Removing duplicate keys from python dictionary but summing the values

Question

I have a dictionary in python

d = {tags[0]: value, tags[1]: value, tags[2]: value, tags[3]: value, tags[4]: value}

imagine that this dict i

Answer 1

This option serves but is done with a list, or best can provide insight

data = []
        for i, j in query.iteritems():
            data.append(int(j))    
        try:
            data.sort()
        except TypeError:
            del data
        data_array = []
        for x in data:
            if x not in data_array:
                data_array.append(x)  
        return data_array

Answer 2

Perhapse what you really want is a tuple of key-value pairs.

[('dog',1), ('cat',2), ('cat',3)]

Answer 3

I'm not sure what you're trying to achieve, but the Counter class might be helpful for what you're trying to do: http://docs.python.org/dev/library/collections.html#collections.Counter

Answer 4

I'd like to improve Paul Seeb's answer:

tps = [('cat',5),('dog',9),('cat',4),('parrot',6),('cat',6)]
result = {}
for k, v in tps:
  result[k] = result.get(k, 0) + v

Answer 5

Instead of just doing dict of those things (can't have multiples of same key in a dict) I assume you can have them in a list of tuple pairs. Then it is just as easy as

tps = [('cat',5),('dog',9),('cat',4),('parrot',6),('cat',6)]
result = {}
for k,v in tps:
    try:
        result[k] += v
    except KeyError:
        result[k] = v

>>> result
{'dog': 9, 'parrot': 6, 'cat': 15}

changed mine to more explicit try-except handling. Alfe's is very concise though

Answer 6

This the perfect situation for using Counter data structure. Lets take a look on what it does on few familiar data structures. Lets start with good old list.

>>> from collections import Counter
>>> list_a = ["A", "A", "B", "C", "C", "A", "D"]
>>> list_b = ["B", "A", "B", "C", "C", "C", "D"]
>>> c1 = Counter(list_a)
>>> c2 = Counter(list_b)
>>> c1
Counter({'A': 3, 'C': 2, 'B': 1, 'D': 1})
>>> c2
Counter({'C': 3, 'B': 2, 'A': 1, 'D': 1})
>>> c1 - c2
Counter({'A': 2})
>>> c1 + c2
Counter({'C': 5, 'A': 4, 'B': 3, 'D': 2})
>>> c_diff = c1 - c2
>>> c_diff.update([77, 77, -99, 0, 0, 0])
>>> c_diff
Counter({0: 3, 'A': 2, 77: 2, -99: 1})

As you can see this behaves as a set that keeps the count of element occurrences as a value. Hm, but what about using a dictionary instead of a list? The dictionary in itself is a set-like structure where for values we don't have to have numbers, so how will that get handled? Lets take a look.

>>> dic1 = {"A":"a", "B":"b"}
>>> cd = Counter(dic1)
>>> cd
Counter({'B': 'b', 'A': 'a'})
>>> cd.update(B='bB123')
>>> cd
Counter({'B': 'bbB123', 'A': 'a'})


>>> dic2 = {"A":[1,2], "B": ("a", 5)}
>>> cd2 = Counter(dic2)
>>> cd2
Counter({'B': ('a', 5), 'A': [1, 2]})
>>> cd2.update(A=[42], B=(2,2))
>>> cd2
Counter({'B': ('a', 5, 2, 2), 'A': [1, 2, 42, 42, 42, 42]})
>>> cd2 = Counter(dic2)
>>> cd2
Counter({'B': ('a', 5), 'A': [1, 2]})
>>> cd2.update(A=[42], B=("new elem",))
>>> cd2
Counter({'B': ('a', 5, 'new elem'), 'A': [1, 2, 42]})

As you can see the value we are adding/changing has to be of the same type in update or it throws TypeError. As for your particular case just go with the flow

>>> d = {'cat': 5, 'dog': 9, 'cat': 4, 'parrot': 6, 'cat': 6}
>>> cd3 = Counter(d)
>>> cd3
Counter({'dog': 9, 'parrot': 6, 'cat': 6})
cd3.update(parrot=123)
cd3
Counter({'parrot': 129, 'dog': 9, 'cat': 6})