在NumPy中计算距离矩阵的有效方法

Question

I have this sample array:

In [38]: arr
Out[38]: array([  0,  44, 121, 154, 191])

The above is just a sample whereas my actual array size is pretty huge. So, what is an efficient way to compute a distance matrix?

The result should be:

In [41]: res
Out[41]: 
array([[   0,   44,  121,  154,  191],
       [ -44,    0,   77,  110,  147],
       [-121,  -77,    0,   33,   70],
       [-154, -110,  -33,    0,   37],
       [-191, -147,  -70,  -37,    0]])

I wrote a for loop based implementation which is too slow. Could this be vectorized for efficiency reasons?

Answer 1

You can use broadcasting :

from numpy import array

arr = array([  0,  44, 121, 154, 191])
arrM = arr.reshape(1, len(arr))
res = arrM - arrM.T

Answer 2

There's subtract . outer , which effectively performs broadcasted subtraction between two arrays.

Apply the ufunc op to all pairs (a, b) with a in A and b in B.

Let M = A.ndim, N = B.ndim. Then the result, C, of op.outer(A, B) is an array of dimension M + N such that:

 C[i_0, ..., i_{M-1}, j_0, ..., j_{N-1}] = op(A[i_0, ..., i_{M-1}],B[j_0, ..., j_{N-1}]) 

np.subtract.outer(arr, arr).T

Or,

arr - arr[:, None] # essentially the same thing as above

array([[   0,   44,  121,  154,  191],
       [ -44,    0,   77,  110,  147],
       [-121,  -77,    0,   33,   70],
       [-154, -110,  -33,    0,   37],
       [-191, -147,  -70,  -37,    0]])