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]
1000000 loops, best of 3: 762 ns per loop
In[4]: %timeit map(lambda _: m[_], l)
1000000 loops, best of 3: 1.66 µs per loop
In[5]: %timeit list(m[_] for _ in l)
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
所以列表推导式和itemgetter是最快的方法来做到这一点。
更新
对于大型随机列表和映射,我的结果有些不同:
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]
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)
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
因此,在这种情况下,明显的优胜者是f = operator.itemgetter(*l); f(m)
,而鲜明的局外人是:map(lambda _: m[_], l)
。
更新 Python 3.6.4
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]
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)
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,Python 3.6.4的结果几乎相同。
mydict
是一个函数调用(返回一个字典),那么这会多次调用该函数,对吗? - endolith