns  472 ns  dict_1.keys() - dict_2.keys()
665 ns  665 ns  668 ns  set(dict_1.keys()) - set(dict_2.keys())
716 ns  716 ns  716 ns  {file for file in dict_1} - {file for file in dict_2}

Benchmark code (Try it online!):

import timeit

setup = '''
dict_1 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_2 = {"file_1":10, "file_2":20, "file_3":30}
'''

codes = [
    '{file for file in dict_1} - {file for file in dict_2}',
    'set(dict_1) - set(dict_2)',
    'set(dict_1.keys()) - set(dict_2.keys())',
    'dict_1.keys() - dict_2',
    'dict_1.keys() - dict_2.keys()',
    '{*dict_1} - {*dict_2}',
    '{file for file in dict_1 if file not in dict_2}',
]

exec(setup)
for code in codes:
    print(eval(code))

tss = [[] for _ in codes]
for _ in range(20):
    print()
    for code, ts in zip(codes, tss):
        number = 10000
        t = min(timeit.repeat(code, setup, number=number)) / number
        ts.append(t)
    for code, ts in sorted(zip(codes, tss), key=lambda cs: sorted(cs[1])):
        print(*('= ns ' % (t * 1e9) for t in sorted(ts)[:3]), code)

CodePudding user response：

The fastest and most efficient way would be to convert your dictionaries to set()

diff_set = set(dict_1) - set(dict_2)

Output:

{'file_4'}

Proof(Execution time comparison):

Method 1(Method discussed in question)

import timeit

start1 = timeit.default_timer()
dict_1 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_2 = {"file_1":10, "file_2":20, "file_3":30}
set_1 = {file for file in dict_1}
set_2 = {file for file in dict_2}
diff_set = set_1 - set_2
stop1 = timeit.default_timer()
execution_time1 = stop1 - start1
print(f"It took {execution_time1} time for method 1")

Method 2(Suggested answer)

import timeit

start2 = timeit.default_timer()
dict_3 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_4 = {"file_1":10, "file_2":20, "file_3":30}
diff_set = set(dict_3) - set(dict_4)
stop2 = timeit.default_timer()
execution_time2 = stop2 - start2
print(f"It took {execution_time2} time for method 2")

Method 3(dict.keys())

import timeit

start3 = timeit.default_timer()
dict_5 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_6 = {"file_1":10, "file_2":20, "file_3":30}
diff_set = set(dict_5.keys()) - set(dict_6.keys())
stop3 = timeit.default_timer()
execution_time3 = stop3 - start3
print(f"It took {execution_time3} time for method 3")

Output:

It took 9.499999578110874e-06 time for method 1
It took 6.399990525096655e-06 time for method 2
It took 8.699993486516178e-06 time for method 3