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finding the inner product of two 3D tensor in keras custom layer
What is the best way to reshape tensor in kerasUsing the K.function() within a custom Keras layerKeras custom layer with no different output_shapeCustom Lambda layer for Kronecker product in Keras - troubles with the dimension reserved for batch_sizeIndexing a Keras TensorHow to create a custom convolution operation with a trainable Kernel in Keras?Multiply multiple tensors pairwise kerasHow to apply convolution on the last three dimensions of a 5D tensor using the Conv2D in Keras?The initial state or constants of an RNN layer cannot be specified with a mix of Keras tensors and non-Keras tensorsUnable to stack keras 1D tensors to form 2D keras tensor
in my keras custom layer, i have to find the inner product of two 3D tensors. For example , x=(?,80,150,12) is the input to the layer and the inner product have to be with all the 4 slices of kernel (80,150,12,4), 4 being the slicing dimension. I am trying to do it with K.dot but i am getting dimension mismatch error.The code snippet is-
kernel1_unpacked = tf.unstack(self.kernel1,axis=3)
all_probs = []
for t in kernel1_unpacked:
t=K.permute_dimensions(t,[2,0,1])
B3=K.sum(K.dot(x,t),[1,2,3])
B3=K.expand_dims(B3,1)
print(B3)
all_probs.append(B3)
B4=tf.concat(all_probs[:],axis=1)
print(B4)
the output should have dimension (?,4). please help
python tensorflow keras inner-product
add a comment |
in my keras custom layer, i have to find the inner product of two 3D tensors. For example , x=(?,80,150,12) is the input to the layer and the inner product have to be with all the 4 slices of kernel (80,150,12,4), 4 being the slicing dimension. I am trying to do it with K.dot but i am getting dimension mismatch error.The code snippet is-
kernel1_unpacked = tf.unstack(self.kernel1,axis=3)
all_probs = []
for t in kernel1_unpacked:
t=K.permute_dimensions(t,[2,0,1])
B3=K.sum(K.dot(x,t),[1,2,3])
B3=K.expand_dims(B3,1)
print(B3)
all_probs.append(B3)
B4=tf.concat(all_probs[:],axis=1)
print(B4)
the output should have dimension (?,4). please help
python tensorflow keras inner-product
@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43
add a comment |
in my keras custom layer, i have to find the inner product of two 3D tensors. For example , x=(?,80,150,12) is the input to the layer and the inner product have to be with all the 4 slices of kernel (80,150,12,4), 4 being the slicing dimension. I am trying to do it with K.dot but i am getting dimension mismatch error.The code snippet is-
kernel1_unpacked = tf.unstack(self.kernel1,axis=3)
all_probs = []
for t in kernel1_unpacked:
t=K.permute_dimensions(t,[2,0,1])
B3=K.sum(K.dot(x,t),[1,2,3])
B3=K.expand_dims(B3,1)
print(B3)
all_probs.append(B3)
B4=tf.concat(all_probs[:],axis=1)
print(B4)
the output should have dimension (?,4). please help
python tensorflow keras inner-product
in my keras custom layer, i have to find the inner product of two 3D tensors. For example , x=(?,80,150,12) is the input to the layer and the inner product have to be with all the 4 slices of kernel (80,150,12,4), 4 being the slicing dimension. I am trying to do it with K.dot but i am getting dimension mismatch error.The code snippet is-
kernel1_unpacked = tf.unstack(self.kernel1,axis=3)
all_probs = []
for t in kernel1_unpacked:
t=K.permute_dimensions(t,[2,0,1])
B3=K.sum(K.dot(x,t),[1,2,3])
B3=K.expand_dims(B3,1)
print(B3)
all_probs.append(B3)
B4=tf.concat(all_probs[:],axis=1)
print(B4)
the output should have dimension (?,4). please help
python tensorflow keras inner-product
python tensorflow keras inner-product
asked Mar 7 at 6:15
Sandeep PandeySandeep Pandey
185
185
@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43
add a comment |
@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43
@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43
@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43
add a comment |
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@giser_yugang Sir, any idea how to get around this ?
– Sandeep Pandey
Mar 8 at 5:43