Question

我刚注意到zeros的{{1}}函数有一个奇怪的行为：

numpy

另一方面，%timeit np.zeros((1000, 1000)) 1.06 ms ± 29.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each) %timeit np.zeros((5000, 5000)) 4 µs ± 66 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)似乎有正常的行为。有人知道为什么用ones函数初始化一个小的numpy数组需要比大数组更多的时间吗？

（Python 3.5，numpy 1.11）

Answer 1

这看起来像calloc达到一个阈值，它会向操作系统请求内存归零，并且不需要手动初始化它。浏览源代码，numpy.zeros最终delegates to calloc获取归零内存块，如果与numpy.empty进行比较，则不执行初始化：

In [15]: %timeit np.zeros((5000, 5000))
The slowest run took 12.65 times longer than the fastest. This could mean that a
n intermediate result is being cached.
100000 loops, best of 3: 10 µs per loop

In [16]: %timeit np.empty((5000, 5000))
The slowest run took 5.05 times longer than the fastest. This could mean that an
 intermediate result is being cached.
100000 loops, best of 3: 10.3 µs per loop

您可以看到np.zeros没有5000x5000阵列的初始化开销。

事实上，操作系统甚至不是真的＆＃34;分配该内存直到您尝试访问它。数TB的数组请求在没有太字节的机器上成功：

In [23]: x = np.zeros(2**40)  # No MemoryError!

Numpy

1 个答案: