Reputation: 8927
I have a list of values, and a dictionary. I want to ensure that each value in the list exists as a key in the dictionary. At the moment I'm using two sets to figure out if any values don't exist in the dictionary
unmapped = set(foo) - set(bar.keys())
Is there a more pythonic way to test this though? It feels like a bit of a hack?
Upvotes: 5
Views: 2760
Reputation: 51155
Your approach will work, however, there will be overhead from the conversion to set
.
Another solution with the same time complexity would be:
all(i in bar for i in foo)
Both of these have time complexity O(len(foo))
bar = {str(i): i for i in range(100000)}
foo = [str(i) for i in range(1, 10000, 2)]
%timeit all(i in bar for i in foo)
462 µs ± 14.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit set(foo) - set(bar)
14.6 ms ± 174 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# The overhead is all the difference here:
foo = set(foo)
bar = set(bar)
%timeit foo - bar
213 µs ± 1.48 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The overhead here makes a pretty big difference, so I would choose all
here.
Upvotes: 4
Reputation: 1875
Try this to see if there is any unmapped item:
has_unmapped = all( (x in bar) for x in foo )
To see the unmapped items:
unmapped_items = [ x for x in foo if x not in bar ]
Upvotes: 1