Python: Dictionary merge by updating but not overwriting if value exists

Question

If I have 2 dicts as follows:

d1 = {(\'unit1\',\'test1\'):2,(\'unit1\',\'test2\'):4}
d2 = {(\'unit1\',\'test1\'):2,(\'unit1\',\'test2\'):\'\'}

Answer 1

Merging Only Non-zero values

to do this we can just create a dict without the empty values and then merge them together this way:

d1 = {'a':1, 'b':1, 'c': '', 'd': ''}
d2 = {'a':2, 'c':2, 'd': ''}
merged_non_zero = {
    k: (d1.get(k) or d2.get(k))
    for k in set(d1) | set(d2)
}
print(merged_non_zero)

outputs:

{'a': 1, 'b': 1, 'c': 2, 'd': ''}

• a -> prefer first value from d1 as 'a' exists on both d1 and d2
• b -> only exists on d1
• c -> non-zero on d2
• d -> empty string on both

Explanation

The above code will create a dictionary using dict comprehension.

if d1 has the value and its non-zero value (i.e. bool(val) is True), it'll use d1[k] value, otherwise it'll take d2[k].

notice that we also merge all keys of the two dicts as they may not have the exact same keys using set union - set(d1) | set(d2).

Python 3.5+ Literal Dict

unless using obsolete version of python you better off using this.

Pythonic & faster way for dict unpacking:

d1 = {'a':1, 'b':1}
d2 = {'a':2, 'c':2}
merged = {**d1, **d2}  # priority from right to left
print(merged)

{'a': 2, 'b': 1, 'c': 2}

its simpler and also faster than the dict(list(d2.items()) + list(d1.items())) alternative:

d1 = {i: 1 for i in range(1000000)}
d2 = {i: 2 for i in range(2000000)}

%timeit dict(list(d1.items()) + list(d2.items())) 
402 ms ± 33.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit {**d1, **d2}
144 ms ± 1.12 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

more on this from PEP448:

The keys in a dictionary remain in a right-to-left priority order, so {**{'a': 1}, 'a': 2, **{'a': 3}} evaluates to {'a': 3}. There is no restriction on the number or position of unpackings.