我使用局部最大滤波器检测到峰值。这是第一个4个爪子的数据集的结果：
我还在9个爪子的第二个数据集上运行了它，效果也很好。
这是你的操作方式：
import numpy as npfrom scipy.ndimage.filters import maximum_filterfrom scipy.ndimage.morphology import generate_binary_structure, binary_erosionimport matplotlib.pyplot as pp#for some reason I had to reshape. Numpy ignored the shape header.paws_data = np.loadtxt("paws.txt").reshape(4,11,14)#getting a list of imagespaws = [p.squeeze() for p in np.vsplit(paws_data,4)]def detect_peaks(image): """ Takes an image and detect the peaks usingthe local maximum filter. Returns a boolean mask of the peaks (i.e. 1 when the pixel's value is the neighborhood maximum, 0 otherwise) """ # define an 8-connected neighborhood neighborhood = generate_binary_structure(2,2) #apply the local maximum filter; all pixel of maximal value #in their neighborhood are set to 1 local_max = maximum_filter(image, footprint=neighborhood)==image #local_max 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 = (image==0) #a little technicality: we must erode the background in order to #successfully subtract it form local_max, otherwise a line will #appear along the background border (artifact of the local maximum filter) eroded_background = binary_erosion(background, structure=neighborhood, border_value=1) #we obtain the final mask, containing only peaks, #by removing the background from the local_max mask (xor operation) detected_peaks = local_max ^ eroded_background return detected_peaks#applying the detection and plotting resultsfor i, paw in enumerate(paws): detected_peaks = detect_peaks(paw) pp.subplot(4,2,(2*i+1)) pp.imshow(paw) pp.subplot(4,2,(2*i+2) ) pp.imshow(detected_peaks)pp.show()你需要做的就是
scipy.ndimage.measurements.label在蒙版上使用以标记所有不同的对象。这样你就可以分别与他们一起玩了。
请注意，该方法效果很好，因为背景不嘈杂。如果是这样，你将在背景中检测到许多其他不需要的峰。另一个重要因素是邻里的大小。如果峰大小发生变化，则需要对其进行调整（应保持大致成比例）。
