Python字典：获取键的值列表

Question

有没有一种内置/快速的方法可以使用字典的键列表来获取相应项的列表？

例如，我有：

>>> mydict = {'one': 1, 'two': 2, 'three': 3}
>>> mykeys = ['three', 'one']

如何使用mykeys以列表形式获取字典中的相应值？

>>> mydict.WHAT_GOES_HERE(mykeys)
[3, 1]

Answer 1

除了列表比较，还有其他几种方法：

• 如果找不到密钥，则生成列表并引发异常：map(mydict.__getitem__, mykeys)
• 使用None生成列表，如果找不到键：map(mydict.get, mykeys)

或者，使用operator.itemgetter可以返回元组：

from operator import itemgetter
myvalues = itemgetter(*mykeys)(mydict)
# use `list(...)` if list is required

注意：在Python3中，map返回一个迭代器而不是一个列表。使用list(map(...))作为列表。

Answer 2

比较一下速度：

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[1]: l = [0,1,2,3,2,3,1,2,0]
In[2]: m = {0:10, 1:11, 2:12, 3:13}
In[3]: %timeit [m[_] for _ in l]  # list comprehension
1000000 loops, best of 3: 762 ns per loop
In[4]: %timeit map(lambda _: m[_], l)  # using 'map'
1000000 loops, best of 3: 1.66 µs per loop
In[5]: %timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
1000000 loops, best of 3: 1.65 µs per loop
In[6]: %timeit map(m.__getitem__, l)
The slowest run took 4.01 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 853 ns per loop
In[7]: %timeit map(m.get, l)
1000000 loops, best of 3: 908 ns per loop
In[33]: from operator import itemgetter
In[34]: %timeit list(itemgetter(*l)(m))
The slowest run took 9.26 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 739 ns per loop

所以列表理解和itemgetter是最快的方法。

更新： 对于大的随机列表和地图，我得到了一些不同的结果：

Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32
In[2]: import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)
%timeit f(m)
%timeit list(itemgetter(*l)(m))
%timeit [m[_] for _ in l]  # list comprehension
%timeit map(m.__getitem__, l)
%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
%timeit map(m.get, l)
%timeit map(lambda _: m[_], l)
1000 loops, best of 3: 1.14 ms per loop
1000 loops, best of 3: 1.68 ms per loop
100 loops, best of 3: 2 ms per loop
100 loops, best of 3: 2.05 ms per loop
100 loops, best of 3: 2.19 ms per loop
100 loops, best of 3: 2.53 ms per loop
100 loops, best of 3: 2.9 ms per loop

在这种情况下，明确的赢家是f = operator.itemgetter(*l); f(m)，而明确的局外人是map(lambda _: m[_], l)。

Python 3.6.4的更新：

import numpy.random as nprnd
l = nprnd.randint(1000, size=10000)
m = dict([(_, nprnd.rand()) for _ in range(1000)])
from operator import itemgetter
import operator
f = operator.itemgetter(*l)
%timeit f(m)
%timeit list(itemgetter(*l)(m))
%timeit [m[_] for _ in l]  # list comprehension
%timeit list(map(m.__getitem__, l))
%timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.
%timeit list(map(m.get, l))
%timeit list(map(lambda _: m[_], l)
1.66 ms ± 74.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
2.1 ms ± 93.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.58 ms ± 88.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.36 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.98 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.7 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
3.14 ms ± 62.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

因此，Python3.6.4的结果几乎相同。

Answer 3

列表理解似乎是一个很好的方法：

>>> [mydict[x] for x in mykeys]
[3, 1]