为什么这些常数函数具有不同的性能？

Question

在以下代码段中，为什么py_sqrt2几乎是np_sqrt2的两倍？

from time import time
from numpy import sqrt as npsqrt
from math import sqrt as pysqrt

NP_SQRT2 = npsqrt(2.0)
PY_SQRT2 = pysqrt(2.0)

def np_sqrt2():
    return NP_SQRT2

def py_sqrt2():
    return PY_SQRT2

def main():
    samples = 10000000

    it = time()
    E = sum(np_sqrt2() for _ in range(samples)) / samples
    print("executed {} np_sqrt2's in {:.6f} seconds E={}".format(samples, time() - it, E))

    it = time()
    E = sum(py_sqrt2() for _ in range(samples)) / samples
    print("executed {} py_sqrt2's in {:.6f} seconds E={}".format(samples, time() - it, E))


if __name__ == "__main__":
    main()

$ python2.7 snippet.py 
executed 10000000 np_sqrt2's in 1.380090 seconds E=1.41421356238
executed 10000000 py_sqrt2's in 0.855742 seconds E=1.41421356238
$ python3.6 snippet.py 
executed 10000000 np_sqrt2's in 1.628093 seconds E=1.4142135623841212
executed 10000000 py_sqrt2's in 0.932918 seconds E=1.4142135623841212

注意，它们是常量函数，它们从具有相同值的预先计算的全局变量中加载，并且常量的不同之处仅在于它们在程序启动时的计算方式。

此外，这些函数的糟糕之处在于，它们按预期运行，并且仅访问全局常量。

In [73]: dis(py_sqrt2)                                                                                                                                                                            
  2           0 LOAD_GLOBAL              0 (PY_SQRT2)
              2 RETURN_VALUE

In [74]: dis(np_sqrt2)                                                                                                                                                                            
  2           0 LOAD_GLOBAL              0 (NP_SQRT2)
              2 RETURN_VALUE

Answer 1

因为您每次仅将其发送给c一个值

尝试以下操作

t0=time.time()
numpy.sqrt([2]*10000)
t1 = time.time()
print("Took %0.3fs to do 10k sqrt(2)"%(t1-t0))

t0 = time.time()
for i in range(10000):
    numpy.sqrt(2)
t1 = time.time()
print("Took %0.3fs to do 10k math.sqrt(2)"%(t1-t0))