假设我在NumPy中有一个包含连续微分函数求值的数组，我想找到局部最小值。没有噪音，因此每个点的值都低于其所有邻居的值都满足我的局部最小值标准。
我有以下列表推导，适用于二维数组，忽略了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。我也想使它适用于任意数量的尺寸。例如，是否有一种简单的方法来获取任何维度数组中的点的所有邻居？还是我完全以错误的方式来解决这个问题？我应该numpy.gradient()改用吗？
numpy.gradient()
可以使用Ivan的detect_peaks函数对任意维度的数组找到局部极小值的位置，并进行少量修改：
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]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]
