是否有一种内置/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mykey在字典中获取相应的值作为列表？
>>> mydict.WHAT_GOES_HERE(mykeys)
[3, 1]
列表理解似乎是一种很好的方法：
>>> [mydict[x] for x in mykeys]
[3, 1]
除了list-comp，还有其他几种方式：
map（mydict.\uu getitem\uuuu，mykeys）或者，使用operator.itemgetter可以返回一个元组：
from operator import itemgetter
myvalues = itemgetter(*mykeys)(mydict)
# use `list(...)` if list is required
注意：在Python3中，map返回迭代器而不是列表。使用列表（映射（…）用于列表。
一点速度比较：
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。
对于大型随机列表和地图，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)
1000 loops, best of 3: 1.14 ms per loop
%timeit list(itemgetter(*l)(m))
1000 loops, best of 3: 1.68 ms per loop
%timeit [m[_] for _ in l] # list comprehension
100 loops, best of 3: 2 ms per loop
%timeit map(m.__getitem__, l)
100 loops, best of 3: 2.05 ms per loop
%timeit list(m[_] for _ in l) # a generator expression passed to a list constructor.
100 loops, best of 3: 2.19 ms per loop
%timeit map(m.get, l)
100 loops, best of 3: 2.53 ms per loop
%timeit map(lambda _: m[_], l)
100 loops, best of 3: 2.9 ms per loop
所以在这种情况下，明显的赢家是f=operator.itemgetter（*l）； f（m），并且明确的局外人：map（lambda_： m[_]， l）。
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)
1.66 ms ± 74.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit list(itemgetter(*l)(m))
2.1 ms ± 93.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit [m[_] for _ in l] # list comprehension
2.58 ms ± 88.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit list(map(m.__getitem__, l))
2.36 ms ± 60.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit list(m[_] for _ in l) # a generator expression passed to a list constructor.
2.98 ms ± 142 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit list(map(m.get, l))
2.7 ms ± 284 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit list(map(lambda _: m[_], l)
3.14 ms ± 62.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
因此，Python3.6的结果。4几乎是一样的。
