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

For example I have two dicts:

Dict A: {'a': 1, 'b': 2, 'c': 3}
Dict B: {'b': 3, 'c': 4, 'd': 5}

I need a pythonic way of 'combining' two dicts such that the result is:

{'a': 1, 'b': 5, 'c': 7, 'd': 5}

That is to say: if a key appears in both dicts, add their values, if it appears in only one dict, keep its value.

Use collections.Counter:

>>> from collections import Counter
>>> A = Counter({'a':1, 'b':2, 'c':3})
>>> B = Counter({'b':3, 'c':4, 'd':5})
>>> A + B
Counter({'c': 7, 'b': 5, 'd': 5, 'a': 1})

Counters are basically a subclass of dict, so you can still do everything else with them you'd normally do with that type, such as iterate over their keys and values.

A more generic solution, which works for non-numeric values as well:

a = {'a': 'foo', 'b':'bar', 'c': 'baz'}
b = {'a': 'spam', 'c':'ham', 'x': 'blah'}

r = dict(a.items() + b.items() +
    [(k, a[k] + b[k]) for k in set(b) & set(a)])

or even more generic:

def combine_dicts(a, b, op=operator.add):
    return dict(a.items() + b.items() +
        [(k, op(a[k], b[k])) for k in set(b) & set(a)])
    

For example:

>>> a = {'a': 2, 'b':3, 'c':4}
>>> b = {'a': 5, 'c':6, 'x':7}

>>> import operator
>>> print combine_dicts(a, b, operator.mul)
{'a': 10, 'x': 7, 'c': 24, 'b': 3}

>>> A = {'a':1, 'b':2, 'c':3}
>>> B = {'b':3, 'c':4, 'd':5}
>>> c = {x: A.get(x, 0) + B.get(x, 0) for x in set(A).union(B)}
>>> print(c)

{'a': 1, 'c': 7, 'b': 5, 'd': 5}

Intro: There are the (probably) best solutions. But you have to know it and remember it and sometimes you have to hope that your Python version isn't too old or whatever the issue could be.

Then there are the most 'hacky' solutions. They are great and short but sometimes are hard to understand, to read and to remember.

There is, though, an alternative which is to to try to reinvent the wheel.

• Why reinventing the wheel?
• Generally because it's a really good way to learn (and sometimes just because the already-existing tool doesn't do exactly what you would like and/or the way you would like it) and the easiest way if you don't know or don't remember the perfect tool for your problem.

So, I propose to reinvent the wheel of the Counter class from the collections module (partially at least):

class MyDict(dict):
    def __add__(self, oth):
        r = self.copy()
        
        try:
            for key, val in oth.items():
                if key in r:
                    r[key] += val  # You can custom it here
                else:
                    r[key] = val
        except AttributeError:  # In case oth isn't a dict
            return NotImplemented  # The convention when a case isn't handled
        
        return r

a = MyDict({'a':1, 'b':2, 'c':3})
b = MyDict({'b':3, 'c':4, 'd':5})

print(a+b)  # Output {'a':1, 'b': 5, 'c': 7, 'd': 5}

There would probably others way to implement that and there are already tools to do that but it's always nice to visualize how things would basically works.

Definitely summing the Counter()s is the most pythonic way to go in such cases but only if it results in a positive value. Here is an example and as you can see there is no c in result after negating the c's value in B dictionary.

In [1]: from collections import Counter

In [2]: A = Counter({'a':1, 'b':2, 'c':3})

In [3]: B = Counter({'b':3, 'c':-4, 'd':5})

In [4]: A + B
Out[4]: Counter({'d': 5, 'b': 5, 'a': 1})

That's because Counters were primarily designed to work with positive integers to represent running counts (negative count is meaningless). But to help with those use cases,python documents the minimum range and type restrictions as follows:

> - The Counter class itself is a dictionary > subclass with no restrictions on its keys and values. The values are > intended to be numbers representing counts, but you could store > anything in the value field. > - The most_common() method requires only > that the values be orderable. > - For in-place operations such as c[key] > += 1, the value type need only support addition and subtraction. So fractions, floats, and decimals would work and negative values are > supported. The same is also true for update() and subtract() which > allow negative and zero values for both inputs and outputs. > - The multiset methods are designed only for use cases with positive values. > The inputs may be negative or zero, but only outputs with positive > values are created. There are no type restrictions, but the value type > needs to support addition, subtraction, and comparison. > - The elements() method requires integer counts. It ignores zero and negative counts.

So for getting around that problem after summing your Counter you can use Counter.update in order to get the desire output. It works like dict.update() but adds counts instead of replacing them.

In [24]: A.update(B)

In [25]: A
Out[25]: Counter({'d': 5, 'b': 5, 'a': 1, 'c': -1})

myDict = {}
for k in itertools.chain(A.keys(), B.keys()):
    myDict[k] = A.get(k, 0)+B.get(k, 0)

The one with no extra imports!

