python – How can I get list of values from dict?

python – How can I get list of values from dict?

dict.values returns a view of the dictionarys values, so you have to wrap it in list:

list(d.values())

You can use * operator to unpack dict_values:

>>> d = {1: a, 2: b}
>>> [*d.values()]
[a, b]

or list object

>>> d = {1: a, 2: b}
>>> list(d.values())
[a, b]

python – How can I get list of values from dict?

There should be one ‒ and preferably only one ‒ obvious way to do it.

Therefore list(dictionary.values()) is the one way.

Yet, considering Python3, what is quicker?

[*L] vs. [].extend(L) vs. list(L)

small_ds = {x: str(x+42) for x in range(10)}
small_df = {x: float(x+42) for x in range(10)}

print(Small Dict(str))
%timeit [*small_ds.values()]
%timeit [].extend(small_ds.values())
%timeit list(small_ds.values())

print(Small Dict(float))
%timeit [*small_df.values()]
%timeit [].extend(small_df.values())
%timeit list(small_df.values())

big_ds = {x: str(x+42) for x in range(1000000)}
big_df = {x: float(x+42) for x in range(1000000)}

print(Big Dict(str))
%timeit [*big_ds.values()]
%timeit [].extend(big_ds.values())
%timeit list(big_ds.values())

print(Big Dict(float))
%timeit [*big_df.values()]
%timeit [].extend(big_df.values())
%timeit list(big_df.values())

Small Dict(str)
256 ns ± 3.37 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
338 ns ± 0.807 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 1.9 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Small Dict(float)
268 ns ± 0.297 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
343 ns ± 15.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
336 ns ± 0.68 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Big Dict(str)
17.5 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.5 ms ± 338 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
16.2 ms ± 19.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Big Dict(float)
13.2 ms ± 41 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
13.1 ms ± 919 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
12.8 ms ± 578 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Done on Intel(R) Core(TM) i7-8650U CPU @ 1.90GHz.

# Name                    Version                   Build
ipython                   7.5.0            py37h24bf2e0_0

The result

1. For small dictionaries * operator is quicker
2. For big dictionaries where it matters list() is maybe slightly quicker