Counting the Number of keywords in a dictionary in python

Question

I have a list of words in a dictionary with the value = 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?

Answer 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()))

Answer 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']

Answer 3

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

Answer 4

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)