BettaMG BettaMG - 1 year ago 330
Python Question

Tensorflow: tensor multiplication row-by-row with more different matrices

I have a matrix A which is defined as a tensor in tensorflow, of n rows and p columns. Moreover, I have say k matrices B1,..., Bk with p rows and q columns. My goal is to obtain a resulting matrix C of n rows and q columns where each row of C is the matrix product of the corresponding row in A with one of the B matrices. Which B to choose is determined by a give index vector I of dimension n that can take values ranging from 1 to k. In my case, the B are weight variables while I is another tensor variable given as input.

An example of code in numpy would look as follows:

A = array([[1, 0, 1],
[0, 0, 1],
[1, 1, 0],
[0, 1, 0]])

B1 = array([[1, 1],
[2, 1],
[3, 6]])

B2 = array([[1, 5],
[3, 2],
[0, 2]])

B = [B1, B2]

I = [1, 0, 0, 1]

n = A.shape[0]
p = A.shape[1]
q = B1.shape[1]

C = np.zeros(shape = (n,q))

for i in xrange(n):
C[i,:] =[i,:],B[I[i]])

How can this be translated in tensor flow?

In my specific case the variables are defined as:

A = tf.placeholder("float", [None, p])
B1 = tf.Variable(tf.random_normal(p,q))
B2 = tf.Variable(tf.random_normal(p,q))
I = tf.placeholder("float",[None])

Answer Source

This is a bit tricky and there are probably better solutions. Taking your first example, my approach computes C as follows:

C = diag([0,1,1,0]) * A * B1 + diag([1,0,0,1]) * A * B2

where diag([0,1,1,0]) is the diagonal matrix having vector [0,1,1,0] in its diagonal. This can be achieved through tf.diag() in TensorFlow.

For convenience, let me assume that k<=n (otherwise some B matrices would remain unused). The following script obtains those diagonal values from vector I and computes C as mentioned above:

k = 2
n = 4
p = 3
q = 2
a = array([[1, 0, 1],
           [0, 0, 1],
           [1, 1, 0],
           [0, 1, 0]])
index_input = [1, 0, 0, 1]

import tensorflow as tf

# Creates a dim·dim tensor having the same vector 'vector' in every row
def square_matrix(vector, dim):
    return tf.reshape(tf.tile(vector,[dim]), [dim,dim])

A = tf.placeholder(tf.float32, [None, p])
B = tf.Variable(tf.random_normal(shape=[k,p,q]))
# For the first example (with k=2): B = tf.constant([[[1, 1],[2, 1],[3, 6]],[[1, 5],[3, 2],[0, 2]]], tf.float32)
C = tf.Variable(tf.zeros((n, q)))
I = tf.placeholder(tf.int32,[None])

# Create a n·n tensor 'indices_matrix' having indices_matrix[i]=I for 0<=i<n (each row vector is I)
indices_matrix = square_matrix(I, n)

# Create a n·n tensor 'row_matrix' having row_matrix[i]=[i,...,i] for 0<=i<n (each row vector is a vector of i's)
row_matrix = tf.transpose(square_matrix(tf.range(0, n, 1), n))

# Find diagonal values by comparing tensors indices_matrix and row_matrix
equal = tf.cast(tf.equal(indices_matrix, row_matrix), tf.float32)

# Compute C
for i in range(k):
    diag = tf.diag(tf.gather(equal, i))
    mul = tf.matmul(diag, tf.matmul(A, tf.gather(B, i)))
    C = C + mul

sess = tf.Session()
print(, feed_dict={A : a, I : index_input}))

As an improvement, C may be computed using a vectorized implementation instead of using a for loop.

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