CODEWITHSUNDEEP
pythonpython-2.7dictionary
user2990084
user2990084

Reputation: 2840

Count items in defaultdict

I have a defaultdict, then converted to a dictionary. Now I am looking for a way to count the items grouped by value. What i mean is:

dict = {1: [103], 2: [9], 3: [3, 4], 4: [103, 106], 10: [3, 4], 11: [1, 9], 28: [1]}

for key in dict.items():
    print key

(1, [103])
(2, [9])
(3, [3, 4])
(4, [103, 106])
(10, [3, 4])
(11, [1, 9])
(28, [1])

item : total
{4 : 2, 106 : 1} ...

How can i do that?

Upvotes: 2

Views: 5990

Answers (3)

Some
Some

Reputation: 514

m_dict = dict
n = {}
for k, v in m_dict.iteritems():
    for j in v:
        n[j] = (n[j] + 1 if n.get(j) != None else 1)
print n

Result:

{1: 2, 3: 2, 4: 2, 9: 2, 103: 2, 106: 1}

Or:

m_dict = dict
n = sum([v for k,v in m_dict.iteritems()], [])
c = {j : n.count(j) for j in n}
print c

Result:

{1: 2, 3: 2, 4: 2, 9: 2, 103: 2, 106: 1}

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239653

You can use collections.Counter to count the number of occurrences of each element in the values lists, like this

>>> from collections import Counter
>>> Counter(value for values in d.itervalues() for value in values)
Counter({1: 2, 3: 2, 4: 2, 103: 2, 9: 2, 106: 1})

Counter is a subclass of dictionary only. So, you can use it just like any other dictionary

The flattening of the values can be done with itertools.chain.from_iterable, like this

>>> from collections import Counter
>>> from itertools import chain
>>> Counter(chain.from_iterable(d.itervalues()))
Counter({1: 2, 3: 2, 4: 2, 103: 2, 9: 2, 106: 1})

Upvotes: 8

alec_djinn
alec_djinn

Reputation: 10819

This is another way to do it without using anything but the basics.

d = {1: [103], 2: [9], 3: [3, 4], 4: [103, 106], 10: [3, 4], 11: [1, 9], 28: [1]}


answer = {} # store the counts here
for key in d:
    for value in d[key]:
        if value in answer:
            # if the key is already in the answer
            answer[value] += 1
        else:
            # if the key has never been counted before
            # then add it to the answer and set its value to 1
            answer.update({value:1}) 


for key in answer:
    print key, answer[key]

It prints:

1 2
3 2
4 2
103 2
9 2
106 1

Upvotes: 0

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