Counting the number of distinct keys in a dictionary in Python
Solution 1
len(yourdict.keys())
or just
len(yourdict)
If you like to count unique words in the file, you could just use set
and do like
len(set(open(yourdictfile).read().split()))
Solution 2
The number of distinct words (i.e. count of entries in the dictionary) can be found using the len()
function.
> a = {'foo':42, 'bar':69}
> len(a)
2
To get all the distinct words (i.e. the keys), use the .keys()
method.
> list(a.keys())
['foo', 'bar']
Solution 3
Calling len()
directly on your dictionary works, and is faster than building an iterator, d.keys()
, and calling len()
on it, but the speed of either will negligible in comparison to whatever else your program is doing.
d = {x: x**2 for x in range(1000)}
len(d)
# 1000
len(d.keys())
# 1000
%timeit len(d)
# 41.9 ns ± 0.244 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
%timeit len(d.keys())
# 83.3 ns ± 0.41 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
Solution 4
If the question is about counting the number of keywords then would recommend something like
def countoccurrences(store, value):
try:
store[value] = store[value] + 1
except KeyError as e:
store[value] = 1
return
in the main function have something that loops through the data and pass the values to countoccurrences function
if __name__ == "__main__":
store = {}
list = ('a', 'a', 'b', 'c', 'c')
for data in list:
countoccurrences(store, data)
for k, v in store.iteritems():
print "Key " + k + " has occurred " + str(v) + " times"
The code outputs
Key a has occurred 2 times
Key c has occurred 2 times
Key b has occurred 1 times
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Dan
Updated on April 28, 2021Comments
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Dan about 3 years
I have a a dictionary mapping keywords to the repetition of the keyword, but I only want a list of distinct words so I wanted to count the number of keywords. Is there a way to count the number of keywords or is there another way I should look for distinct words?
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Flimm about 3 yearsThe keys in a Python dictionary are already distinct from each other. You can't have the exact some keyword as a key twice in a Python dictionary. Therefore, counting the number of keys is the same as counting the number of distinct keys.
-
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john_science over 11 yearsI know this post is old, but I was curious. Is this the fastest method? Or: is it a reasonably fast method for large dictionaries?
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Chih-Hsuan Yen about 8 yearsBoth
len(yourdict.keys())
andlen(yourdict)
are O(1). The latter is slightly faster. See my tests below. -
ntk4 almost 8 yearsI'd like to note that you can also go for the values (I know the question didn't ask it) with
len(yourdict.values())
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Graham almost 6 yearsPEP 8 naming conventions dictate that
countoccurrences()
should instead becount_occurrences()
. Also, if you importcollections.Counter
, there's a much better way to do it:from collections import Counter; store = Counter(); for data in list: store[list] += 1
.