检查两个3D numpy数组是否包含重叠的2D数组 [英] Check if two 3D numpy arrays contain overlapping 2D arrays

问题描述

我有两个非常大的numpy数组，它们都是3D的.我需要找到一种有效的方法来检查它们是否重叠，因为首先将它们都转换为集合会花费很长时间.我尝试使用在此找到的另一个解决方案来解决相同的问题，但适用于2D阵列，但是我没有设法使其适用于3D. 这是2D解决方案:

I have two very large numpy arrays, which are both 3D. I need to find an efficient way to check if they are overlapping, because turning them both into sets first takes too long. I tried to use another solution I found here for this same problem but for 2D arrays, but I didn't manage to make it work for 3D. Here is the solution for 2D:

nrows, ncols = A.shape
dtype={'names':['f{}'.format(i) for i in range(ndep)],
       'formats':ndep * [A.dtype]}
C = np.intersect1d(A.view(dtype).view(dtype), B.view(dtype).view(dtype))
# This last bit is optional if you're okay with "C" being a structured array...
C = C.view(A.dtype).reshape(-1, ndep)

(其中A和B是2D数组) 我需要找到重叠的numpy数组的数量，而不是特定的数组.

(where A and B are the 2D arrays) I need to find the number of overlapping numpy arrays, but not the specific ones.

推荐答案

我们可以使用我在几个Q& A中使用的辅助函数来利用views.为了获得子数组的存在，我们可以在视图上使用np.isin，或者在np.searchsorted上使用更费力的视图.

We could leverage views using a helper function that I have used across few Q&As. To get the presence of subarrays, we could use np.isin on the views or use a more laborious one with np.searchsorted.

方法1:使用np.isin-

# https://stackoverflow.com/a/45313353/ @Divakar
def 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::我们还可以在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b中a中每个子阵列的存在的掩码.因此，要获得所需的计数，应为-isin_nd(a,b).sum()或isin_nd_searchsorted(a,b).sum().

So, these two solutions give us the mask of presence of each of the subarrays from a in b. Hence, to get our desired count, it would be - isin_nd(a,b).sum() or 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]: 3

In [73]: isin_nd_searchsorted(a,b).sum()
Out[73]: 3

大型数组上的计时-

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 output
In [82]: np.allclose(isin_nd(a,b),isin_nd_searchsorted(a,b))
Out[82]: True

In [75]: %timeit isin_nd(a,b).sum()
10 loops, best of 3: 31.2 ms per loop

In [76]: %timeit isin_nd_searchsorted(a,b).sum()
100 loops, best of 3: 1.98 ms per loop

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