Count occurrences of each key in python dictionary

Question

I have a python dictionary object that looks somewhat like this:

[{"house": 4,  "sign": "Aquarius"},
 {"house": 2,  "sign": "Sagittarius"},
 {"house": 8,  "sign": "Gemini"},
 {"house": 3,  "sign": "Capricorn"},
 {"house": 2,  "sign": "Sagittarius"},
 {"house": 3,  "sign": "Capricorn"},
 {"house": 10, "sign": "Leo"},
 {"house": 4,  "sign": "Aquarius"},
 {"house": 10, "sign": "Leo"},
 {"house": 1,  "sign": "Scorpio"}]

Now for each 'sign' key, I'd like to count how many times each value occurs.

def predominant_sign(data):
    signs = [k['sign'] for k in data if k.get('sign')]
    print len(signs)

This however, prints number of times 'sign' appears in the dictionary, instead of getting the value of the sign and counting the number of times a particular value appears.

For example, the output I'd like to see is:

Aquarius: 2
Sagittarius: 2
Gemini: 1
...

And so on. What should I change to get the desired output?

Answer 1

Use collections.Counter and its most_common method:

from collections import Counter
def predominant_sign(data):
    signs = Counter(k['sign'] for k in data if k.get('sign'))
    for sign, count in signs.most_common():
        print(sign, count)

Answer 2

You can use collections.Counter module, with a simple generator expression, like this

>>> from collections import Counter
>>> Counter(k['sign'] for k in data if k.get('sign'))
Counter({'Sagittarius': 2, 'Capricorn': 2, 'Aquarius': 2, 'Leo': 2, 'Scorpio': 1, 'Gemini': 1}) 

This will give you a dictionary which has the signs as keys and their number of occurrences as the values.

---

You can do the same with a normal dictionary, like this

>>> result = {}
>>> for k in data:
...     if 'sign' in k:
...         result[k['sign']] = result.get(k['sign'], 0) + 1
>>> result
{'Sagittarius': 2, 'Capricorn': 2, 'Aquarius': 2, 'Leo': 2, 'Scorpio': 1, 'Gemini': 1}

The dictionary.get method, accepts a second parameter, which will be the default value to be returned if the key is not found in the dictionary. So, if the current sign is not in result, it will give 0 instead.