Frequency count of values in a python dict

Question

I have a python dict that I am trying to work out the frequency of which a value is listed.

I cannot import any additional classes so effectively need to write it from scratch. So I have a dict() of

1: Dog, 2: Cat, 3: Dog, 4: Elephant

And I want to return something like:-

2: Dog, 1: Cat, 1:Elephant

This is what I have so far but I cannot work out how to get a count to work. I keep getting

0: Dog, 0: Cat, 0: Elephant

This is what I have so far, could someonle please tell me where I am going wrong?

It's obviously something to do with my count statement (this is just something that I found online but it clearly doesn't work.

I create a empty dict and then add values (animals) to it along with a corresponding key. I then need to go through all the values in all the keys to return the frequency.

The input is a dict() object that contains the values listed of 1: Dog, 2: Cat, 3: Dog, 4: Elephant

def frequency(self):  

    result = set()
    for i in self.items:
        result.add((self.count(i), self.items[i]))
    return sorted(result)

Answer 1

suppose your dictionary is named d and you want a count of the values:

from collections import Counter
d = {1: "dog", 2: "cat", 3: "dog", 4: "elephant"}
counts = Counter(d.values())

Now you can use the counts Counter:

counts['dog'] # 2
counts['elephant'] # 1
counts['fish'] # 0

If you must use custom code and classes for this instead of the standard library, I think your error is here:

result.add((self.count(i), self.items[i]))

What you may want is:

animal_name = self.items[i] #?
result.add((self.count(animal_name), animal_name))

Otherwise you may want to share with us what is in self.items ...

Answer 2

d = {1:"Dog",2:"Cat", 3:"Dog", 4:"Elephant"}
count = {}
for v in d.values():
    if(not(v in count)):
        count[v] = 0
    count[v] += 1
print(count)

{'Dog': 2, 'Cat': 1, 'Elephant': 1}