This vignette details the simulations tools provided with the selectboost package by providing five examples of use.
If you are a Linux/Unix or a Macos user, you can install a version of SelectBoost with support for doMC
from github with:
We want to creates \(NDatasets=200\) datasets with \(\textrm{length}(group)=10\) variables and \(N=10\) observations. In that example we want \(9\) groups:
The correlation structure of the explanatory variables of the dataset is provided by group
and the intra-group Pearson correlation value for each of the groups by cor_group
. A value must be provided even for single variable groups and the number of variables is length of the group
vector. Use the simulation_cor
function to create the correlation matrix (CM
).
Then generation of an explanatory dataset with \(N=10\) observations is made by the simulation_X
function.
A response can now be added to the dataset by the simulation_Data
function. We have to specifiy the support of the response, i.e. the explanatory variables that will be used in the linear model created to compute the response. The support is given by the supp
vector whose entries are either \(0\) or \(1\). The length of the supp
vector must be equal to the number of explanatory variables and if the \(i\)entry is equal to \(1\), it means that the \(i\)variable will be used to derive the response value, whereas if the \(i\)entry is equal to \(0\), it means that the \(i\)variable will not be used to derive the response value (beta<-rep(0,length(supp))
). The values of the coefficients for the explanatory variables that are in the support of the response are random (either absolute value and sign) and given by beta[which(supp==1)]<-runif(sum(supp),minB,maxB)*(rbinom(sum(supp),1,.5)*2-1)
. Hence, the user can specify their minimal absolute value with the minB
option and their maximal absolute value with the maxB
option. The stn
option is a scaling factor for the noise added to the response vector ((t(beta)%*%var(X)%*%beta)/stn
, with X
the matrix of explanatory variables). The higher the stn
value, the smaller the noise: for instance for a given X
dataset, an stn
value \(\alpha\) times larger will result in a noise exactly \(\sqrt{\alpha}\) times smaller.
To generate multiple datasets, repeat steps 2 and 3, for instance use a for
loop. We create \(NDatasets=200\) datasets and assign them to the objects DATA_exemple1_nb_1
to DATA_exemple1_nb_200
.
set.seed(3141)
NDatasets=200
for(i in 1:NDatasets){
X<-simulation_X(N,CM)
assign(paste("DATA_exemple1_nb_",i,sep=""),simulation_DATA(X,supp,minB,maxB,stn))
}
We now check the correlation structure of the explanatory variable. First we compute the mean correlation matrix.
corr_sum=matrix(0,length(group),length(group))
for(i in 1:NDatasets){
corr_sum=corr_sum+cor(get(paste("DATA_exemple1_nb_",i,sep=""))$X)
}
corr_mean=corr_sum/NDatasets
Then we display and plot that the mean correlation matrix.
coef_sum=rep(0,length(group))
names(coef_sum)<-paste("x",1:length(group),sep="")
error_counter=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple1_nb_",i,sep=""))$Y,
get(paste("DATA_exemple1_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter=error_counter+1
} else{
coef_sum=coef_sum+abs(tempcoef)
}
}
error_counter
coef_mean=coef_sum/NDatasets
All fits were sucessful. Then we display and plot that the mean coefficient vector values.
Reduce the noise in the response for the new responses by a factor \(\sqrt{5000/50}=10\). \(1/stn\cdot \beta_{support}^t\mathrm{Var}(X)\beta_{support}\) where \(\beta_{support}\) is the vector of coefficients wh
set.seed(3141)
stn <- 5000
for(i in 1:NDatasets){
X<-simulation_X(N,CM)
assign(paste("DATA_exemple1_bis_nb_",i,sep=""),simulation_DATA(X,supp,minB,maxB,stn))
}
Since it is the same explanatory dataset for response generation, we can compare the \(\sigma\) between those \(NDatasets=200\) datasets.
stn_ratios=rep(0,NDatasets)
for(i in 1:NDatasets){
stn_ratios[i]<-get(paste("DATA_exemple1_nb_",i,sep=""))$sigma/
get(paste("DATA_exemple1_bis_nb_",i,sep=""))$sigma
}
all(sapply(stn_ratios,all.equal,10))
All the ratios are equal to 10 as anticipated.
