我们可以利用
views我在一些问答中使用过的辅助功能来发挥作用。要获得子数组的存在，我们可以
np.isin在视图上使用或使用更加费力的视图
np.searchsorted。
方法1： 使用
np.isin-
# https://stackoverflow.com/a/45313353/ @Divakardef view1D(a, b): # a, b are arrays a = np.ascontiguousarray(a) b = np.ascontiguousarray(b) void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1])) return a.view(void_dt).ravel(), b.view(void_dt).ravel()def isin_nd(a,b): # a,b are the 3D input arrays to give us "isin-like" functionality across them A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1)) return np.isin(A,B)
方法2： 我们也可以利用
np.searchsorted于
views-
def isin_nd_searchsorted(a,b): # a,b are the 3D input arrays A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1)) sidx = A.argsort() sorted_index = np.searchsorted(A,B,sorter=sidx) sorted_index[sorted_index==len(A)] = len(A)-1 idx = sidx[sorted_index] return A[idx] == B
因此，这两个解决方案为我们提供了
ain中每个子数组的存在掩码
b。因此，为了获得所需的计数，它应该是-
isin_nd(a,b).sum()或
isin_nd_searchsorted(a,b).sum()。
样品运行-
In [71]: # Setup with 3 common "subarrays" ...: np.random.seed(0) ...: a = np.random.randint(0,9,(10,4,5)) ...: b = np.random.randint(0,9,(7,4,5)) ...: ...: b[1] = a[4] ...: b[3] = a[2] ...: b[6] = a[0]In [72]: isin_nd(a,b).sum()Out[72]: 3In [73]: isin_nd_searchsorted(a,b).sum()Out[73]: 3
大型阵列上的时间-
In [74]: # Setup ...: np.random.seed(0) ...: a = np.random.randint(0,9,(100,100,100)) ...: b = np.random.randint(0,9,(100,100,100)) ...: idxa = np.random.choice(range(len(a)), len(a)//2, replace=False) ...: idxb = np.random.choice(range(len(b)), len(b)//2, replace=False) ...: a[idxa] = b[idxb]# Verify outputIn [82]: np.allclose(isin_nd(a,b),isin_nd_searchsorted(a,b))Out[82]: TrueIn [75]: %timeit isin_nd(a,b).sum()10 loops, best of 3: 31.2 ms per loopIn [76]: %timeit isin_nd_searchsorted(a,b).sum()100 loops, best of 3: 1.98 ms per loop
