CODEWITHSUNDEEP
pythonlistpython-2.7dictionarycollections
Donato Perconti
Donato Perconti

Reputation: 813

Python Collections Counter for a List of Dictionaries

I have a dynamically growing list of arrays that I would like to add like values together. Here's an example:

{"something" : [{"one":"200"}, {"three":"400"}, {"one":"100"}, {"two":"800"} ... ]}

I'd like to be able to add together the dictionaries inside the list. So, in this case for the key "something", the result would be:

["one":400, "three": 400, "two": 800]

or something to that effect. I'm familiar with the Python's collection counter, but since the "something" list contains dicts, it will not work (unless I'm missing something). The dict is also being dynamically created, so I can't build the list without the dicts. EG: 

Counter({'b':3, 'c':4, 'd':5, 'b':2})

Would normally work, but as soon as I try to add an element, the previous value will be overwritten. I've noticed other questions such as these:

Is there any pythonic way to combine two dicts (adding values for keys that appear in both)?

Python count of items in a dictionary of lists

But again, the objects within the list are dicts.

Upvotes: 5

Views: 7577

Answers (2)

Steve Jessop
Steve Jessop

Reputation: 279285

I think this does what you want, but I'm not sure because I don't know what "The dict is also being dynamically created, so I can't build the list without the dicts" means. Still:

input = {
    "something" : [{"one":"200"}, {"three":"400"}, {"one":"100"}, {"two":"800"}], 
    "foo" : [{"a" : 100, "b" : 200}, {"a" : 300, "b": 400}],
}

def counterize(x):
    return Counter({k : int(v) for k, v in x.iteritems()})

counts = {
    k : sum((counterize(x) for x in v), Counter()) 
    for k, v in input.iteritems()
}

Result:

{
    'foo': Counter({'b': 600, 'a': 400}), 
    'something': Counter({'two': 800, 'three': 400, 'one': 300})
}

I expect using sum with Counter is inefficient (in the same way that using sum with strings is so inefficient that Guido banned it), but I might be wrong. Anyway, if you have performance problems, you could write a function that creates a Counter and repeatedly calls += or update on it:

def makeints(x):
    return {k : int(v) for k, v in x.iteritems()}

def total(seq):
    result = Counter()
    for s in seq:
        result.update(s)
    return result

counts = {k : total(makeints(x) for x in v) for k, v in input.iteritems()}

Upvotes: 10

Marcin
Marcin

Reputation: 238309

One way would be do as follows:

from collections import defaultdict

d = {"something" :
     [{"one":"200"}, {"three":"400"}, {"one":"100"}, {"two":"800"}]}

dd = defaultdict(list)

# first get and group values from the original data structure
# and change strings to ints
for inner_dict in d['something']:
    for k,v in inner_dict.items():
        dd[k].append(int(v))


# second. create output dictionary by summing grouped elemetns
# from the first step.
out_dict =  {k:sum(v) for k,v in dd.items()}

print(out_dict)
# {'two': 800, 'one': 300, 'three': 400}

In here I don't use counter, but defaultdict. Its a two step approach.

Upvotes: 1

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