Since, the correlation structure is the same as before, we do not need to check it again. As befor, we infer the coefficients values of a linear model using the lm
function.
coef_sum_bis=rep(0,length(group))
names(coef_sum_bis)<-paste("x",1:length(group),sep="")
error_counter_bis=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple1_bis_nb_",i,sep=""))$Y,
get(paste("DATA_exemple1_bis_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter_bis=error_counte_bisr+1
} else{
coef_sum_bis=coef_sum_bis+abs(tempcoef)
}
}
error_counter_bis
coef_mean_bis=coef_sum_bis/NDatasets
All fits were sucessful. Then we display and plot that the mean coefficient vector values. As expected, the noise reduction enhances the retrieval of the true
mean coefficient absolute values by the models.
The simulation process looks sucessfull. What are the confidence indices for those variables?
We want to creates \(NDatasets=200\) datasets with \(\textrm{length}(group)=50\) variables and \(N=20\) observations. In that example we want \(1\) group:
We now check the correlation structure of the explanatory variable. First we compute the mean correlation matrix.
corr_sum=matrix(0,length(group),length(group))
for(i in 1:NDatasets){
corr_sum=corr_sum+cor(get(paste("DATA_exemple2_nb_",i,sep=""))$X)
}
corr_mean=corr_sum/NDatasets
Then we display and plot that the mean correlation matrix.
coef_sum=rep(0,length(group))
coef_lasso_sum=rep(0,length(group))
names(coef_sum)<-paste("x",1:length(group),sep="")
names(coef_lasso_sum)<-paste("x",1:length(group),sep="")
error_counter=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple2_nb_",i,sep=""))$Y,
get(paste("DATA_exemple2_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
require(lars)
lasso.1 <- lars::lars(x=get(paste("DATA_exemple2_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple2_nb_",i,sep=""))$Y, type="lasso",
trace=FALSE, normalize=FALSE, intercept = FALSE)
# cv.lars() uses crossvalidation to estimate optimal position in path
cv.lasso.1 <- lars::cv.lars(x=get(paste("DATA_exemple2_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple2_nb_",i,sep=""))$Y,
plot.it=FALSE, type="lasso")
# Use the "+1SE rule" to find best model:
# Take the min CV and add its SE ("limit").
# Find smallest model that has its own CV within this limit (at "s.cv.1")
limit <- min(cv.lasso.1$cv) + cv.lasso.1$cv.error[which.min(cv.lasso.1$cv)]
s.cv.1 <- cv.lasso.1$index[min(which(cv.lasso.1$cv < limit))]
# Print out coefficients at optimal s.
coef_lasso_sum=coef_lasso_sum+abs(coef(lasso.1, s=s.cv.1, mode="fraction"))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter=error_counter+1
} else{
coef_sum=coef_sum+abs(tempcoef)
}
}
error_counter
coef_mean=coef_sum/NDatasets
coef_lasso_mean=coef_lasso_sum/NDatasets
With regular least squares and lasso estimators all fits were sucessful, yet only 20 variables coefficients could be estimated with regular least squares estimates for the linear model. Then we display and plot that the mean coefficient vector values for the least squares estimates.
coef_lasso_mean
barplot(coef_lasso_mean,ylim=c(0,1.5))
abline(h=(minB+maxB)/2,lwd=2,lty=2,col="blue")
The simulation process looks sucessfull: the lasso estimates retrives mostly the correct variables, yet the other ones are also selected sometimes. What are the confidence indices for those variables?
We want to creates \(NDatasets=200\) datasets with \(\textrm{length}(supp)=100\) variables and \(N=24\) observations. In that example we use real data for the X
variables that we sample from all the \(1650\) probesets that are differentially expressed between the two conditions US and S. The main interest of that simulation is that the correlation structure of the X
dataset will be a real one.
First retrieve the datasets and get the differentially expressed probesets. Run the code to get additionnal plots.
require(CascadeData)
data(micro_S)
data(micro_US)
require(Cascade)
micro_US<-as.micro_array(micro_US,c(60,90,240,390),6)
micro_S<-as.micro_array(micro_S,c(60,90,240,390),6)
S<-geneSelection(list(micro_S,micro_US),list("condition",c(1,2),1),-1)
Sel<-micro_S@microarray[S@name,]
Generates the datasets sampling for each of them 100 probesets expressions among the 1650 that were selected and linking the response to the expressions of the first five probesets.
