Autoencoder and SVD: Matrix apllications












0















in my study, I am using the so-called Lee Carter Model (Mortality model) in which you can get the model parameters by using Singular Values Decomposition on the matrix of (log mortality rate- the average age-specific pattern of mortality).
I am trying to find a substitution of Singular Value Decomposition, I saw that a good choice could be an autoencoding applied by a Recurrent Neural network. In fact, an SVD could be converging to autoencoder in which the activation function is a linear function. At this purpose, I would try using a nonlinear activation function in order to obtain the same items obtained by SVD with a nonlinear shape.
Let's use this steps in order to obtain data: mortality rates for ages and years



rm(list = ls())

library(MortalitySmooth)

ages <- 0:100

years <- 1960:2009

D <- as.matrix(selectHMDdata("Japan", "Deaths",
"Females", ages,
years))

D[D==0] <- 1

E <- as.matrix(selectHMDdata("Japan", "Exposures",
"Females", ages,
years))

E[E==0] <- 1


lMX <- log(D/E)

alpha <- apply(lMX, 1, mean)`

cent.logMXMatrix <- sweep(lMX, 1, alpha)


Now we apply SVD on cent.logMXMatrix
when I use SVD in R I get this:



SVD <- svd(cent.logMXMatrix)


and I need to get the components of SVD:



SVD$d
SVD$v
SVD$u


I would like to get SVD component using Autoencoder...Is it possible?
I would like to get your opinion, some suggestion from you and whether is possible I need a basic python code formulation for autoencoder on the "cent.logMXMatrix"



Thank a lot,
Andrea










share|improve this question

























  • SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

    – Matthieu Brucher
    Nov 19 '18 at 10:30











  • Thanks, can you show me?

    – an.dr.ea
    Nov 20 '18 at 11:16
















0















in my study, I am using the so-called Lee Carter Model (Mortality model) in which you can get the model parameters by using Singular Values Decomposition on the matrix of (log mortality rate- the average age-specific pattern of mortality).
I am trying to find a substitution of Singular Value Decomposition, I saw that a good choice could be an autoencoding applied by a Recurrent Neural network. In fact, an SVD could be converging to autoencoder in which the activation function is a linear function. At this purpose, I would try using a nonlinear activation function in order to obtain the same items obtained by SVD with a nonlinear shape.
Let's use this steps in order to obtain data: mortality rates for ages and years



rm(list = ls())

library(MortalitySmooth)

ages <- 0:100

years <- 1960:2009

D <- as.matrix(selectHMDdata("Japan", "Deaths",
"Females", ages,
years))

D[D==0] <- 1

E <- as.matrix(selectHMDdata("Japan", "Exposures",
"Females", ages,
years))

E[E==0] <- 1


lMX <- log(D/E)

alpha <- apply(lMX, 1, mean)`

cent.logMXMatrix <- sweep(lMX, 1, alpha)


Now we apply SVD on cent.logMXMatrix
when I use SVD in R I get this:



SVD <- svd(cent.logMXMatrix)


and I need to get the components of SVD:



SVD$d
SVD$v
SVD$u


I would like to get SVD component using Autoencoder...Is it possible?
I would like to get your opinion, some suggestion from you and whether is possible I need a basic python code formulation for autoencoder on the "cent.logMXMatrix"



Thank a lot,
Andrea










share|improve this question

























  • SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

    – Matthieu Brucher
    Nov 19 '18 at 10:30











  • Thanks, can you show me?

    – an.dr.ea
    Nov 20 '18 at 11:16














0












0








0








in my study, I am using the so-called Lee Carter Model (Mortality model) in which you can get the model parameters by using Singular Values Decomposition on the matrix of (log mortality rate- the average age-specific pattern of mortality).
I am trying to find a substitution of Singular Value Decomposition, I saw that a good choice could be an autoencoding applied by a Recurrent Neural network. In fact, an SVD could be converging to autoencoder in which the activation function is a linear function. At this purpose, I would try using a nonlinear activation function in order to obtain the same items obtained by SVD with a nonlinear shape.
Let's use this steps in order to obtain data: mortality rates for ages and years



rm(list = ls())

library(MortalitySmooth)

ages <- 0:100

years <- 1960:2009

D <- as.matrix(selectHMDdata("Japan", "Deaths",
"Females", ages,
years))

D[D==0] <- 1

E <- as.matrix(selectHMDdata("Japan", "Exposures",
"Females", ages,
years))

E[E==0] <- 1


lMX <- log(D/E)

alpha <- apply(lMX, 1, mean)`

cent.logMXMatrix <- sweep(lMX, 1, alpha)


Now we apply SVD on cent.logMXMatrix
when I use SVD in R I get this:



