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?
This question is related to
python
dictionary
count
keyword
In order to count the number of keywords in a dictionary:
def dict_finder(dict_finders):
x=input("Enter the thing you want to find: ")
if x in dict_finders:
print("Element found")
else:
print("Nothing found:")
Some modifications were made on posted answer UnderWaterKremlin to make it python3 proof. A surprising result below as answer.
System specs:
import timeit
d = {x: x**2 for x in range(1000)}
#print (d)
print (len(d))
# 1000
print (len(d.keys()))
# 1000
print (timeit.timeit('len({x: x**2 for x in range(1000)})', number=100000)) # 1
print (timeit.timeit('len({x: x**2 for x in range(1000)}.keys())', number=100000)) # 2
Result:
1) = 37.0100378
2) = 37.002148899999995
So it seems that len(d.keys())
is currently faster than just using len()
.
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 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
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']
Source: Stackoverflow.com