Here are the plots of an example of correlation structure, namely for DATA_exemple3_nb_200$X
. Run the code to get the graphics.
coef_sum=rep(0,length(supp))
coef_lasso_sum=rep(0,length(supp))
names(coef_sum)<-paste("x",1:length(supp),sep="")
names(coef_lasso_sum)<-paste("x",1:length(supp),sep="")
error_counter=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple3_nb_",i,sep=""))$Y,
get(paste("DATA_exemple3_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
require(lars)
lasso.1 <- lars::lars(x=get(paste("DATA_exemple3_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple3_nb_",i,sep=""))$Y, type="lasso",
trace=FALSE, normalize=FALSE, intercept = FALSE)
# cv.lars() uses crossvalidation to estimate optimal position in path
cv.lasso.1 <- lars::cv.lars(x=get(paste("DATA_exemple3_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple3_nb_",i,sep=""))$Y,
plot.it=FALSE, normalize=FALSE, intercept = FALSE, type="lasso")
# Use the "+1SE rule" to find best model:
# Take the min CV and add its SE ("limit").
# Find smallest model that has its own CV within this limit (at "s.cv.1")
limit <- min(cv.lasso.1$cv) + cv.lasso.1$cv.error[which.min(cv.lasso.1$cv)]
s.cv.1 <- cv.lasso.1$index[min(which(cv.lasso.1$cv < limit))]
# Print out coefficients at optimal s.
coef_lasso_sum=coef_lasso_sum+abs(coef(lasso.1, s=s.cv.1, mode="fraction"))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter=error_counter+1
} else{
coef_sum=coef_sum+abs(tempcoef)
}
}
error_counter
coef_mean=coef_sum/NDatasets
coef_lasso_mean=coef_lasso_sum/NDatasets
With regular least squares and lasso estimators all fits were sucessful, yet only 20 variables coefficients could be estimated with regular least squares estimates for the linear model. Then we display and plot that the mean coefficient vector values for the least squares estimates.
coef_lasso_mean
barplot(coef_lasso_mean,ylim=c(0,1.5))
abline(h=(minB+maxB)/2,lwd=2,lty=2,col="blue")
The simulation process looks sucessfull: the lasso estimates retrives mostly the correct variables, yet the other ones are also selected sometimes. What are the confidence indices for those variables?
We want to creates \(NDatasets=101\) datasets with \(\textrm{length}(supp)=100\) variables and \(N=18\) observations. In that example we use real data for the variables that are the \(101\) probesets that are the more differentially expressed between the two conditions US and S. We create \(101\) datasets by leaving one of the variables out each time and using it as the response that shall be predicted. We also only use for the explanatory variables the observations that are the measurements for the 1st, 2nd and 3rd timepoints and for the responses the observations that are the measurements of the same variables for the 2nd, 3rd and 4th timepoints. The main interest of that simulation is that the correlation structure of the X
dataset will be a real one and that it is a typical setting for cascade network reverse-engineering in genomics or proteomics, see the Cascade
package for more details.
First retrieve the datasets and get the differentially expressed probesets. Run the code to get additionnal plots.
require(CascadeData)
data(micro_S)
data(micro_US)
require(Cascade)
micro_US<-as.micro_array(micro_US,c(60,90,240,390),6)
micro_S<-as.micro_array(micro_S,c(60,90,240,390),6)
S<-geneSelection(list(micro_S,micro_US),list("condition",c(1,2),1),101)
Sel<-micro_S@microarray[S@name,]
suppt<-rep(1:4,6)
supp<-c(1,1,1,1,1,rep(0,95)) #not used since we use one of the probeset expressions as response
minB<-1 #not used since we use one of the probeset expressions as response
maxB<-2 #not used since we use one of the probeset expressions as response
stn<-50 #not used since we use one of the probeset expressions as response
NDatasets<-101
set.seed(3141)
for(i in 1:NDatasets){
#the explanatory variables are the values for the 1st, 2nd and 3rd timepoints
X<-t(as.matrix(Sel[-i,suppt!=4]))
Xnorm<-t(t(X)/sqrt(colSums(X*X)))
DATA<-simulation_DATA(Xnorm,supp,minB,maxB,stn)
#the reponses are the values for the 2nd, 3rd and 4th timepoints
DATA$Y<-as.vector(t(Sel[i,suppt!=1]))
assign(paste("DATA_exemple4_nb_",i,sep=""),DATA)
rm(DATA)
}
Here are the plots of an example of correlation structure, namely for DATA_exemple3_nb_200$X
. Run the code to get the graphics.
