CODEWITHSUNDEEP
pythonpython-2.7dictionary
Newtt
Newtt

Reputation: 6200

Count occurrences of each key in python dictionary

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?

Upvotes: 23

Views: 53315

Answers (3)

Vaibhav Hiwase
Vaibhav Hiwase

Reputation: 557

def counter(my_list):
    my_list = sorted(my_list)
    first_val, *all_val = my_list
    p_index = my_list.index(first_val)
    my_counter = {}
    for item in all_val:
         c_index = my_list.index(item)
         diff = abs(c_index-p_index)
         p_index = c_index
         my_counter[first_val] = diff 
         first_val = item
    c_index = my_list.index(item)
    diff = len(my_list) - c_index
    my_counter[first_val] = diff 
    return my_counter

>>> counter([list(i.values())[1] for i in my_list])
{'Aquarius': 2,
 'Capricorn': 2,
 'Gemini': 1,
 'Leo': 2,
 'Sagittarius': 2,
 'Scorpio': 1}

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239683

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.

Upvotes: 10

vaultah
vaultah

Reputation: 46603

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)

Upvotes: 23

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