访问与正在循环的数组的当前位置匹配的移动窗口数组的索引

Question

我已经编写了一个邻域平滑过滤器，适用于用户提供的2D数组-它可以正常工作，但它可以更快/更少浪费内存，因为目前我每次循环运行时都要复制整个输入数组。当传入大型数组时，这将证明是一个真正的问题。

过滤器定义为:

    import numpy as np
    import os
    def conservative_smooth(array2D, kernel_size = 3):
        stepsize = 1    
        if(kernel_size % 2 != 0 and kernel_size >= 3):
            window = np.ones([kernel_size,kernel_size])
        elif(kernel_size % 2 == 0 or kernel_size < 3):
            print "kernel is even - it needs to be odd and at least of a value of 3"
            os._exit(1)
        nxwind, nywind = array2D.shape
        for i in range(0, nxwind, stepsize):
            for j in range(0, nywind, stepsize):
            # CALCULATE MAX AND MIN RANGES OF ROWS AND COLS THAT CAN BE ACCESSED 
            # BY THE WINDOW
            imin=max(0,i-((kernel_size-1)/2)) 
            imax=min(nxwind-1,i+((kernel_size-1)/2))+1
            jmin=max(0,j-((kernel_size-1)/2))
            jmax=min(nywind-1,j+((kernel_size-1)/2))+1
            # THIS IS THE MOST INEFFICIENT PART OF THE CODE
            array2D_temp = array2D.copy() 
            array2D_temp[i,j] = np.nan
            data_wind=array2D_temp[imin:imax,jmin:jmax]
            centre_value = array2D[i,j]
            max_value = np.nanmax(data_wind) 
            min_value = np.nanmin(data_wind) 
            if(centre_value > max_value):
                centre_value = max_value
            elif(centre_value < min_value):
                centre_value = min_value
            else:
                centre_value = centre_value
            ## Append new centre value to output array
            array2D[i,j] = centre_value         
        return array2D

复制整个数组，以便数组中位置[i,j]的值可以临时改为NaN -我不能只复制数组的移动窗口区域(这更好)，因为主数组的[i,j]不会是移动窗口数组的[i,j]。

一个简单的"if value at position in moving window == value in main array"条件也不会起作用，因为如果值重复，这个条件会失败。

我一直在使用一个简单的随机10x10数组(a = np.random.rand(10,10))测试函数

有人有什么建议吗?

Answer 1

据我所知，这似乎与您的原始功能相同，不需要复制。

def conservative_smooth(array2D, kernel_size = 3):
    stepsize = 1    
    if(kernel_size % 2 != 0 and kernel_size >= 3):
        window = np.ones([kernel_size,kernel_size])
    elif(kernel_size % 2 == 0 or kernel_size < 3):
        print "kernel is even - it needs to be odd and at least of a value of 3"
        os._exit(1)
    nxwind, nywind = array2D.shape
    for i in range(0, nxwind, stepsize):
        for j in range(0, nywind, stepsize):
            # CALCULATE MAX AND MIN RANGES OF ROWS AND COLS THAT CAN BE ACCESSED 
            # BY THE WINDOW
            imin=max(0,i-((kernel_size-1)/2)) 
            imax=min(nxwind-1,i+((kernel_size-1)/2))+1
            jmin=max(0,j-((kernel_size-1)/2))
            jmax=min(nywind-1,j+((kernel_size-1)/2))+1
            centre_value = array2D[i,j]
            array2D[i,j] = np.nan
            max_value = np.nanmax(array2D[imin:imax,jmin:jmax]) 
            min_value = np.nanmin(array2D[imin:imax,jmin:jmax]) 
            if(centre_value > max_value):
                centre_value = max_value
            elif(centre_value < min_value):
                centre_value = min_value
            else:
                centre_value = centre_value
            ## Append new centre value to output array
            array2D[i,j] = centre_value      
    return array2D