Keras : is it possible to give lamda layer a name?
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I have following code inside my model definition
# (None, 6) -> (None, 8)
def affine_to_projective(x):
y = tf.pad(x,
paddings=tf.constant([[0, 0], [0, 2]]),
mode='CONSTANT',
constant_values=0
)
return y
projective_transform = Lambda(affine_to_projective)(affine_transform)
In model summary it's seen as
lambda_1 (Lambda) (None, 8) 0 affine_transform[0][0]
Is it possible to give this layer a name instead of lambda_1?
python keras deep-learning
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
add a comment |
0
I have following code inside my model definition
# (None, 6) -> (None, 8)
def affine_to_projective(x):
y = tf.pad(x,
paddings=tf.constant([[0, 0], [0, 2]]),
mode='CONSTANT',
constant_values=0
)
return y
projective_transform = Lambda(affine_to_projective)(affine_transform)
In model summary it's seen as
lambda_1 (Lambda) (None, 8) 0 affine_transform[0][0]
Is it possible to give this layer a name instead of lambda_1?
python keras deep-learning
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57
add a comment |
0
0
0
I have following code inside my model definition
# (None, 6) -> (None, 8)
def affine_to_projective(x):
y = tf.pad(x,
paddings=tf.constant([[0, 0], [0, 2]]),
mode='CONSTANT',
constant_values=0
)
return y
projective_transform = Lambda(affine_to_projective)(affine_transform)
In model summary it's seen as
lambda_1 (Lambda) (None, 8) 0 affine_transform[0][0]
Is it possible to give this layer a name instead of lambda_1?
python keras deep-learning
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
I have following code inside my model definition
# (None, 6) -> (None, 8)
def affine_to_projective(x):
y = tf.pad(x,
paddings=tf.constant([[0, 0], [0, 2]]),
mode='CONSTANT',
constant_values=0
)
return y
projective_transform = Lambda(affine_to_projective)(affine_transform)
In model summary it's seen as
lambda_1 (Lambda) (None, 8) 0 affine_transform[0][0]
Is it possible to give this layer a name instead of lambda_1?
python keras deep-learning
python keras deep-learning
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
asked Nov 26 '18 at 16:54
mrgloommrgloom
5,6151164140
5,6151164140
Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57
add a comment |
Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57
Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57
Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57
add a comment |
1 Answer
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projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
add a comment |
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1 Answer
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projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
add a comment |
1
projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
add a comment |
1
1
1
projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
answered Nov 26 '18 at 17:12
Daniel MöllerDaniel Möller
38k673109
38k673109
add a comment |
add a comment |
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Yes, did you try to give it a name?
– Matias Valdenegro
Nov 26 '18 at 16:57