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This is mostly a question of good/pythonic style. I have a dictionary which has lists for values, i.e.

my_dict = {"a": a_list, "b": b_list, "c": c_list}

and so on. I also have an empty dictionary with the same keys, in which I want to store the mean of these lists against their key. If instead of using a second dictionary, I used a nested list, I could do

mean_lists = [[key, sum(l)/len(l)] for key, l in my_dict.items() if l]

giving an output

[["a", a_mean], ["b", b_mean], ["c", c_mean]]

which seems neat to me. Is there a way to do this nicely outputting as a dictionary, or is something like this:

mean_dict = {key: [] for key in my_dict}
for key, l in my_dict.items():
    if l:
        mean_dict[key] = sum(l)/len(l)

the best I can do?

Idle_92
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  • I think I was getting hung up on this because I was convinced I had to initialise my dictionary beforehand. I am also not proud of my ability to search for previously answered questions... – Idle_92 Oct 09 '18 at 15:18

2 Answers2

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Yes, by using a dictionary comprehension; you already have the parts to build each key and value together in your list comprehension, just alter the syntax a little:

{key: sum(l) / len(l) for key, l in my_dict.items() if l}

The syntax is closely related to list comprehensions, but instead of a single expression for each list element value, you have two expressions: one for the key (just key in the above example) and one for the associated value (sum(l) / len(l) in the above).

You already appear to be familiar with the syntax, as you created a dictionary mapping from keys to empty lists:

mean_dict = {key: [] for key in my_dict}

before using a separate loop to replace those lists with the sum(l) / len(l) calculations for the exact same keys.

Martijn Pieters
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Use a dictionary comprehension!

mean_lists = {key:sum(l)/len(l) for key, l in my_dict.items() if l}
Joe Iddon
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