如何在 Python 中获取 dict 中的值列表？
在 Java 中，将 Map 的值作为 List 获取就像做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。
问问题
1249179 次
5 回答
783
dict.values返回字典值的视图，因此您必须将其包装在list：
list(d.values())
于 2013-04-26T03:27:37.010 回答
67
你可以使用* 操作符来解压 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']
于 2019-04-02T18:59:31.393 回答
46
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())是一种方式。
然而，考虑到 Python3，什么更快？
[*L]对比[].extend(L)对比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) Core(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
* operator更快 - 对于重要的大字典
list()可能会稍微快一些
于 2019-06-24T12:32:28.130 回答
16
按照下面的例子——
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))))
于 2019-02-14T11:24:42.813 回答
-3
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...仅当您希望以段落类型的形式打印长字典时才有用。
于 2019-07-13T01:10:04.313 回答