解决方案

数据文件:paw.txt。源代码:

from scipy import *
from operator import itemgetter

n = 5  # how many fingers are we looking for

d = loadtxt("paw.txt")
width, height = d.shape

# Create an array where every element is a sum of 2x2 squares.

fourSums = d[:-1,:-1] + d[1:,:-1] + d[1:,1:] + d[:-1,1:]

# Find positions of the fingers.

# Pair each sum with its position number (from 0 to width*height-1),

pairs = zip(arange(width*height), fourSums.flatten())

# Sort by descending sum value, filter overlapping squares

def drop_overlapping(pairs):
    no_overlaps = []
    def does_not_overlap(p1, p2):
        i1, i2 = p1[0], p2[0]
        r1, col1 = i1 / (width-1), i1 % (width-1)
        r2, col2 = i2 / (width-1), i2 % (width-1)
        return (max(abs(r1-r2),abs(col1-col2)) >= 2)
    for p in pairs:
        if all(map(lambda prev: does_not_overlap(p,prev), no_overlaps)):
            no_overlaps.append(p)
    return no_overlaps

pairs2 = drop_overlapping(sorted(pairs, key=itemgetter(1), reverse=True))

# Take the first n with the heighest values

positions = pairs2[:n]

# Print results

print d, "\n"

for i, val in positions:
    row = i / (width-1)
    column = i % (width-1)
    print "sum = %f @ %d,%d (%d)" % (val, row, column, i)
    print d[row:row+2,column:column+2], "\n"

输出不重叠方块。似乎选择的区域与您的示例中相同。

一些评论

棘手的部分是计算所有2x2平方的和。我猜你们都需要，所以可能会有重叠。我使用切片从原始2D数组中切割第一个/最后一个列和行，然后将它们全部重叠在一起并计算总和。

为了更好地理解它，对一个3x3数组进行成像:

>>> a = arange(9).reshape(3,3) ; a
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

然后你可以把它切片:

>>> a[:-1,:-1]
array([[0, 1],
       [3, 4]])
>>> a[1:,:-1]
array([[3, 4],
       [6, 7]])
>>> a[:-1,1:]
array([[1, 2],
       [4, 5]])
>>> a[1:,1:]
array([[4, 5],
       [7, 8]])

现在想象一下，你把它们一个叠在另一个上面，然后在相同的位置上求和。这些和将与左上角位置相同的2x2平方的和完全相同:

>>> sums = a[:-1,:-1] + a[1:,:-1] + a[:-1,1:] + a[1:,1:]; sums
array([[ 8, 12],
       [20, 24]])

当你有2x2平方的和时，你可以用max来求最大值，或者排序，或者排序来求峰值。

为了记住峰值的位置，我将每个值(总和)与其在扁平数组中的序数位置(参见zip)耦合起来。然后在打印结果时再次计算行/列位置。

笔记

我允许2x2平方重叠。编辑后的版本过滤掉其中的一些，这样只有不重叠的正方形出现在结果中。

选择手指(一个想法)

另一个问题是如何从所有的峰中选择可能是手指的峰。我有个主意，可能有用也可能没用。我现在没有时间实现它，所以只是伪代码。

我注意到，如果前面的手指几乎保持在一个完美的圆圈上，后面的手指应该在这个圆圈里。此外，食指的间距也差不多相等。我们可以尝试使用这些启发式属性来检测手指。

伪代码:

select the top N finger candidates (not too many, 10 or 12)
consider all possible combinations of 5 out of N (use itertools.combinations)
for each combination of 5 fingers:
    for each finger out of 5:
        fit the best circle to the remaining 4
        => position of the center, radius
        check if the selected finger is inside of the circle
        check if the remaining four are evenly spread
        (for example, consider angles from the center of the circle)
        assign some cost (penalty) to this selection of 4 peaks + a rear finger
        (consider, probably weighted:
             circle fitting error,
             if the rear finger is inside,
             variance in the spreading of the front fingers,
             total intensity of 5 peaks)
choose a combination of 4 peaks + a rear peak with the lowest penalty

这是一种蛮力方法。如果N相对较小，那么我认为它是可行的。对于N=12，有C_12^5 = 792种组合，乘以5种选择后指的方法，所以每个爪子要评估3960种情况。

我用局部最大滤波器检测了峰值。下面是你的第一个4个爪子数据集的结果:

我还在9个爪子的第二个数据集上运行了它，效果也很好。

你可以这样做:

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()

之后你所需要做的就是在蒙版上使用scipy. nmage .measurements.label来标记所有不同的对象。然后你就可以单独和他们玩了。

注意，该方法工作得很好，因为背景没有噪声。如果是的话，你会在背景中检测到一堆其他不想要的峰。另一个重要因素是社区的大小。如果峰值大小发生变化，您将需要调整它(应该保持大致成比例)。

这里有一个想法:你计算图像的(离散)拉普拉斯量。我希望它在最大值处是(负的)大的，以一种比原始图像更引人注目的方式。因此，极大值更容易找到。

这里有另一个想法:如果你知道高压点的典型大小，你可以首先通过用相同大小的高斯函数卷积来平滑你的图像。这可能会让你更容易处理图像。

也许一个简单的方法在这里就足够了:建立一个平面上所有2x2正方形的列表，按它们的和排序(降序)。

首先，在你的“爪子列表”中选择价值最高的方块。然后，迭代地选择4个次优正方形，这些正方形不与之前找到的任何正方形相交。

我脑子里有几个想法:

取扫描的梯度(导数)，看看是否消除了错误的调用 取局部极大值的最大值

你可能还想看看OpenCV，它有一个相当不错的Python API，可能有一些你会发现有用的函数。

谢谢你的原始数据。我在火车上，这是我到的最远的地方(我的站就要到了)。我用regexps按摩了你的txt文件，并将其放入一个html页面，使用一些javascript进行可视化。我在这里分享它是因为一些人，比如我自己，可能会发现它比python更容易被破解。

我认为一个很好的方法是尺度和旋转不变，我的下一步将是研究高斯的混合。(每个爪垫是高斯分布的中心)。

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