7

I have a problem with concatenating dictionaries. Have so much code so I show in example what my problem is.

d1 = {'the':3, 'fine':4, 'word':2}
+
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
+
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
=
finald = {'the':10, 'fine':16, 'word':6, 'knight':1, 'orange':1, 'sequel':1, 'jimbo':1}

It is prepering wordcounts for wordcloud. I dont know how to concatenate values of the keys it is puzzle for me. Please help. Best regards

rafaelc
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MTG
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3 Answers3

8

I would use a Counter from collections for this.

from collections import Counter

d1 = {'the':3, 'fine':4, 'word':2}
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}

c = Counter()
for d in (d1, d2, d3):
    c.update(d)
print(c)

Outputs:

Counter({'fine': 16, 'the': 10, 'word': 6, 'orange': 1, 'jimbo': 1, 'sequel': 1, 'knight': 1})
Patrick Haugh
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  • Counters seems to solve a lot of dict questions these days! For this problem I would do `reduce(lambda x,y:Counter(x)+Counter(y),[d1,d2, d3])` . Is update() quicker/better than adding the counters? – themistoklik Dec 08 '16 at 21:33
  • @themistoklik Probably not by much, if at all, and `reduce` is deprecated in python 3. I had originally written this with `map(c.update, (d1, d2, d3))` but then I my inner functional programmer refused to allow me to abuse side effects like that. – Patrick Haugh Dec 08 '16 at 23:57
2
import itertools

d1 = {'the':3, 'fine':4, 'word':2}
d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
dicts = [d1, d2, d3]

In [31]: answer = {k:sum(d[k] if k in d else 0 for d in dicts) for k in itertools.chain.from_iterable(dicts)}

In [32]: answer
Out[32]: 
{'sequel': 1,
 'the': 10,
 'fine': 16,
 'jimbo': 1,
 'word': 6,
 'orange': 1,
 'knight': 1}
inspectorG4dget
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2
def sumDicts(*dicts):
    summed = {}
    for subdict in dicts:
        for (key, value) in subdict.items():
            summed[key] = summed.get(key, 0) + value
    return summed

Shell example:

>>> d1 = {'the':3, 'fine':4, 'word':2}
>>> d2 = {'the':2, 'fine':4, 'word':1, 'knight':1, 'orange':1}
>>> d3 = {'the':5, 'fine':8, 'word':3, 'sequel':1, 'jimbo':1}
>>> sumDicts(d1, d2, d3)
{'orange': 1, 'the': 10, 'fine': 16, 'jimbo': 1, 'word': 6, 'knight': 1, 'sequel': 1}
rassar
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