Building a keras model to apply a Dense network to every column in a 3-D array and return a 2-D array
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I have a large number of nxmxm arrays. I would like to train a keras model that learns a single Dense NN to apply to all of the nx1 column vectors. As a concrete example, suppose A is a 6x10x10 tensor; it therefore has 100 6x1 column vectors.
I have a keras model to train a Dense neural network:
import keras as K
import keras.layers as L
def column_nn():
layers=[12,36,12,1]
columns=L.Input(shape=(6,))
x=L.Dense(layers[0],activation='relu')(columns)
for l in layers[1:]:
x=L.Dense(l,activation='relu')(x)
return K.models.Model(inputs=columns, outputs=x)
I'd like to apply this to each of the 100 column vectors, but I want to return a 2-D 10x10 tensor that I can do other things to, such as pass through Conv2D layers.
One way that comes to mind is to use reshaping and a keras shared layer.
data=L.Input(shape=(6,10,10))
column_nn=column()
x=L.Permute((2,3,1))(data)
x=L.Reshape((-1,6))(x)
new_layer=column_nn()(x)
x=L.Reshape((10,10))(new_layer)
# now do a bunch of stuff to the 2-D new_layer, such as
x=L.Conv2d(filters=5,kernel_size=[3,3])(x)
x=L.MaxPooling2D(pool_size=(2,2),strides=(2,2))(x)
x=L.Flatten(x)
output=L.Dense(x)
return K.models.Model(inputs=A,outputs=output)
Seem ok? I'd love to know if if there were a slicker way to pull this off?
python tensorflow keras
asked Jan 4 at 14:35
AstroBenAstroBen
14619
add a comment |
I have a large number of nxmxm arrays. I would like to train a keras model that learns a single Dense NN to apply to all of the nx1 column vectors. As a concrete example, suppose A is a 6x10x10 tensor; it therefore has 100 6x1 column vectors.
I have a keras model to train a Dense neural network:
import keras as K
import keras.layers as L
def column_nn():
layers=[12,36,12,1]
columns=L.Input(shape=(6,))
x=L.Dense(layers[0],activation='relu')(columns)
for l in layers[1:]:
x=L.Dense(l,activation='relu')(x)
return K.models.Model(inputs=columns, outputs=x)
I'd like to apply this to each of the 100 column vectors, but I want to return a 2-D 10x10 tensor that I can do other things to, such as pass through Conv2D layers.
One way that comes to mind is to use reshaping and a keras shared layer.
data=L.Input(shape=(6,10,10))
column_nn=column()
x=L.Permute((2,3,1))(data)
x=L.Reshape((-1,6))(x)
new_layer=column_nn()(x)
x=L.Reshape((10,10))(new_layer)
# now do a bunch of stuff to the 2-D new_layer, such as
x=L.Conv2d(filters=5,kernel_size=[3,3])(x)
x=L.MaxPooling2D(pool_size=(2,2),strides=(2,2))(x)
x=L.Flatten(x)
output=L.Dense(x)
return K.models.Model(inputs=A,outputs=output)
Seem ok? I'd love to know if if there were a slicker way to pull this off?
python tensorflow keras
asked Jan 4 at 14:35
AstroBenAstroBen
14619
add a comment |
I have a large number of nxmxm arrays. I would like to train a keras model that learns a single Dense NN to apply to all of the nx1 column vectors. As a concrete example, suppose A is a 6x10x10 tensor; it therefore has 100 6x1 column vectors.
I have a keras model to train a Dense neural network:
import keras as K
import keras.layers as L
def column_nn():
layers=[12,36,12,1]
columns=L.Input(shape=(6,))
x=L.Dense(layers[0],activation='relu')(columns)
for l in layers[1:]:
x=L.Dense(l,activation='relu')(x)
return K.models.Model(inputs=columns, outputs=x)
I'd like to apply this to each of the 100 column vectors, but I want to return a 2-D 10x10 tensor that I can do other things to, such as pass through Conv2D layers.
One way that comes to mind is to use reshaping and a keras shared layer.
data=L.Input(shape=(6,10,10))
column_nn=column()
x=L.Permute((2,3,1))(data)
x=L.Reshape((-1,6))(x)
new_layer=column_nn()(x)
x=L.Reshape((10,10))(new_layer)
# now do a bunch of stuff to the 2-D new_layer, such as
x=L.Conv2d(filters=5,kernel_size=[3,3])(x)
x=L.MaxPooling2D(pool_size=(2,2),strides=(2,2))(x)
x=L.Flatten(x)
output=L.Dense(x)
return K.models.Model(inputs=A,outputs=output)
Seem ok? I'd love to know if if there were a slicker way to pull this off?
python tensorflow keras
asked Jan 4 at 14:35
AstroBenAstroBen
14619
I have a large number of nxmxm arrays. I would like to train a keras model that learns a single Dense NN to apply to all of the nx1 column vectors. As a concrete example, suppose A is a 6x10x10 tensor; it therefore has 100 6x1 column vectors.
