How to sum dict elements

A little ugly, but a one-liner:

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

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}

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

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))