Dot product of two vectors in tensorflow

问题

I was wondering if there is an easy way to calculate the dot product of two vectors (i.e. 1-d tensors) and return a scalar value in tensorflow.

Given two vectors X=(x1,...,xn) and Y=(y1,...,yn), the dot product is dot(X,Y) = x1 * y1 + ... + xn * yn

I know that it is possible to achieve this by first broadcasting the vectors X and Y to a 2-d tensor and then using tf.matmul. However, the result is a matrix, and I am after a scalar.

Is there an operator like tf.matmul that is specific to vectors?

回答1:

One of the easiest way to calculate dot product between two tensors (vector is 1D tensor) is using tf.tensordot

a = tf.placeholder(tf.float32, shape=(5))
b = tf.placeholder(tf.float32, shape=(5))

dot_a_b = tf.tensordot(a, b, 1)

with tf.Session() as sess:
    print(dot_a_b.eval(feed_dict={a: [1, 2, 3, 4, 5], b: [6, 7, 8, 9, 10]}))
# results: 130.0

回答2:

In addition to tf.reduce_sum(tf.multiply(x, y)), you can also do tf.matmul(x, tf.reshape(y, [-1, 1])).

回答3:

you can use tf.matmul and tf.transpose

tf.matmul(x,tf.transpose(y))

tf.matmul(tf.transpose(x),y)

depending on the dimensions of x and y

回答4:

import tensorflow as tf

x = tf.Variable([1, -2, 3], tf.float32, name='x')
y = tf.Variable([-1, 2, -3], tf.float32, name='y')

dot_product = tf.reduce_sum(tf.multiply(x, y))

sess = tf.InteractiveSession()
init_op = tf.global_variables_initializer()
sess.run(init_op)

dot_product.eval()

Out[46]: -14

Here, x and y are both vectors. We can do element wise product and then use tf.reduce_sum to sum the elements of the resulting vector. This solution is easy to read and does not require reshaping.

Interestingly, it does not seem like there is a built in dot product operator in the docs.

Note that you can easily check intermediate steps:

In [48]: tf.multiply(x, y).eval()
Out[48]: array([-1, -4, -9], dtype=int32)

回答5:

You can do tf.mul(x,y), followed by tf.reduce_sum()

回答6:

In newer versions (I think since 0.12), you should be able to do

tf.einsum('i,i->', x, y)

(Before that, the reduction to a scalar seemed not to be allowed/possible.)

回答7:

Maybe with the new docs you can just set the transpose option to true for either the first argument of the dot product or the second argument:

tf.matmul(a, b, transpose_a=False, transpose_b=False, adjoint_a=False, adjoint_b=False, a_is_sparse=False, b_is_sparse=False, name=None)

leading:

tf.matmul(a, b, transpose_a=True, transpose_b=False)
tf.matmul(a, b, transpose_a=False, transpose_b=True)

回答8:

Just use * and reduce_sum

ab = tf.reduce_sum(a*b)

Take a simple example as follows:

import tensorflow as tf
a = tf.constant([1,2,3])
b = tf.constant([2,3,4])

print(a.get_shape())
print(b.get_shape())

c = a*b
ab = tf.reduce_sum(c)

with tf.Session() as sess:
    print(c.eval())
    print(ab.eval())

# output
# (3,)
# (3,)
# [2 6 12]
# 20

回答9:

Let us assume that you have two column vectors

u = tf.constant([[2.], [3.]])
v = tf.constant([[5.], [7.]])

If you want a 1x1 matrix you can use

tf.einsum('ij,ik->jk',x,y)

If you are interested in a scalar you can use

tf.einsum('ij,ik->',x,y)

来源：https://stackoverflow.com/questions/40670370/dot-product-of-two-vectors-in-tensorflow

标签

python

tensorflow

dot-product