There should be one - and preferably only one - obvious way to do it.
Therefore list(dictionary.values()) is the one way.
[*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
* operator is quickerlist() is maybe slightly quicker