is tf.matmul equivalent to the operations the Dense layer makes in tensorflow in some cases?

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I have created this model

num_items = 1250
num_users = 1453
emb_size = 64

input_userID = Input(shape=[1], name='user_ID')
input_itemID = Input(shape=[1], name='item_ID')

user_emb_GMF = Embedding(num_users, emb_size, name='user_emb_GMF')(input_userID)
item_emb_GMF = Embedding(num_items, emb_size, name='item_emb_GMF')(input_itemID)

interraction_map = tf.expand_dims(Dot(axes=1)([user_emb_GMF,item_emb_GMF]), -1)
print(interraction_map)
conv = Conv2D(32, 2, strides=2, activation='relu', padding="SAME", input_shape=interraction_map.shape[1:], name='conv1')(interraction_map)

for i in range(2,7):#les autres conv layer
    conv = Conv2D(32, 2, strides=2, activation='relu', padding="SAME",name='conv%d'%(i))(conv)

reshaped_conv = Flatten()(conv)
# c'est la que je doit agir et ajouter creer la prédiction


out = Dense(1, name='output' )(reshaped_conv)

#out = Dense(1,activation='sigmoid',name='output')(layer)


oncf_model = Model([input_userID, input_itemID], out)

tf.keras.utils.plot_model(oncf_model, show_shapes=True)

and i want the output layer to be the result of this operation:

output_layer = tf.matmul(reshaped_conv, W) + b

with W is tensor of shape(32,1)(weights) and b is tensor of shape(1) (bias).

I wonder if in this particular case the operation done with matmul is equivalent to the one the Dense layer does

out = Dense(1, name='output' )(reshaped_conv)
keras tensorflow deep-learning machine-learning python
1 Answers

yes, they are the same... you can test it on your own

X = np.random.uniform(0,1, (32,10)).astype('float32')

x = Dense(1)
pred = x(X)

W, b = x.get_weights()

(pred == (tf.matmul(X, W) + b)).numpy().all() # TRUE
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