我将如何在 C++ 中编写 numpy.tensordot?
How would I write numpy.tensordot in c++?
我正在尝试用 C++ 复制 numpy.tensordot。 numpy 文档中的示例显示了一个我可以开始工作的嵌套循环,但是如果不是
呢?
c = np.tensordot(a,b, axes=([1,0],[0,1]))
我想做:
c = np.tensordot(a,b, axes=([1,2],[0,1]))
python 中的新嵌套循环会是什么样子?在 C++ 中是否有 easier/faster 方法来执行此操作?现在我在 C++ 中使用与 std::vector 相同的嵌套“for”循环。我看过一些可能有用的库,但我正在尝试仅使用 c++ 标准库。
这是那个 numpy 示例,link 文档:https://numpy.org/doc/stable/reference/generated/numpy.tensordot.html
Examples
A “traditional” example:
>>>
a = np.arange(60.).reshape(3,4,5)
b = np.arange(24.).reshape(4,3,2)
c = np.tensordot(a,b, axes=([1,0],[0,1]))
c.shape
(5, 2)
c
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])
# A slower but equivalent way of computing the same...
d = np.zeros((5,2))
for i in range(5):
for j in range(2):
for k in range(3):
for n in range(4):
d[i,j] += a[k,n,i] * b[n,k,j]
c == d
array([[ True, True],
[ True, True],
[ True, True],
[ True, True],
[ True, True]])
谢谢
我发现首先重写 np.einsum
很有帮助,因为生成的 for 循环代码在概念上看起来非常相似:
a = np.random.rand(16, 8, 2)
b = np.random.rand(8, 2, 1)
c = np.tensordot(a, b, axes=([1,2],[0,1]))
# same thing written with einsum
c_ein = np.einsum("ijk,jko->io", a, b)
# same thing done with for loops,
# notice how we can use the same letters and indexing as einsum
c_manual = np.zeros((16, 1))
for i in range(16):
for o in range(1):
# j and k are summed since they don't appear in output
total = 0
for j in range(8):
for k in range(2):
total += a[i, j, k] * b[j, k, o]
c_manual[i, o] = total
assert np.allclose(c, c_ein, c_manual)
我正在尝试用 C++ 复制 numpy.tensordot。 numpy 文档中的示例显示了一个我可以开始工作的嵌套循环,但是如果不是
呢?c = np.tensordot(a,b, axes=([1,0],[0,1]))
我想做:
c = np.tensordot(a,b, axes=([1,2],[0,1]))
python 中的新嵌套循环会是什么样子?在 C++ 中是否有 easier/faster 方法来执行此操作?现在我在 C++ 中使用与 std::vector 相同的嵌套“for”循环。我看过一些可能有用的库,但我正在尝试仅使用 c++ 标准库。
这是那个 numpy 示例,link 文档:https://numpy.org/doc/stable/reference/generated/numpy.tensordot.html
Examples
A “traditional” example:
>>>
a = np.arange(60.).reshape(3,4,5)
b = np.arange(24.).reshape(4,3,2)
c = np.tensordot(a,b, axes=([1,0],[0,1]))
c.shape
(5, 2)
c
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])
# A slower but equivalent way of computing the same...
d = np.zeros((5,2))
for i in range(5):
for j in range(2):
for k in range(3):
for n in range(4):
d[i,j] += a[k,n,i] * b[n,k,j]
c == d
array([[ True, True],
[ True, True],
[ True, True],
[ True, True],
[ True, True]])
谢谢
我发现首先重写 np.einsum
很有帮助,因为生成的 for 循环代码在概念上看起来非常相似:
a = np.random.rand(16, 8, 2)
b = np.random.rand(8, 2, 1)
c = np.tensordot(a, b, axes=([1,2],[0,1]))
# same thing written with einsum
c_ein = np.einsum("ijk,jko->io", a, b)
# same thing done with for loops,
# notice how we can use the same letters and indexing as einsum
c_manual = np.zeros((16, 1))
for i in range(16):
for o in range(1):
# j and k are summed since they don't appear in output
total = 0
for j in range(8):
for k in range(2):
total += a[i, j, k] * b[j, k, o]
c_manual[i, o] = total
assert np.allclose(c, c_ein, c_manual)