Python字典：获取键列表的值列表

Question

是否有内置/快速方法使用字典键列表来获取相应项的列表？

比如我有：

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

>>> mykeys = ['three', 'one']

如何使用mykeys字典中的相应值作为列表？

>>> mydict.WHAT_GOES_HERE(mykeys)

[3, 1]

Answer 1

除list-comp之外的其他几种方式：如果找不到密钥，则构建列表并抛出异常： map(mydict.__getitem__, mykeys)None如果未找到密钥，则构建列表：map(mydict.get, mykeys)或者，使用operator.itemgetter可以返回一个元组：from operator import itemgettermyvalues = 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list(map(...))了列表。

Answer 2

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 win32In[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 comprehension1000000 loops, best of 3: 762 ns per loopIn[4]: %timeit map(lambda _: m[_], l)  # using 'map'1000000 loops, best of 3: 1.66 µs per loopIn[5]: %timeit list(m[_] for _ in l)  # a generator expression passed to a list constructor.1000000 loops, best of 3: 1.65 µs per loopIn[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 loopIn[7]: %timeit map(m.get, l)1000000 loops, best of 3: 908 ns per loopIn[33]: from operator import itemgetterIn[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：对于大型随机列表和地图，我有一些不同的结果：Python 2.7.11 |Anaconda 2.4.1 (64-bit)| (default, Dec  7 2015, 14:10:42) [MSC v.1500 64 bit (AMD64)] on win32In[2]: import numpy.random as nprndl = nprnd.randint(1000, size=10000)m = dict([(_, nprnd.rand()) for _ in range(1000)])from operator import itemgetterimport operatorf = 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 loop1000 loops, best of 3: 1.68 ms per loop100 loops, best of 3: 2 ms per loop100 loops, best of 3: 2.05 ms per loop100 loops, best of 3: 2.19 ms per loop100 loops, best of 3: 2.53 ms per loop100 loops, best of 3: 2.9 ms per loop因此，在这种情况下，明确的赢家是f = operator.itemgetter(*l); f(m)明确的局外人：map(lambda _: m[_], l)。Python 3.6.4的更新：import numpy.random as nprndl = nprnd.randint(1000, size=10000)m = dict([(_, nprnd.rand()) for _ in range(1000)])from operator import itemgetterimport operatorf = 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，Python 3.6.4的结果几乎相同。