Their is a pythonic standard called EAFP(Easier to Ask for Forgiveness than Permission). Below code is based on that python standard.

# The A and B dictionaries
A = {'a': 1, 'b': 2, 'c': 3}
B = {'b': 3, 'c': 4, 'd': 5}

# The final dictionary. Will contain the final outputs.
newdict = {}

# Make sure every key of A and B get into the final dictionary 'newdict'.
newdict.update(A)
newdict.update(B)

# Iterate through each key of A.
for i in A.keys():

    # If same key exist on B, its values from A and B will add together and
    # get included in the final dictionary 'newdict'.
    try:
        addition = A[i] + B[i]
        newdict[i] = addition

    # If current key does not exist in dictionary B, it will give a KeyError,
    # catch it and continue looping.
    except KeyError:
        continue

EDIT: thanks to jerzyk for his improvement suggestions.

import itertools
import collections

dictA = {'a':1, 'b':2, 'c':3}
dictB = {'b':3, 'c':4, 'd':5}

new_dict = collections.defaultdict(int)
# use dict.items() instead of dict.iteritems() for Python3
for k, v in itertools.chain(dictA.iteritems(), dictB.iteritems()):
    new_dict[k] += v

print dict(new_dict)

# OUTPUT
{'a': 1, 'c': 7, 'b': 5, 'd': 5}

OR

Alternative you can use Counter as @Martijn has mentioned above.

For a more generic and extensible way check mergedict. It uses singledispatch and can merge values based on its types.

Example:

from mergedict import MergeDict

class SumDict(MergeDict):
    @MergeDict.dispatch(int)
    def merge_int(this, other):
        return this + other

d2 = SumDict({'a': 1, 'b': 'one'})
d2.merge({'a':2, 'b': 'two'})

assert d2 == {'a': 3, 'b': 'two'}

From python 3.5: merging and summing

Thanks to @tokeinizer_fsj that told me in a comment that I didn't get completely the meaning of the question (I thought that add meant just adding keys that eventually where different in the two dictinaries and, instead, i meant that the common key values should be summed). So I added that loop before the merging, so that the second dictionary contains the sum of the common keys. The last dictionary will be the one whose values will last in the new dictionary that is the result of the merging of the two, so I thing the problem is solved. The solution is valid from python 3.5 and following versions.

a = {
    "a": 1,
    "b": 2,
    "c": 3
}

b = {
    "a": 2,
    "b": 3,
    "d": 5
}

# Python 3.5

for key in b:
    if key in a:
        b[key] = b[key] + a[key]

c = {**a, **b}
print(c)

>>> c
{'a': 3, 'b': 5, 'c': 3, 'd': 5}

Reusable code

a = {'a': 1, 'b': 2, 'c': 3}
b = {'b': 3, 'c': 4, 'd': 5}


def mergsum(a, b):
    for k in b:
        if k in a:
            b[k] = b[k] + a[k]
    c = {**a, **b}
    return c


print(mergsum(a, b))

Additionally, please note a.update( b ) is 2x faster than a + b

from collections import Counter
a = Counter({'menu': 20, 'good': 15, 'happy': 10, 'bar': 5})
b = Counter({'menu': 1, 'good': 1, 'bar': 3})

%timeit a + b;
## 100000 loops, best of 3: 8.62 µs per loop
## The slowest run took 4.04 times longer than the fastest. This could mean that an intermediate result is being cached.

%timeit a.update(b)
## 100000 loops, best of 3: 4.51 µs per loop

def merge_with(f, xs, ys):
    xs = a_copy_of(xs) # dict(xs), maybe generalizable?
    for (y, v) in ys.iteritems():
        xs[y] = v if y not in xs else f(xs[x], v)

merge_with((lambda x, y: x + y), A, B)

You could easily generalize this:

def merge_dicts(f, *dicts):
    result = {}
    for d in dicts:
        for (k, v) in d.iteritems():
            result[k] = v if k not in result else f(result[k], v)

Then it can take any number of dicts.

This is a simple solution for merging two dictionaries where += can be applied to the values, it has to iterate over a dictionary only once

a = {'a':1, 'b':2, 'c':3}
    
dicts = [{'b':3, 'c':4, 'd':5},         {'c':9, 'a':9, 'd':9}]

def merge_dicts(merged,mergedfrom):
    for k,v in mergedfrom.items():
        if k in merged:
            merged[k] += v
        else:
            merged[k] = v
    return merged

for dct in dicts:
    a = merge_dicts(a,dct)
print (a)
#{'c': 16, 'b': 5, 'd': 14, 'a': 10}

One line solution is to use dictionary comprehension.