SVD <- svd(cent.logMXMatrix)


and I need to get the components of SVD:



SVD$d
SVD$v
SVD$u


I would like to get SVD component using Autoencoder...Is it possible?
I would like to get your opinion, some suggestion from you and whether is possible I need a basic python code formulation for autoencoder on the "cent.logMXMatrix"



Thank a lot,
Andrea










share|improve this question
















in my study, I am using the so-called Lee Carter Model (Mortality model) in which you can get the model parameters by using Singular Values Decomposition on the matrix of (log mortality rate- the average age-specific pattern of mortality).
I am trying to find a substitution of Singular Value Decomposition, I saw that a good choice could be an autoencoding applied by a Recurrent Neural network. In fact, an SVD could be converging to autoencoder in which the activation function is a linear function. At this purpose, I would try using a nonlinear activation function in order to obtain the same items obtained by SVD with a nonlinear shape.
Let's use this steps in order to obtain data: mortality rates for ages and years



rm(list = ls())

library(MortalitySmooth)

ages <- 0:100

years <- 1960:2009

D <- as.matrix(selectHMDdata("Japan", "Deaths",
"Females", ages,
years))

D[D==0] <- 1

E <- as.matrix(selectHMDdata("Japan", "Exposures",
"Females", ages,
years))

E[E==0] <- 1


lMX <- log(D/E)

alpha <- apply(lMX, 1, mean)`

cent.logMXMatrix <- sweep(lMX, 1, alpha)


Now we apply SVD on cent.logMXMatrix
when I use SVD in R I get this:



SVD <- svd(cent.logMXMatrix)


and I need to get the components of SVD:



SVD$d
SVD$v
SVD$u


I would like to get SVD component using Autoencoder...Is it possible?
I would like to get your opinion, some suggestion from you and whether is possible I need a basic python code formulation for autoencoder on the "cent.logMXMatrix"



Thank a lot,
Andrea







python matrix autoencoder svd






share|improve this question















share|improve this question













share|improve this question




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edited Nov 18 '18 at 12:58









Dmitriy Fialkovskiy

1,59521425




1,59521425










asked Nov 18 '18 at 12:05









an.dr.eaan.dr.ea

44




44













  • SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

    – Matthieu Brucher
    Nov 19 '18 at 10:30











  • Thanks, can you show me?

    – an.dr.ea
    Nov 20 '18 at 11:16



















  • SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

    – Matthieu Brucher
    Nov 19 '18 at 10:30











  • Thanks, can you show me?

    – an.dr.ea
    Nov 20 '18 at 11:16

















SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

– Matthieu Brucher
Nov 19 '18 at 10:30





SVD is a linear algebra problem. You can use a pseudo autoencoder with just 2 linear layers to get it, of course.

– Matthieu Brucher
Nov 19 '18 at 10:30













Thanks, can you show me?

– an.dr.ea
Nov 20 '18 at 11:16





Thanks, can you show me?

– an.dr.ea
Nov 20 '18 at 11:16












1 Answer
1






active

oldest

votes


















1














A one-layer autoencoder linearly maps a datapoint to a low-dimensional latent space, then applies a non-linear activation to project the result to the original space while minimizing a reconstruction error.

If we replace the non-linear activation by a linear one (identity) and use the L2 norm as a reconstruction error, you will be performing the same operation as an SVD.



# use keras with tensorflow backend
# This is a vanilla autoencoder with one hidden layer
from keras.layers import Input, Dense
from keras.models import Model

input_dim = Input(shape = (nfeat, )) # nfeat=the number of initial features
encoded1 = Dense(layer_size1, activation='linear')(input_dim) # layer_size1:size of your encoding layer
decoded1 = Dense(nfeat, activation='linear')
autoencoder = Model(inputs = input_dim, outputs = decoded1)
autoencoder.compile(loss='mean_squared_error', optimizer='adam')





share|improve this answer





















  • 1





    No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

    – Matthieu Brucher
    Nov 20 '18 at 14:37













  • @MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

    – Akihiko
    Nov 20 '18 at 15:09













  • Thanks for fixing this!

    – Matthieu Brucher
    Nov 20 '18 at 15:23











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A one-layer autoencoder linearly maps a datapoint to a low-dimensional latent space, then applies a non-linear activation to project the result to the original space while minimizing a reconstruction error.

If we replace the non-linear activation by a linear one (identity) and use the L2 norm as a reconstruction error, you will be performing the same operation as an SVD.



