如何从字典中获取值列表？

如何在Python中获取字典中的值列表？

在Java中，将Map的值作为List变得容易list = map.values();。我想知道Python中是否有类似的简单方法可以从字典中获取值列表。

是的，这与Python 2完全相同：

d.values()

在Python 3中（在其中dict.values返回字典值的视图）：

list(d.values())

您可以使用*运算符解压缩dict_values：

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

或列出对象

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

应该有一种方法，最好只有一种方法。

因此list(dictionary.values())是一种方法。

但是，考虑到Python3，更快的方法是什么？

[*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)

在Intel®Core™i7-8650U CPU @ 1.90GHz上完成。

# Name                    Version                   Build
ipython                   7.5.0            py37h24bf2e0_0

结果

1. 对于小词典* operator更快
2. 对于重要的大字典来说，list()可能会更快

请按照以下示例-

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))))

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

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

纯粹出于视觉目的。不会产生有用的产品...仅在您希望长字典以段落类型形式打印时才有用。