2

Consider two numpy arrays having the same shape, A and B, composed of 1s and 0s. A small example is shown:

A = [[1 0 0 1]         B = [[0 0 0 0]
     [0 0 1 0]              [0 0 0 0]
     [0 0 0 0]              [1 1 0 0]
     [0 0 0 0]              [0 0 1 0]
     [0 0 1 1]]             [0 1 0 1]]

I now want to assign values to the two Boolean variables test1 and test2 as follows:

test1: Is there at least one instance where a 1 in an A column and a 1 in the SAME B column have row differences of exactly 1 or 2? If so, then test1 = True, otherwise False.

In the example above, column 0 of both arrays have 1s that are 2 rows apart, so test1 = True. (there are other instances in column 2 as well, but that doesn't matter - we only require one instance.)

test2: Do the 1 values in A and B all have different array addresses? If so, then test2 = True, otherwise False.

In the example above, both arrays have [4,3] = 1, so test2 = False.

I'm struggling to find an efficient way to do this and would appreciate some assistance.

4

3 回答 3

1

您可以使用 masked_arrays 并且对于第二个任务，您可以执行以下操作：

A_m = np.ma.masked_equal(A, 0)
B_m = np.ma.masked_equal(B, 0)

test2 = np.any((A_m==B_m).compressed())

完成第一项任务的一种天真的方式是：

test1 = np.any((np.vstack((A_m[:-1],A_m[:-2],A_m[1:],A_m[2:]))==np.vstack((B_m[1:],B_m[2:],B_m[:-1],B_m[:-2]))).compressed())

输出：

True
True
于 2020-08-18T01:14:09.710 回答
1

Here is a simple way to test if two arrays have an entry one element apart in the same column (only in one direction):

(A[1:, :] * B[:-1, :]).any(axis=None)

So you can do

test1 = (A[1:, :] * B[:-1, :] + A[:-1, :] * B[1:, :]).any(axis=None) or (A[2:, :] * B[:-2, :] + A[:-2, :] * B[2:, :]).any(axis=None)

The second test can be done by converting the locations to indices, stacking them together, and using np.unique to count the number of duplicates. Duplicates can only come from the same index in two arrays since an array will never have duplicate indices. We can further speed up the calculation by using flatnonzero instead of nonzero:

test2 = np.all(np.unique(np.concatenate((np.flatnonzero(A), np.flatnonzero(B))), return_counts=True)[1] == 1)

A more efficient test would use np.intersect1d in a similar manner:

test2 = not np.intersect1d(np.flatnonzero(A), np.flatnonzero(B)).size
于 2020-08-18T02:53:53.323 回答
0

对于 Test2：您可以检查他们是否找到任何类似的索引值 1。

A =  np.array([[1, 0, 0, 1],[0, 0, 1, 0],[0, 0, 0, 0],[0, 0, 0, 0],[0, 0, 1, 1]])
B = np.array([[0, 0, 0, 0],[0, 0, 0, 0],[1, 1, 0, 0],[0, 0, 1, 0],[0, 1, 0, 1]])
print(len(np.intersect1d(np.flatnonzero(A==1),np.flatnonzero(B==1)))>0))
于 2020-08-18T02:56:23.553 回答