# use keras with tensorflow backend
# This is a vanilla autoencoder with one hidden layer
from keras.layers import Input, Dense
from keras.models import Model

input_dim = Input(shape = (nfeat, )) # nfeat=the number of initial features
encoded1 = Dense(layer_size1, activation='linear')(input_dim) # layer_size1:size of your encoding layer
decoded1 = Dense(nfeat, activation='linear')
autoencoder = Model(inputs = input_dim, outputs = decoded1)
autoencoder.compile(loss='mean_squared_error', optimizer='adam')





share|improve this answer





















  • 1





    No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

    – Matthieu Brucher
    Nov 20 '18 at 14:37













  • @MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

    – Akihiko
    Nov 20 '18 at 15:09













  • Thanks for fixing this!

    – Matthieu Brucher
    Nov 20 '18 at 15:23
















1














A one-layer autoencoder linearly maps a datapoint to a low-dimensional latent space, then applies a non-linear activation to project the result to the original space while minimizing a reconstruction error.

If we replace the non-linear activation by a linear one (identity) and use the L2 norm as a reconstruction error, you will be performing the same operation as an SVD.



# use keras with tensorflow backend
# This is a vanilla autoencoder with one hidden layer
from keras.layers import Input, Dense
from keras.models import Model

input_dim = Input(shape = (nfeat, )) # nfeat=the number of initial features
encoded1 = Dense(layer_size1, activation='linear')(input_dim) # layer_size1:size of your encoding layer
decoded1 = Dense(nfeat, activation='linear')
autoencoder = Model(inputs = input_dim, outputs = decoded1)
autoencoder.compile(loss='mean_squared_error', optimizer='adam')





share|improve this answer





















  • 1





    No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

    – Matthieu Brucher
    Nov 20 '18 at 14:37













  • @MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

    – Akihiko
    Nov 20 '18 at 15:09













  • Thanks for fixing this!

    – Matthieu Brucher
    Nov 20 '18 at 15:23














1












1








1







A one-layer autoencoder linearly maps a datapoint to a low-dimensional latent space, then applies a non-linear activation to project the result to the original space while minimizing a reconstruction error.

If we replace the non-linear activation by a linear one (identity) and use the L2 norm as a reconstruction error, you will be performing the same operation as an SVD.



# use keras with tensorflow backend
# This is a vanilla autoencoder with one hidden layer
from keras.layers import Input, Dense
from keras.models import Model

input_dim = Input(shape = (nfeat, )) # nfeat=the number of initial features
encoded1 = Dense(layer_size1, activation='linear')(input_dim) # layer_size1:size of your encoding layer
decoded1 = Dense(nfeat, activation='linear')
autoencoder = Model(inputs = input_dim, outputs = decoded1)
autoencoder.compile(loss='mean_squared_error', optimizer='adam')





share|improve this answer















A one-layer autoencoder linearly maps a datapoint to a low-dimensional latent space, then applies a non-linear activation to project the result to the original space while minimizing a reconstruction error.

If we replace the non-linear activation by a linear one (identity) and use the L2 norm as a reconstruction error, you will be performing the same operation as an SVD.



# use keras with tensorflow backend
# This is a vanilla autoencoder with one hidden layer
from keras.layers import Input, Dense
from keras.models import Model

input_dim = Input(shape = (nfeat, )) # nfeat=the number of initial features
encoded1 = Dense(layer_size1, activation='linear')(input_dim) # layer_size1:size of your encoding layer
decoded1 = Dense(nfeat, activation='linear')
autoencoder = Model(inputs = input_dim, outputs = decoded1)
autoencoder.compile(loss='mean_squared_error', optimizer='adam')






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 20 '18 at 15:13

























answered Nov 20 '18 at 12:46









AkihikoAkihiko

557




557








  • 1





    No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

    – Matthieu Brucher
    Nov 20 '18 at 14:37













  • @MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

    – Akihiko
    Nov 20 '18 at 15:09













  • Thanks for fixing this!

    – Matthieu Brucher
    Nov 20 '18 at 15:23














  • 1





    No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

    – Matthieu Brucher
    Nov 20 '18 at 14:37













  • @MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

    – Akihiko
    Nov 20 '18 at 15:09













  • Thanks for fixing this!

    – Matthieu Brucher
    Nov 20 '18 at 15:23








1




1





No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

– Matthieu Brucher
Nov 20 '18 at 14:37







No need for more layers in a linear encoder. But thanks for showing the code for OP. Even here, you should have only 2 layers, not three. Which is why I'm not up voting.

– Matthieu Brucher
Nov 20 '18 at 14:37















@MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

– Akihiko
Nov 20 '18 at 15:09







@MatthieuBrucher You're right. Since we have linear activation, no need to add more layers. Sorry for this !

– Akihiko
Nov 20 '18 at 15:09















Thanks for fixing this!

– Matthieu Brucher
Nov 20 '18 at 15:23





Thanks for fixing this!

– Matthieu Brucher
Nov 20 '18 at 15:23


















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