I have a keras model to train a Dense neural network:
import keras as K
import keras.layers as L
def column_nn():
layers=[12,36,12,1]
columns=L.Input(shape=(6,))
x=L.Dense(layers[0],activation='relu')(columns)
for l in layers[1:]:
x=L.Dense(l,activation='relu')(x)
return K.models.Model(inputs=columns, outputs=x)
I'd like to apply this to each of the 100 column vectors, but I want to return a 2-D 10x10 tensor that I can do other things to, such as pass through Conv2D layers.
One way that comes to mind is to use reshaping and a keras shared layer.
data=L.Input(shape=(6,10,10))
column_nn=column()
x=L.Permute((2,3,1))(data)
x=L.Reshape((-1,6))(x)
new_layer=column_nn()(x)
x=L.Reshape((10,10))(new_layer)
# now do a bunch of stuff to the 2-D new_layer, such as
x=L.Conv2d(filters=5,kernel_size=[3,3])(x)
x=L.MaxPooling2D(pool_size=(2,2),strides=(2,2))(x)
x=L.Flatten(x)
output=L.Dense(x)
return K.models.Model(inputs=A,outputs=output)
Seem ok? I'd love to know if if there were a slicker way to pull this off?
python tensorflow keras
python tensorflow keras
asked Jan 4 at 14:35
AstroBenAstroBen
14619
asked Jan 4 at 14:35
AstroBenAstroBen
14619
edited Jan 4 at 15:31
AstroBen
asked Jan 4 at 14:35
AstroBenAstroBen
14619
asked Jan 4 at 14:35
AstroBenAstroBen
14619
asked Jan 4 at 14:35
AstroBenAstroBen
14619
14619
add a comment |
add a comment |
1 Answer
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votes
If you reshape and transpose your input data to (m*m, n), you can use Dense(k) in conjunction with TimeDistributed to apply the same weights to the m^2 vectors separately. The output shape would be (m*m, k), after which you can reshape again to suit your needs
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on usingTimeDistributed. I will definitely look into it.
– AstroBen
Jan 4 at 15:26
This answer is ok, butTimeDistributed(Dense(...))is equal to justDense(...). -- The layer will be applied to the last dimension.
– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you reshape and transpose your input data to (m*m, n), you can use Dense(k) in conjunction with TimeDistributed to apply the same weights to the m^2 vectors separately. The output shape would be (m*m, k), after which you can reshape again to suit your needs
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on usingTimeDistributed. I will definitely look into it.
– AstroBen
Jan 4 at 15:26
This answer is ok, butTimeDistributed(Dense(...))is equal to justDense(...). -- The layer will be applied to the last dimension.
– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
add a comment |
If you reshape and transpose your input data to (m*m, n), you can use Dense(k) in conjunction with TimeDistributed to apply the same weights to the m^2 vectors separately. The output shape would be (m*m, k), after which you can reshape again to suit your needs
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on usingTimeDistributed. I will definitely look into it.
– AstroBen
Jan 4 at 15:26
This answer is ok, butTimeDistributed(Dense(...))is equal to justDense(...). -- The layer will be applied to the last dimension.
– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
add a comment |
If you reshape and transpose your input data to (m*m, n), you can use Dense(k) in conjunction with TimeDistributed to apply the same weights to the m^2 vectors separately. The output shape would be (m*m, k), after which you can reshape again to suit your needs
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
If you reshape and transpose your input data to (m*m, n), you can use Dense(k) in conjunction with TimeDistributed to apply the same weights to the m^2 vectors separately. The output shape would be (m*m, k), after which you can reshape again to suit your needs
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
answered Jan 4 at 15:02
BlackBearBlackBear
15.5k83368
15.5k83368
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on usingTimeDistributed. I will definitely look into it.
– AstroBen
Jan 4 at 15:26
This answer is ok, butTimeDistributed(Dense(...))is equal to justDense(...). -- The layer will be applied to the last dimension.
– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
add a comment |
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on usingTimeDistributed. I will definitely look into it.
– AstroBen
Jan 4 at 15:26
This answer is ok, butTimeDistributed(Dense(...))is equal to justDense(...). -- The layer will be applied to the last dimension.
– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on using
TimeDistributed. I will definitely look into it.– AstroBen
Jan 4 at 15:26
I was just looking into reshape and transpose when you wrote this. :-) Great pointer on using
TimeDistributed. I will definitely look into it.– AstroBen
Jan 4 at 15:26
This answer is ok, but
TimeDistributed(Dense(...)) is equal to just Dense(...). -- The layer will be applied to the last dimension.– Daniel Möller
Jan 4 at 15:37
This answer is ok, but
TimeDistributed(Dense(...)) is equal to just Dense(...). -- The layer will be applied to the last dimension.– Daniel Möller
Jan 4 at 15:37
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
@DanielMöller I didn't know that, thank you for pointing out
– BlackBear
Jan 5 at 8:15
add a comment |
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