英文:
how omit axis on outer substaract numpy
问题
我正在计算欧几里德（或其他）距离的表格。这是一个空间距离的查找表。对于二维点，查找表格如下 table[x1, y1, x2, y2]。例如 table[0, 0, 0, 10] 应该等于 10，table[0, 0, 10, 10] 应该是 14.142135623730951。
我有一个可工作的示例，我想重写它，使其适用于具有任意维度的点。这段代码仅适用于二维情况。
import numpy as npto = np.indices((5, 5)).Tx_shape, y_shape, n_dim = to.shapex_sub = np.subtract.outer(to[..., 0], to[..., 0])y_sub = np.subtract.outer(to[..., 1], to[..., 1])distances = np.sqrt(x_sub ** 2 + y_sub ** 2)n_test = 5for i in np.random.choice(x_shape, n_test):for j in np.random.choice(y_shape, n_test):for k in np.random.choice(x_shape, n_test):for l in np.random.choice(y_shape, n_test):d = np.sqrt((i - k) ** 2 + (j - l) ** 2)assert (distances[i, j, k, l] == d)print('3[92m TEST PASSES')
现在，我可以简单地运行一个循环，并在每个维度上执行 np.subtract.outer，但我想知道是否有一种方法可以直接在索引上执行外部减法（就像 np.subtract.outer(to, to)）。当我尝试这样做时，结果的形状是 (5, 5, 2, 5, 5, 2)，而我需要的是 (5, 5, 5, 5, 2)。有人知道如何做到这一点吗？
从被接受的答案派生的解决方案：
import numpy as npto = np.indices((5, 5, 5)).Tsh = list(to.shape)n_dims = sh[-1]t = to.reshape(sh[:-1] + [1] * n_dims + [n_dims]) - to.reshape([1] * n_dims + sh[:-1] + [n_dims])distances = np.sqrt(np.sum(np.power(t, 2), axis=-1))n_test = 100idxs = np.indices(distances.shape).T.reshape(-1, n_dims * 2)for idx in np.random.permutation(idxs)[:n_test]:assert distances[tuple(idx)] == np.linalg.norm(idx[:n_dims] - idx[n_dims:])print('3[92m TEST PASSES')
英文:
I am computing a table of euclidean (or other) distances. This is a lookup for distances in a space. For 2-d points the lookup is done thusly table[x1,y1,x2,y2]. For instance table[0,0,0,10] should equal 10 and table[0,0,10,10] would be 14.142135623730951.
I have a working example that I would like to rewrite so that it works on points with an arbitrary number of dimensions. This code works only for two.
import numpy as npto = np.indices((5, 5)).Tx_shape, y_shape, n_dim = to.shapex_sub = np.subtract.outer(to[..., 0], to[..., 0])y_sub = np.subtract.outer(to[..., 1], to[..., 1])distances = np.sqrt(x_sub ** 2 + y_sub ** 2)n_test = 5for i in np.random.choice(x_shape, n_test):for j in np.random.choice(y_shape, n_test):for k in np.random.choice(x_shape, n_test):for l in np.random.choice(y_shape, n_test):d = np.sqrt((i - k) ** 2 + (j - l) ** 2)assert (distances[i, j, k, l] == d)print('3[92m TEST PASSES')
Now, I could simply run a loop and do np.subtract.outer on each dimm, but I would like to know if there is a way to do outer subtract on the indices directly. (As in np.subtract.outer(to, to))
When I try this the result has a shape of (5, 5, 2, 5, 5, 2) while what i need is (5, 5, 5, 5, 2). Anyone know how to do this?
Solution derived from accepted answer:
import numpy as npto = np.indices((5, 5, 5)).Tsh = list(to.shape)n_dims = sh[-1]t = to.reshape(sh[:-1] + [1] * n_dims + [n_dims]) - to.reshape([1] * n_dims + sh[:-1] + [n_dims])distances = np.sqrt(np.sum(np.power(t, 2), axis=-1))n_test = 100idxs = np.indices(distances.shape).T.reshape(-1, n_dims * 2)for idx in np.random.permutation(idxs)[:n_test]:assert distances[tuple(idx)] == np.linalg.norm(idx[:n_dims] - idx[n_dims:])print('3[92m TEST PASSES')
答案1
得分: 0
I'll illustrate with a (3,4) array:
在一个 (3,4) 的数组中进行演示：
In [9]: x = np.arange(12).reshape(3,4)
在 [9]: x = np.arange(12).reshape(3,4)
In [10]: np.subtract.outer(x,x).shape
在 [10]: np.subtract.outer(x,x).shape
Out[10]: (3, 4, 3, 4)
Out[10]: (3, 4, 3, 4)
But with broadcasting, we can do an 'outer' on first dimensions, while "sharing" the second. That can be generalized.
但是通过 broadcasting，我们可以在第一维上执行 'outer'，同时"共享"第二维。这可以泛化。
In [11]: (x[:,None,:]-x[None,:,:]).shape
在 [11]: (x[:,None,:]-x[None,:,:]).shape
Out[11]: (3, 3, 4)
Out[11]: (3, 3, 4)
'outer' can be done with 'broadcasting'
'outer' 可以使用 'broadcasting' 完成
In [12]: np.allclose(np.subtract.outer(x,x), x[:,:,None,None]-x[None,None,:,:])
在 [12]: np.allclose(np.subtract.outer(x,x), x[:,:,None,None]-x[None,None,:,:])
Out[12]: True
Out[12]: True
英文:
I'll illustrate with a (3,4) array:
In [9]: x = np.arange(12).reshape(3,4)In [10]: np.subtract.outer(x,x).shapeOut[10]: (3, 4, 3, 4)
But with broadcasting, we can do an 'outer' on first dimensions, while "sharing" the second. That can be generalized.
In [11]: (x[:,None,:]-x[None,:,:]).shapeOut[11]: (3, 3, 4)
outer can be done with broadcasting
In [12]: np.allclose(np.subtract.outer(x,x), x[:,:,None,None]-x[None,None,:,:])Out[12]: True
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