13

假设我在 NumPy 中有一个包含连续可微函数评估的数组，我想找到局部最小值。没有噪音，所以每个点的值都低于其所有邻居的值，都符合我的局部最小值标准。

我有以下列表理解，它适用于二维数组，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，这很慢。我也想让它适用于任意数量的维度。例如，有没有一种简单的方法可以在任意维度的数组中获取一个点的所有邻居？还是我完全以错误的方式解决了这个问题？我应该numpy.gradient()改用吗？

4

2 回答 2

19

可以使用Ivan的detect_peaks 函数为任意维度的数组找到局部最小值的位置，只需稍作修改：

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)

您可以像这样使用它：

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] 处：

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]
于 2010-10-21T11:22:30.347 回答
5

试试这个 2D：

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)))

这将为您返回一个具有 True/False 的类似 array2d 的数组，其中局部最小值（四个邻居）所在的位置。

于 2010-10-21T10:29:04.413 回答