I tried to sum up element-wise two tensors but, failed. My code is as below. I want to assign elements of 'a' as [x[0]*y[0]-x[1]*y[1], x[0]*y[1]+x[1]*y[0], x[2]*y[2]-x[3]*y[3], x[2]*y[3]+x[3]*y[2]]. Since I'm a beginner in Keras and tensorflow, I don't wanna use 'session' in tensorflow. How can I assign the above value merely using keras backend?
from keras import backend as K
def rayleigh_fading(x):
global noise_std, n_channel
y = K.random_normal((2*n_channel,), mean=0, stddev=1.0)*(1/np.sqrt(2))
a = K.placeholder(shape=(4,))
a[0] = x[0]*y[0]-x[1]*y[1]
a[1] = x[0]*y[1]+x[1]*y[0]
a[2] = x[2]*y[2]-x[3]*y[3]
a[3] = x[2]*y[3]+x[3]*y[2]
z = a + K.random_normal((2*n_channel,), mean=0, sttdev=noise_std)
return z
Try something like the following:
import tensorflow as tf
import numpy as np
keras = tf.keras
from tensorflow.keras import backend as K
noise_std = 1
n_channel = 2
def rayleigh_fading(x):
global noise_std, n_channel
y = K.random_normal((K.shape(x)[0], 2*n_channel,), mean=0, stddev=1.0)*(1/np.sqrt(2))
a0 = x[:, 0]*y[:, 0]-x[:, 1]*y[:, 1]
a1 = x[:, 0]*y[:, 1]+x[:, 1]*y[:, 0]
a2 = x[:, 2]*y[:, 2]-x[:, 3]*y[:, 3]
a3 = x[:, 2]*y[:, 3]+x[:, 3]*y[:, 2]
a_tensors = [K.expand_dims(t) for t in [a0, a1, a2, a3]]
a = K.concatenate(a_tensors, axis=1)
z = a + K.random_normal((2*n_channel,), mean=0, stddev=noise_std)
return z
def make_model():
inp = keras.layers.Input(shape=(4,))
rf = keras.layers.Lambda(rayleigh_fading)(inp)
out = keras.layers.Dense(1)(rf)
model = keras.models.Model(inp, out)
model.compile('adam', 'mse')
return model
model = make_model()
model.summary()
I assume that y should sample from a random variable on a per batch basis; z should most likely have a shape (batch_size, 2*n_channel) also... where batch_size is K.shape(x)[0].
The main issue with your attempt is that a placeholder is not a variable that can be modified. It is a placeholder for input data to the model. Also you should take into account that at the tensor level x has a batch dimension. So to access the first column of x across all batches you need to use x[:, 0].
I avoided creating a variable for a by just concatenating the vectors in your equation.