使用scipy，您可以将这样的点描述为既是其邻域的最大值又是最小值的点：
import numpy as npimport scipy.ndimage.filters as filtersdef using_filters(data): return np.where(np.logical_and.reduce( [data == f(data, footprint=np.ones((3,3)), mode='constant', cval=np.inf) for f in (filters.maximum_filter, filters.minimum_filter)]))using_filters(data)# (array([2, 3]), array([5, 9]))
仅使用numpy，您可以将
data其自身的8个移位切片进行比较以找到相等的点：
def using_eight_shifts(data): h, w = data.shape data2 = np.empty((h+2, w+2)) data2[(0,-1),:] = np.nan data2[:,(0,-1)] = np.nan data2[1:1+h,1:1+w] = data result = np.where(np.logical_and.reduce([ (data2[i:i+h,j:j+w] == data) for i in range(3) for j in range(3) if not (i==1 and j==1)])) return result
如您在上面所看到的，此策略创建了一个扩展数组，该数组在数据周围具有NaN边界。这允许将移位的切片表示为
data2[i:i+h,j:j+w]。
如果您知道要与邻居进行比较，则可能应该
data从一开始就用NaN的边界进行定义，这样就不必像上面那样做第二个数组了。
使用八移位（和比较）比循环遍历每个单元格
data并将其与相邻单元格进行比较要快得多：
def using_quadratic_loop(data): return np.array([[i,j] for i in range(1,np.shape(data)[0]-1) for j in range(1,np.shape(data)[1]-1) if np.all(data[i-1:i+2,j-1:j+2]==data[i,j])]).T
这是一个基准：
using_filters : 0.130using_eight_shifts : 0.340using_quadratic_loop : 18.794
这是用于产生基准的代码：
import timeitimport operatorimport numpy as npimport scipy.ndimage.filters as filtersimport matplotlib.pyplot as pltdata = np.array([ [0,1,2,3,4,7,6,7,8,9,10], [3,3,3,4,7,7,7,8,11,12,11], [3,3,3,5,7,7,7,9,11,11,11], [3,4,3,6,7,7,7,10,11,11,11], [4,5,6,7,7,9,10,11,11,11,11] ])data = np.tile(data, (50,50))def using_filters(data): return np.where(np.logical_and.reduce( [data == f(data, footprint=np.ones((3,3)), mode='constant', cval=np.inf) for f in (filters.maximum_filter, filters.minimum_filter)]))def using_eight_shifts(data): h, w = data.shape data2 = np.empty((h+2, w+2)) data2[(0,-1),:] = np.nan data2[:,(0,-1)] = np.nan data2[1:1+h,1:1+w] = data result = np.where(np.logical_and.reduce([ (data2[i:i+h,j:j+w] == data) for i in range(3) for j in range(3) if not (i==1 and j==1)])) return resultdef using_quadratic_loop(data): return np.array([[i,j] for i in range(1,np.shape(data)[0]-1) for j in range(1,np.shape(data)[1]-1) if np.all(data[i-1:i+2,j-1:j+2]==data[i,j])]).Tnp.testing.assert_equal(using_quadratic_loop(data), using_filters(data))np.testing.assert_equal(using_eight_shifts(data), using_filters(data))timing = dict()for f in ('using_filters', 'using_eight_shifts', 'using_quadratic_loop'): timing[f] = timeit.timeit('{f}(data)'.format(f=f), 'from __main__ import data, {f}'.format(f=f), number=10)for f, t in sorted(timing.items(), key=operator.itemgetter(1)): print('{f:25}: {t:.3f}'.format(f=f, t=t))
