Computing the sum of all unique values in a numpy array containing rows of dicts

Question

I have a large numpy array, with each row containing a dict of words, in a similar format to below:

data = [{'a': 1, 'c': 2}, {'ba': 3, 'a': 4}, ... }

Could someone please point me in the right direction for how would I go about computing the sum of all the unique values of the dicts in each row of the numpy array? From the example above, I would hope to obtain something like this:

result = {'a': 5, 'c': 2, 'ba': 3, ...}

At the moment, the only way I can think to do it is iterating through each row of the data, and then each key of the dict, if a unique key is found then append it to the new dict and set the value, if a key that's already contained in the dict is found then add the value of that key to the key in the 'result'. Although this seems like an inefficient way to do it.

You might want to check out some of the ideas in here: stackoverflow.com/questions/16458340/… — Pablo Oliva
– Pablo Oliva, Commented Nov 21, 2017 at 19:41
Would it be possible to use Counters instead? Then this could be as simple as sum(data, Counter()) — Patrick Haugh
– Patrick Haugh, Commented Nov 21, 2017 at 19:43
That looks like a list, not a numpy array. If it is indeed an array, why are you using an array to hold dicts? — Mad Physicist
– Mad Physicist, Commented Nov 21, 2017 at 19:47

Reblochon Masque · Accepted Answer · 2017-11-21 19:54:26Z

3

You could use a Counter() and update it with each dictionary contained in data, in a loop:

from collections import Counter

data = [{'a': 1, 'c': 2}, {'ba': 3, 'a': 4}]
c = Counter()
for d in data:
    c.update(d)

output:

Counter({'a': 5, 'ba': 3, 'c': 2})

alternate one liner:

(as proposed by @AntonVBR in the comments)

sum((Counter(dict(x)) for x in data), Counter())

edited Nov 21, 2017 at 19:54

answered Nov 21, 2017 at 19:43

Reblochon Masque

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1 Comment

Anton vBR Over a year ago

Can be written in one line like this: sum((Counter(dict(x)) for x in data),Counter()) according to stackoverflow.com/questions/11290092/…

Joe Iddon · Accepted Answer · 2017-11-21 19:43:05Z

2

A pure Python solution using for-loops:

data = [{'a': 1, 'c': 2}, {'ba': 3, 'a': 4}]
result = {}
for d in data:
    for k, v in d.items():
        if k in result:
            result[k] += v
        else:
            result[k] = v

output:

{'c': 2, 'a': 5, 'ba': 3}

answered Nov 21, 2017 at 19:43

Joe Iddon

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