用numpy滚动最大 [英] Rolling maximum with numpy

问题描述

这将在长度为K的滑动窗口上计算A的滚动最大值"(类似于滚动平均值):

This computes the "rolling max" of A (similar to rolling average) over a sliding window of length K:

import numpy as np
A = np.random.rand(100000)
K = 10
rollingmax = np.array([max(A[j:j+K]) for j in range(len(A)-K)])

但是我认为在性能方面远非最佳.

but I think it is far from optimal in terms of performance.

我知道pandas库具有

I know that the pandas library has rolling_max, but in my project, I don't want to use this new dependance.

问题:是否有一种简单的方法可以仅使用numpy计算滚动最大值?

推荐答案

我猜想使用

I guess this little trick using strides and as_strided will do the job:

def max_rolling1(a, window,axis =1):
        shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
        strides = a.strides + (a.strides[-1],)
        rolling = np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
        return np.max(rolling,axis=axis)

出于比较目的，我根据您的算法定义了另一个函数:

and for comaprison purpose I defined another function based on your algorithm :

def max_rolling2(A,K):
    rollingmax = np.array([max(A[j:j+K]) for j in range(len(A)-K)])
    return rollingmax

，并且在我的笔记本电脑上通过timeit进行的比较是:

and the comparison by timeit on my laptop is :

与:

A = np.random.rand(100000)
K = 10


%timeit X = max_rolling2(A,K)

170 ms ± 19.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit X = max_rolling1(A,K)

> 3.75 ms ± 479 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

这篇关于用numpy滚动最大的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！