Merge values of a dictionary by key based on custom function

Question

Assume you have two dictionaries and you want to merge the two dictionaries by applying a function to the values that have matching keys. here I use the + operator as binary function.

x = { 1: "a", 2: "b", 3: "c" }
y = { 1: "A", 2: "B", 3: "C" }

result = { t[0][0]: t[0][1] + t[1][1] for t in zip(sorted(x.items()), sorted(y.items())) }

print result # gives { 1: "aA", 2: "bB", 3: "cC" }

I would prefer a self contained expression instead of statements, but this is unreadable.

so far I'm doing:

def dzip(f, a, b):
    least_keys = set.intersection(set(a.keys()), set(b.keys()))
    copy_dict = dict()
    for i in least_keys.keys():
        copy_dict[i] = f(a[i], b[i])
    return copy_dict

print dzip(lambda a,b: a+b,x,y)

Is there a more readable solution to this than the expression I gave?

Answer 1

In the first case, you can directly use a dict comprehension:

>>> x = { 1: "a", 2: "b", 3: "c" }
>>> y = { 1: "A", 2: "B", 3: "C" }
>>> {key: x.get(key, "") + y.get(key, "") for key in set.intersection(set(x.keys()), set(y.keys()))}
{1: 'aA', 2: 'bB', 3: 'cC'}

So that in your second piece of code, you can simplify it to a simple one liner:

def dzip(f, a, b):
    return {key: f(a.get(key, ""), b.get(key, "")) for key in set.inersection(set(a.keys()) + set(b.keys()))}

You can even define dzip as a lambda:

dzip = lambda f, a, b: {key: f(a.get(key, ""), b.get(key, "")) 
    for key in set.intersection(set(a.keys()), set(b.keys()))}

In a single run, this becomes:

>>> dzip = lambda f, a, b: {key: f(a.get(key, ""), b.get(key, "")) 
...         for key in set.intersection(set(a.keys()), set(b.keys()))}    
>>> 
>>> print dzip(lambda a,b: a+b,x,y)
{1: 'aA', 2: 'bB', 3: 'cC'}

Note that this will work even if x and y have different sets of keys (just something that can break in your first version of the code).

Answer 2

You can use Counter for this type of dict merging

from collections import Counter
>>>Counter(x)+Counter(y)
Counter({3: 'cC', 2: 'bB', 1: 'aA'})