Pythonic way to merge two lists

Question

I want to append value to an existing dictionary. Here is my code:

tmp_result = [{'M': 8}, {'N': 16},]
cross_configs = [({'device': 'cpu'},), ({'device': 'cuda'},)]

import copy
generated_configs = []
for config in cross_configs:
    for value in config:
            new_value = copy.deepcopy(tmp_result)
            new_value.append(value)
            generated_configs.append(new_value)

print (generated_configs)

Output: 
[[{'M': 8}, {'N': 16}, {'device': 'cpu'}], [{'M': 8}, {'N': 16}, {'device': 'cuda'}]]

I don't like the inner loop which does deepcopy and append. What is a pythonic way to do that?

This does not "append" values to dictionaries; this appends additional single-entry dictionaries to a list. Your result is two lists of three independent dictionaries each. What problem are you trying to solve with this data structure? — Prune
– Prune, Commented Jul 25, 2019 at 18:37
I don't see how you're appending to a dictionary. You seem to be dealing with lists of lists who elements just happen to be single-key dictionaries, or sets/tuples of one dictionary each. Are you ultimately trying to take [a,b,c] and [(c),(d)] and return [ [a,b,c], [a,b,d] ]? — Bill M.
– Bill M., Commented Jul 25, 2019 at 18:39
That does not really explain why someone thought it might be a good idea to pack a single entry dict into a list into a list. Maybe suggesting a refactoring of the interface is worthwhile as it can help avoid poor performance and hard to maintain source. Furthermore it is not really clear why you would need a deep copy. Are the lists altered afterwards? — Christian B.
– Christian B., Commented Jul 25, 2019 at 18:49

Austin · Accepted Answer · 2019-07-25 18:38:57Z

1

You could do a list-comprehension:

[tmp_result + list(x) for x in cross_configs]

Example:

tmp_result = [{'M': 8}, {'N': 16},]
cross_configs = [({'device': 'cpu'},), ({'device': 'cuda'},)]

print([tmp_result + list(x) for x in cross_configs])
# [[{'M': 8}, {'N': 16}, {'device': 'cpu'}], [{'M': 8}, {'N': 16}, {'device': 'cuda'}]]

answered Jul 25, 2019 at 18:38

Austin

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chepner · Accepted Answer · 2019-07-25 18:40:37Z

0

A nested list comprehension will suffice; the explicit call to dict is sufficient to avoid the need for deepcopy given the example shown.

generated_configs = [[dict(y) for y in tmp_result + list(x)] for x in cross_configs]

If you object to tmp_result + list(x), use itertools.chain instead.

from itertools import chain
generated_configs = [[dict(y) for y in chain(tmp_result, x)] for x in cross_configs]

answered Jul 25, 2019 at 18:40

chepner

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3 Comments

Austin Over a year ago

Couldn't nested list-comprehension be avoided as in my answer?

chepner Over a year ago

list(x) will only make a shallow copy of the tuple, meaning each sublist will share a reference to the same device dict. The call to dict ensures that each dict in the result is distinct. (It's not equivalent to a deep copy of each list, in the event that the dicts themselves contain references to mutable objects, but it is sufficient for the dicts shown in the question.)

chepner Over a year ago

It's almost certainly not an issue for the example show, since the two dict literals will create separate objects. But consider something like d1 = {'device': 'cpu'}; d2 = {'device': 'cuda'}; cross_configs = [(d1,), (d1,), (d2,), (d1,), (d2,)].

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