How can I get list of values from dict?

Question or problem about Python programming:

How can I get a list of the values in a dict in Python?

In Java, getting the values of a Map as a List is as easy as doing list = map.values();. I’m wondering if there is a similarly simple way in Python to get a list of values from a dict.

How to solve the problem:

Solution 1:

Yes it’s the exact same thing in Python 2:

d.values()

In Python 3 (where dict.values returns a view of the dictionary’s values instead):

list(d.values())

Solution 2:

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']

Solution 3:

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

Solution 4:

Follow the below example —

songs = [
{"title": "happy birthday", "playcount": 4},
{"title": "AC/DC", "playcount": 2},
{"title": "Billie Jean", "playcount": 6},
{"title": "Human Touch", "playcount": 3}
]

print("====================")
print(f'Songs --> {songs} \n')
title = list(map(lambda x : x['title'], songs))
print(f'Print Title --> {title}')

playcount = list(map(lambda x : x['playcount'], songs))
print(f'Print Playcount --> {playcount}')
print (f'Print Sorted playcount --> {sorted(playcount)}')

# Aliter -
print(sorted(list(map(lambda x: x['playcount'],songs))))

Solution 5:

out: dict_values([{1:a, 2:b}])

in:  str(dict.values())[14:-3]    
out: 1:a, 2:b

Purely for visual purposes. Does not produce a useful product… Only useful if you want a long dictionary to print in a paragraph type form.

Hope this helps!