coef_sum=rep(0,length(supp)+1)
coef_lasso_sum=rep(0,length(supp)+1)
names(coef_sum)<-paste("x",1:(length(supp)+1),sep="")
names(coef_lasso_sum)<-paste("x",1:(length(supp)+1),sep="")
error_counter=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple4_nb_",i,sep=""))$Y,
get(paste("DATA_exemple4_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
require(lars)
lasso.1 <- lars::lars(x=get(paste("DATA_exemple4_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple4_nb_",i,sep=""))$Y, type="lasso",
trace=FALSE, normalize=FALSE, intercept = FALSE)
# cv.lars() uses crossvalidation to estimate optimal position in path
cv.lasso.1 <- lars::cv.lars(x=get(paste("DATA_exemple4_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple4_nb_",i,sep=""))$Y,
plot.it=FALSE, normalize=FALSE, intercept = FALSE, type="lasso")
# Use the "+1SE rule" to find best model:
# Take the min CV and add its SE ("limit").
# Find smallest model that has its own CV within this limit (at "s.cv.1")
limit <- min(cv.lasso.1$cv) + cv.lasso.1$cv.error[which.min(cv.lasso.1$cv)]
s.cv.1 <- cv.lasso.1$index[min(which(cv.lasso.1$cv < limit))]
# Print out coefficients at optimal s.
coef_lasso_sum[-i]=coef_lasso_sum[-i]+abs(coef(lasso.1, s=s.cv.1, mode="fraction"))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter=error_counter+1
} else{
coef_sum[-i]=coef_sum[-i]+abs(tempcoef)
}
}
error_counter
coef_mean=coef_sum/NDatasets
coef_lasso_mean=coef_lasso_sum/NDatasets
With regular least squares and lasso estimators all fits were sucessful, yet only 20 variables coefficients could be estimated with regular least squares estimates for the linear model. Then we display and plot that the mean coefficient vector values for the least squares estimates.
Some probesets seem explanatory for many other ones (=hubs). What are the confidence indices for those variables?
We want to creates \(NDatasets=200\) datasets with \(\textrm{length}(group)=500\) variables and \(N=25\) observations. In that example we want \(1\) group:
We now check the correlation structure of the explanatory variable. First we compute the mean correlation matrix.
corr_sum=matrix(0,length(group),length(group))
for(i in 1:NDatasets){
corr_sum=corr_sum+cor(get(paste("DATA_exemple5_nb_",i,sep=""))$X)
}
corr_mean=corr_sum/NDatasets
Then we display and plot that the mean correlation matrix.
coef_sum=rep(0,length(group))
coef_lasso_sum=rep(0,length(group))
names(coef_sum)<-paste("x",1:length(group),sep="")
names(coef_lasso_sum)<-paste("x",1:length(group),sep="")
error_counter=0
for(i in 1:NDatasets){
tempdf=data.frame(cbind(Y=get(paste("DATA_exemple5_nb_",i,sep=""))$Y,
get(paste("DATA_exemple5_nb_",i,sep=""))$X))
tempcoef=coef(lm(Y~.-1,data=tempdf))
require(lars)
lasso.1 <- lars::lars(x=get(paste("DATA_exemple5_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple5_nb_",i,sep=""))$Y, type="lasso",
trace=FALSE, normalize=FALSE, intercept = FALSE)
# cv.lars() uses crossvalidation to estimate optimal position in path
cv.lasso.1 <- lars::cv.lars(x=get(paste("DATA_exemple5_nb_",i,sep=""))$X,
y=get(paste("DATA_exemple5_nb_",i,sep=""))$Y,
plot.it=FALSE, type="lasso")
# Use the "+1SE rule" to find best model:
# Take the min CV and add its SE ("limit").
# Find smallest model that has its own CV within this limit (at "s.cv.1")
limit <- min(cv.lasso.1$cv) + cv.lasso.1$cv.error[which.min(cv.lasso.1$cv)]
s.cv.1 <- cv.lasso.1$index[min(which(cv.lasso.1$cv < limit))]
# Print out coefficients at optimal s.
coef_lasso_sum=coef_lasso_sum+abs(coef(lasso.1, s=s.cv.1, mode="fraction"))
if(is.null(tempcoef)){
cat("Error in lm fit, skip coefficients\n")
error_counter=error_counter+1
} else{
coef_sum=coef_sum+abs(tempcoef)
}
}
error_counter
coef_mean=coef_sum/NDatasets
coef_lasso_mean=coef_lasso_sum/NDatasets
With regular least squares and lasso estimators all fits were sucessful, yet only 20 variables coefficients could be estimated with regular least squares estimates for the linear model. Then we display and plot that the mean coefficient vector values for the least squares estimates.
head(coef_lasso_mean, 40)
barplot(coef_lasso_mean,ylim=c(0,1.5))
abline(h=(minB+maxB)/2,lwd=2,lty=2,col="blue")
The simulation process looks sucessfull: the lasso estimates retrives mostly the correct variables, yet the other ones are also selected sometimes. What are the confidence indices for those variables?