Counting the number of distinct keys in a dictionary in Python

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()))

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']

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)

If the question is about counting the number of keywords then I 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 the 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