字典Values返回一个字典值的视图，所以你必须将它包装在list中:

list(d.values())

遵循下面的例子——

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))))

你可以使用*操作符解包dict_values:

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

或列表对象

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

应该有一种——最好只有一种——明显的方法来做到这一点。

因此list(dictionary.values())是一种方法。

但是，考虑到Python3，哪个更快?

[*L]和[].extend(L)和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(R)酷睿(TM) i7-8650U CPU @ 1.90GHz上完成。

# Name                    Version                   Build
ipython                   7.5.0            py37h24bf2e0_0

结果

对于小字典*运算符更快 对于重要的大字典，list()可能稍微快一些

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

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

纯粹是为了视觉效果。没有生产出有用的产品…只有当你想要以段落形式打印一本长字典时才有用。

获取字典中特定键的值列表

最直接的方法是通过迭代list_of_keys来使用理解式。如果list_of_keys包含不是d的键，.get()方法可用于返回默认值(默认为None，但可以更改)。

res = [d[k] for k in list_of_keys] 
# or
res = [d.get(k) for k in list_of_keys]

通常情况下，Python中内置了一个方法，可以从内置操作符模块获取键下的值:itemgetter()。

from operator import itemgetter
res = list(itemgetter(*list_of_keys)(d))

演示:

d = {'a':2, 'b':4, 'c':7}
list_of_keys = ['a','c']
print([d.get(k) for k in list_of_keys])
print(list(itemgetter(*list_of_keys)(d)))
# [2, 7]
# [2, 7]

从字典列表中获取相同键的值

同样，这里使用了理解(遍历字典列表)。在列表上映射itemgetter()以获取特定键的值也是如此。

list_of_dicts = [ {"title": "A", "body": "AA"}, {"title": "B", "body": "BB"} ]

list_comp = [d['title'] for d in list_of_dicts]
itmgetter = list(map(itemgetter('title'), list_of_dicts))
print(list_comp)
print(itmgetter)
# ['A', 'B']
# ['A', 'B']