C = { k: A.get(k,0) + B.get(k,0) for k in list(B.keys()) + list(A.keys()) }

This solution is easy to use, it is used as a normal dictionary, but you can use the sum function.

class SumDict(dict):
    def __add__(self, y):
        return {x: self.get(x, 0) + y.get(x, 0) for x in set(self).union(y)}

A = SumDict({'a': 1, 'c': 2})
B = SumDict({'b': 3, 'c': 4})  # Also works: B = {'b': 3, 'c': 4}
print(A + B)  # OUTPUT {'a': 1, 'b': 3, 'c': 6}

What about:

def dict_merge_and_sum( d1, d2 ):
	ret = d1
	ret.update({ k:v + d2[k] for k,v in d1.items() if k in d2 })
	ret.update({ k:v for k,v in d2.items() if k not in d1 })
	return ret

A = {'a': 1, 'b': 2, 'c': 3}
B = {'b': 3, 'c': 4, 'd': 5}

print( dict_merge_and_sum( A, B ) )

Output:

{'d': 5, 'a': 1, 'c': 7, 'b': 5}

The above solutions are great for the scenario where you have a small number of Counters. If you have a big list of them though, something like this is much nicer:

from collections import Counter

A = Counter({'a':1, 'b':2, 'c':3})
B = Counter({'b':3, 'c':4, 'd':5}) 
C = Counter({'a': 5, 'e':3})
list_of_counts = [A, B, C]

total = sum(list_of_counts, Counter())

print(total)
# Counter({'c': 7, 'a': 6, 'b': 5, 'd': 5, 'e': 3})

The above solution is essentially summing the Counters by:

total = Counter()
for count in list_of_counts:
    total += count
print(total)
# Counter({'c': 7, 'a': 6, 'b': 5, 'd': 5, 'e': 3})

This does the same thing but I think it always helps to see what it is effectively doing underneath.

More conventional way to combine two dict. Using modules and tools are good but understanding the logic behind it will help in case you don't remember the tools.

Program to combine two dictionary adding values for common keys.

def combine_dict(d1,d2):

for key,value in d1.items():
  if key in d2:
    d2[key] += value
  else:
      d2[key] = value
return d2

combine_dict({'a':1, 'b':2, 'c':3},{'b':3, 'c':4, 'd':5})

output == {'b': 5, 'c': 7, 'd': 5, 'a': 1}

Here's a very general solution. You can deal with any number of dict + keys that are only in some dict + easily use any aggregation function you want:

def aggregate_dicts(dicts, operation=sum):
    """Aggregate a sequence of dictionaries using `operation`."""
    all_keys = set().union(*[el.keys() for el in dicts])
    return {k: operation([dic.get(k, None) for dic in dicts]) for k in all_keys}

example:

dicts_same_keys = [{'x': 0, 'y': 1}, {'x': 1, 'y': 2}, {'x': 2, 'y': 3}]
aggregate_dicts(dicts_same_keys, operation=sum)
#{'x': 3, 'y': 6}

example non-identical keys and generic aggregation:

dicts_diff_keys = [{'x': 0, 'y': 1}, {'x': 1, 'y': 2}, {'x': 2, 'y': 3, 'c': 4}]

def mean_no_none(l):
    l_no_none = [el for el in l if el is not None]
    return sum(l_no_none) / len(l_no_none)

aggregate_dicts(dicts_diff_keys, operation=mean_no_none)
# {'x': 1.0, 'c': 4.0, 'y': 2.0}

Here's yet another option using dictionary comprehensions combined with the behavior of dict():

dict3 = dict(dict1, **{ k: v + dict1.get(k, 0) for k, v in dict2.items() })
# {'a': 4, 'b': 2, 'c': 7, 'g': 1}

From https://docs.python.org/3/library/stdtypes.html#dict:

> https://docs.python.org/3/library/stdtypes.html#dict

and also

> If keyword arguments are given, the keyword arguments and their values are added to the dictionary created from the positional argument.

The dict comprehension

**{ k: v + dict1.get(v, 0), v in dict2.items() }

handles adding dict1[1] to v. We don't need an explicit if here because the default value for our dict1.get can be set to 0 instead.

dict1 = {'a':1, 'b':2, 'c':3}
dict2 = {'a':3, 'g':1, 'c':4}
dict3 = {} # will store new values

for x in dict1:
    
    if x in dict2: #sum values with same key
        dict3[x] = dict1[x] +dict2[x]
    else: #add the values from x to dict1
        dict3[x] = dict1[x]
        
#search for new values not in a
for x in dict2:
    if x not in dict1:
        dict3[x] = dict2[x]
print(dict3) # {'a': 4, 'b': 2, 'c': 7, 'g': 1}

Merging three dicts a,b,c in a single line without any other modules or libs

If we have the three dicts

a = {"a":9}
b = {"b":7}
c = {'b': 2, 'd': 90}

Merge all with a single line and return a dict object using

c = dict(a.items() + b.items() + c.items())

Returning

{'a': 9, 'b': 2, 'd': 90}