在TensorFlow中计算Kronecker产品的最有效方法是什么？

作者: 林老爷的日常
发布时间: 2024-12-29 11:16:09 (5天前)
转自：

5 条回复

0#
回复此人
biu~ | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 如果您将阅读数学定义 <a href="https://www.tensorflow.org/api_docs/python/tf/nn/conv2d_transpose" rel="nofollow noreferrer"> <code> conv2d_transpose </code> </A> 看看有什么 <a href="https://en.wikipedia.org/wiki/Kronecker_product" rel="nofollow noreferrer"> Kronecker产品 </A> 计算，你会看到适当大小的stides <code> conv2d_tranpose </code> （第二个矩阵的宽度，高度），它做同样的事情。 </p> <P> 此外，您甚至可以将Kronecker的产品开箱即用 <code> conv2d_transpose </code> 。 </p> <HR /> <P> 以下是您从wiki计算Kronecker的矩阵乘积的示例。 </p> <pre> <code> import tensorflow as tf a = [[1, 2], [3, 4]] b = [[0, 5], [6, 7]] i, k, s = len(a), len(b), len(b) o = s * (i - 1) + k a_tf = tf.reshape(tf.constant(a, dtype=tf.float32), [1, i, i, 1]) b_tf = tf.reshape(tf.constant(b, dtype=tf.float32), [k, k, 1, 1]) res = tf.squeeze(tf.nn.conv2d_transpose(a_tf, b_tf, (1, o, o, 1), [1, s, s, 1], "VALID")) with tf.Session() as sess: print sess.run(res) </code> </pre> <P> 请注意，对于非方形矩阵，您需要在行中计算更多维度： </p> <pre> <code> i, k, s = len(a), len(b), len(b) o = s * (i - 1) + k </code> </pre> <P> 并正确使用它们作为你的步幅/输出参数。 </p> </DIV>

编辑
1#
回复此人
谦成 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 这是我用于此的实用程序。看到 <code> kronecker_test </code> 例如用法 </p> <pre> <code> def fix_shape(tf_shape): return tuple(int(dim) for dim in tf_shape) def concat_blocks(blocks, validate_dims=True): """Takes 2d grid of blocks representing matrices and concatenates to single matrix (aka ArrayFlatten)""" if validate_dims: col_dims = np.array([[int(b.shape[1]) for b in row] for row in blocks]) col_sums = col_dims.sum(1) assert (col_sums[0] == col_sums).all() row_dims = np.array([[int(b.shape[0]) for b in row] for row in blocks]) row_sums = row_dims.sum(0) assert (row_sums[0] == row_sums).all() block_rows = [tf.concat(row, axis=1) for row in blocks] return tf.concat(block_rows, axis=0) def chunks(l, n): """Yield successive n-sized chunks from l.""" for i in range(0, len(l), n): yield l[i:i + n] from tensorflow.python.framework import ops original_shape_func = ops.set_shapes_for_outputs def disable_shape_inference(): ops.set_shapes_for_outputs = lambda _: _ def enable_shape_inference(): ops.set_shapes_for_outputs = original_shape_func def kronecker(A, B, do_shape_inference=True): """Kronecker product of A,B. turn_off_shape_inference: if True, makes 10x10 kron go 2.4 sec -> 0.9 sec """ Arows, Acols = fix_shape(A.shape) Brows, Bcols = fix_shape(B.shape) Crows, Ccols = Arows*Brows, Acols*Bcols temp = tf.reshape(A, [-1, 1, 1])*tf.expand_dims(B, 0) Bshape = tf.constant((Brows, Bcols)) # turn off shape inference if not do_shape_inference: disable_shape_inference() # [1, n, m] => [n, m] slices = [tf.reshape(s, Bshape) for s in tf.split(temp, Crows)] # import pdb; pdb.set_trace() grid = list(chunks(slices, Acols)) assert len(grid) == Arows result = concat_blocks(grid, validate_dims=do_shape_inference) if not do_shape_inference: enable_shape_inference() result.set_shape((Arows*Brows, Acols*Bcols)) return result def kronecker_test(): A0 = [[1,2],[3,4]] B0 = [[6,7],[8,9]] A = tf.constant(A0) B = tf.constant(B0) C = kronecker(A, B) sess = tf.Session() C0 = sess.run(C) Ct = [[6, 7, 12, 14], [8, 9, 16, 18], [18, 21, 24, 28], [24, 27, 32, 36]] Cnp = np.kron(A0, B0) check_equal(C0, Ct) check_equal(C0, Cnp) </code> </pre> </DIV>

编辑
2#
回复此人
没身份别烦我 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> TensorFlow 1.7+提供了该功能 <a href="https://www.tensorflow.org/api_docs/python/tf/contrib/kfac/utils/kronecker_product" rel="nofollow noreferrer"> <code> kronecker_product </code> </A> 在 <code> tf.contrib.kfac.utils.kronecker_product </code> ： </p> <pre> <code> a = tf.eye(3) b = tf.constant([[1., 2.], [3., 4.]]) kron = tf.contrib.kfac.utils.kronecker_product(a, b) tf.Session().run(kron) </code> </pre> <P> 输出： </p> <pre> <code> array([[1., 2., 0., 0., 0., 0.], [3., 4., 0., 0., 0., 0.], [0., 0., 1., 2., 0., 0.], [0., 0., 3., 4., 0., 0.], [0., 0., 0., 0., 1., 2.], [0., 0., 0., 0., 3., 4.]], dtype=float32) </code> </pre> </DIV>

编辑
3#
回复此人
不易青年。 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 尝试以下解决方案，看看它是否适合您： </p> <pre> <code> def tf_kron(a,b): a_shape = [a.shape[0].value,a.shape[1].value] b_shape = [b.shape[0].value,b.shape[1].value] return tf.reshape(tf.reshape(a,[a_shape[0],1,a_shape[1],1])*tf.reshape(b,[1,b_shape[0],1,b_shape[1]]),[a_shape[0]*b_shape[0],a_shape[1]*b_shape[1]]) </code> </pre> </DIV>

编辑

登录后才能参与评论