在TensorFlow中計算Kronecker產品的最有效方法是什麼？

Question

我有興趣使用TensorFlow中的Kronecker循環單元執行this paper。

這涉及克羅內克產品的計算。 TensorFlow沒有Kronecker產品的操作。我正在尋找一種有效且可靠的方法來計算這一點。

這是否存在，還是我需要手動定義TensorFlow操作？

Answer 1

這是我用於此的實用程序。見kronecker_test例如使用

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) 

Answer 2

的如果你願意讀的conv2d_transpose數學定義，看看Kronecker product計算，你會看到，與stides適當大小conv2d_tranpose（寬度，第二矩陣的高度），它做同樣的事情。

此外，您甚至可以批量訂購Kronecker的產品conv2d_transpose。

---

下面是一個計算Kronecker產品的維基矩陣產品的例子。

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) 

注意，在非方陣的情況下，你需要在線路calulcate多個維度：

i, k, s = len(a), len(b), len(b) 
o = s * (i - 1) + k 

，並正確地使用它們作爲你的進步/輸出參數。

Answer 3

請嘗試以下解決方案，看看它是否適合你：

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]])