Keras : is it possible to give lamda layer a name?












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







share|improve this question























  • Yes, did you try to give it a name?

    – Matias Valdenegro
    Nov 26 '18 at 16:57
















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







share|improve this question























  • Yes, did you try to give it a name?

    – Matias Valdenegro
    Nov 26 '18 at 16:57














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







share|improve this question














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






share|improve this question













share|improve this question











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asked Nov 26 '18 at 16:54









mrgloommrgloom

5,2851063133




5,2851063133













  • 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





Yes, did you try to give it a name?

– Matias Valdenegro
Nov 26 '18 at 16:57












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projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)





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    1 Answer
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    projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)





    share|improve this answer




























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      projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)





      share|improve this answer


























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        projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)





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        projective_transform = Lambda(affine_to_projective, name='lambda_name')(affine_transform)






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        answered Nov 26 '18 at 17:12









        Daniel MöllerDaniel Möller

        35.7k667104




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