How to sum dict elements

Answer #1:

A little ugly, but a one-liner:

dictf = reduce(lambda x, y: dict((k, v + y[k]) for k, v in x.iteritems()), dict1)

Answer #2:

You can use the collections.Counter

counter = collections.Counter()
for d in dict1: 
    counter.update(d)

Or, if you prefer oneliners:

functools.reduce(operator.add, map(collections.Counter, dict1))

Answer #3:

Leveraging sum() should get better performance when adding more than a few dicts

>>> dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
>>> from operator import itemgetter
>>> {k:sum(map(itemgetter(k), dict1)) for k in dict1[0]}        # Python2.7+
{'a': 5, 'b': 7}
>>> dict((k,sum(map(itemgetter(k), dict1))) for k in dict1[0])  # Python2.6
{'a': 5, 'b': 7}

adding Stephan’s suggestion

>>> {k: sum(d[k] for d in dict1) for k in dict1[0]}            # Python2.7+
{'a': 5, 'b': 7}
>>> dict((k, sum(d[k] for d in dict1)) for k in dict1[0])      # Python2.6
{'a': 5, 'b': 7}

I think Stephan’s version of the Python2.7 code reads really nicely

Answer #4:

This might help:

def sum_dict(d1, d2):
    for key, value in d1.items():
        d1[key] = value + d2.get(key, 0)
    return d1

>>> dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]
>>> reduce(sum_dict, dict1)
{'a': 5, 'b': 7}

Answer #5:

The following code shows one way to do it:

dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]

final = {}
for k in dict1[0].keys():           # Init all elements to zero.
    final[k] = 0
for d in dict1:
    for k in d.keys():
        final[k] = final[k] + d[k]  # Update the element.

print final

This outputs:

{'a': 5, 'b': 7}

as you desired.

Or, as inspired by kriss, better but still readable:

dict1 = [{'a':2, 'b':3},{'a':3, 'b':4}]

final = {}
for d in dict1:
    for k in d.keys():
        final[k] = final.get(k,0) + d[k]

print final

I pine for the days of the original, readable Python 🙂

Answer #6:

I was interested in the performance of the proposed Counter, reduce and sum methods for large lists. Maybe someone else is interested in this as well.
You can have a look here: https://gist.github.com/torstenrudolf/277e98df296f23ff921c

I tested the three methods for this list of dictionaries:

dictList = [{'a': x, 'b': 2*x, 'c': x**2} for x in xrange(10000)]

the sum method showed the best performance, followed by reduce and Counter was the slowest. The time showed below is in seconds.

In [34]: test(dictList)
Out[34]: 
{'counter': 0.01955194902420044,
 'reduce': 0.006518083095550537,
 'sum': 0.0018319153785705566}

But this is dependent on the number of elements in the dictionaries. the sum method will slow down faster than the reduce.

l = [{y: x*y for y in xrange(100)} for x in xrange(10000)]

In [37]: test(l, num=100)
Out[37]: 
{'counter': 0.2401433277130127,
 'reduce': 0.11110662937164306,
 'sum': 0.2256883692741394}

Answer #7:

In Python 2.7 you can replace the dict with a collections.Counter object. This supports addition and subtraction of Counters.

Answer #8:

Here is a reasonable beatiful one.

final = {}
for k in dict1[0].Keys():
    final[k] = sum(x[k] for x in dict1)
return final