如何在 NumPy 中找到平滑多维数组的局部最小值

Question

假设我在 NumPy 中有一个数组，其中包含连续可微函数的评估，并且我想找到局部最小值。没有噪声，因此值低于所有邻居值的每个点都满足我的局部最小值标准。

我有以下列表理解，适用于二维数组，忽略边界上的潜在最小值：

import numpy as N

def local_minima(array2d):
    local_minima = [ index 
                     for index in N.ndindex(array2d.shape)
                     if index[0] > 0
                     if index[1] > 0
                     if index[0] < array2d.shape[0] - 1
                     if index[1] < array2d.shape[1] - 1
                     if array2d[index] < array2d[index[0] - 1, index[1] - 1]
                     if array2d[index] < array2d[index[0] - 1, index[1]]
                     if array2d[index] < array2d[index[0] - 1, index[1] + 1]
                     if array2d[index] < array2d[index[0], index[1] - 1]
                     if array2d[index] < array2d[index[0], index[1] + 1]
                     if array2d[index] < array2d[index[0] + 1, index[1] - 1]
                     if array2d[index] < array2d[index[0] + 1, index[1]]
                     if array2d[index] < array2d[index[0] + 1, index[1] + 1]
                   ]
    return local_minima

但是，这非常慢。我还想让它适用于任意数量的维度。例如，是否有一种简单的方法可以获取任意维度数组中某个点的所有邻居？或者我完全以错误的方式处理这个问题？我应该使用 numpy.gradient() 来代替吗？

Answer 1

可以为任意维度的数组找到局部最小值的位置
使用伊万的detect_peaks 函数，稍加修改：

import numpy as np
import scipy.ndimage.filters as filters
import scipy.ndimage.morphology as morphology

def detect_local_minima(arr):
    # https://stackoverflow.com/questions/3684484/peak-detection-in-a-2d-array/3689710#3689710
    """
    Takes an array and detects the troughs using the local maximum filter.
    Returns a boolean mask of the troughs (i.e. 1 when
    the pixel's value is the neighborhood maximum, 0 otherwise)
    """
    # define an connected neighborhood
    # http://www.scipy.org/doc/api_docs/SciPy.ndimage.morphology.html#generate_binary_structure
    neighborhood = morphology.generate_binary_structure(len(arr.shape),2)
    # apply the local minimum filter; all locations of minimum value 
    # in their neighborhood are set to 1
    # http://www.scipy.org/doc/api_docs/SciPy.ndimage.filters.html#minimum_filter
    local_min = (filters.minimum_filter(arr, footprint=neighborhood)==arr)
    # local_min is a mask that contains the peaks we are 
    # looking for, but also the background.
    # In order to isolate the peaks we must remove the background from the mask.
    # 
    # we create the mask of the background
    background = (arr==0)
    # 
    # a little technicality: we must erode the background in order to 
    # successfully subtract it from local_min, otherwise a line will 
    # appear along the background border (artifact of the local minimum filter)
    # http://www.scipy.org/doc/api_docs/SciPy.ndimage.morphology.html#binary_erosion
    eroded_background = morphology.binary_erosion(
        background, structure=neighborhood, border_value=1)
    # 
    # we obtain the final mask, containing only peaks, 
    # by removing the background from the local_min mask
    detected_minima = local_min ^ eroded_background
    return np.where(detected_minima)       

您可以像这样使用：

arr=np.array([[[0,0,0,-1],[0,0,0,0],[0,0,0,0],[0,0,0,0],[-1,0,0,0]],
              [[0,0,0,0],[0,-1,0,0],[0,0,0,0],[0,0,0,-1],[0,0,0,0]]])
local_minima_locations = detect_local_minima(arr)
print(arr)
# [[[ 0  0  0 -1]
#   [ 0  0  0  0]
#   [ 0  0  0  0]
#   [ 0  0  0  0]
#   [-1  0  0  0]]

#  [[ 0  0  0  0]
#   [ 0 -1  0  0]
#   [ 0  0  0  0]
#   [ 0  0  0 -1]
#   [ 0  0  0  0]]]

这表示最小值出现在索引 [0,0,3]、[0,4,0]、[1,1,1] 和 [1,3,3]：

print(local_minima_locations)
# (array([0, 0, 1, 1]), array([0, 4, 1, 3]), array([3, 0, 1, 3]))
print(arr[local_minima_locations])
# [-1 -1 -1 -1]

Answer 2

尝试进行 2D 操作：

import numpy as N

def local_minima(array2d):
    return ((array2d <= N.roll(array2d,  1, 0)) &
            (array2d <= N.roll(array2d, -1, 0)) &
            (array2d <= N.roll(array2d,  1, 1)) &
            (array2d <= N.roll(array2d, -1, 1)))

这将返回一个类似于 array2d 的数组，其中包含 True/False，其中局部最小值（四个邻居）所在的位置。