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 np
from scipy.ndimage.filters import maximum_filter
from scipy.ndimage.morphology import generate_binary_structure, binary_erosion
import 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 images
paws = [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 results
for 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）。
602392714
清零编程群