R:从n个子集中建立pls校准模型,并用它们来预测不同的测试集 [英] R: make pls calibration models from n number of subset and use them to predict different test sets
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问题描述
cpo<-function(data,newdata1,newdata2,newdata3,newdata4,newdata5,newdata6,newdata7,newdata8,newdata9){
data.pls<-plsr(protein~.,8,data=data,validation="LOO")#making a pls model
newdata1.pred<-predict(data.pls,8,newdata=newdata1) #using the model to predict test sets
newdata2.pred<-predict(data.pls,8,newdata=newdata2)
newdata3.pred<-predict(data.pls,8,newdata=newdata3)
newdata4.pred<-predict(data.pls,8,newdata=newdata4)
newdata5.pred<-predict(data.pls,8,newdata=newdata5)
newdata6.pred<-predict(data.pls,8,newdata=newdata6)
newdata7.pred<-predict(data.pls,8,newdata=newdata7)
newdata8.pred<-predict(data.pls,8,newdata=newdata8)
newdata9.pred<-predict(data.pls,8,newdata=newdata9)
pred.bias1<-mean(newdata1.pred-newdata1[742]) #calculating the prediction bias
pred.bias2<-mean(newdata2.pred-newdata2[742])
pred.bias3<-mean(newdata3.pred-newdata3[742]) #[742] reference values in column742
pred.bias4<-mean(newdata4.pred-newdata4[742])
pred.bias5<-mean(newdata5.pred-newdata5[742])
pred.bias6<-mean(newdata6.pred-newdata6[742])
pred.bias7<-mean(newdata7.pred-newdata7[742])
pred.bias8<-mean(newdata8.pred-newdata8[742])
pred.bias9<-mean(newdata9.pred-newdata9[742])
r<-c(R2(data.pls,"train"),RMSEP(data.pls,"train"),pred.bias1,
pred.bias2,pred.bias3,pred.bias4,pred.bias5,pred.bias6,
pred.bias7,pred.bias8,pred.bias9)
return(r)
}
选择n个子集(基于我问题[1]的答案:Select several subsets by taking different row interval and appy function to all subsets
并将CPO函数应用于我尝试的每个子集
根据@Gavin建议编辑
FO03 <- function(data, nSubsets, nSkip){
outList <- vector("list", 11)
names(outList) <- c("R2train","RMSEPtrain", paste("bias", 1:9, sep = ""))
sub <- vector("list", length = nSubsets) # sub is the n number subsets created by selecting rows
names(sub) <- c( paste("sub", 1:nSubsets, sep = ""))
totRow <- nrow(data)
for (i in seq_len(nSubsets)) {
rowsToGrab <- seq(i, totRow, nSkip)
sub[[i]] <- data[rowsToGrab ,]
}
for(i in sub) { #for every subset in sub i want to apply cpo
outList[[i]] <- cpo(data=sub,newdata1=gag11p,newdata2=gag12p,newdata3=gag13p,
newdata4=gag21p,newdata5=gag22p,newdata6=gag23p,
newdata7=gag31p,newdata8=gag32p,newdata9=gag33p) #new data are test sets loaded in the workspace
}
return(outlist)
}
FOO3(GAGp,10,10)
当我尝试这样做时,我一直收到‘ERROR in val(expr,envir,enclos):对象’蛋白质‘未找到’未找到。
蛋白质在CPO的PLSR公式中使用,并且在数据集中。
然后,我尝试直接使用plsr函数,如下所示
FOO4 <- function(data, nSubsets, nSkip){
outList <- vector("list", 11)
names(outList) <- c("R2train","RMSEPtrain", paste("bias", 1:9, sep = ""))
sub <- vector("list", length = nSubsets)
names(sub) <- c( paste("sub", 1:nSubsets, sep = ""))
totRow <- nrow(data)
for (i in seq_len(nSubsets)) {
rowsToGrab <- seq(i, totRow, nSkip)
sub[[i]] <- data[rowsToGrab ,]
}
cal<-vector("list", length=nSubsets) #for each subset in sub make a pls model for protein
names(cal)<-c(paste("cal",1:nSubsets, sep=""))
for(i in sub) {
cal[[i]] <- plsr(protein~.,8,data=sub,validation="LOO")
}
return(outlist) # return is just used to end script and check if error still occurs
}
FOO4(gagpm,10,10)
当我尝试这样做时,我得到了相同的错误‘ERROR in val(expr,envir,enclos):Object’Protein‘Not Found’。
如能就如何处理此问题并使该功能正常工作提供任何建议,我们将不胜感激。
推荐答案
我已经成功地使用此函数实现了我想要的结果,如果有更好的方法(我相信肯定有),我很想学习。此函数将执行以下任务
1.从数据帧中选择n个子集
2.对于创建的每个子集,建立一个PLSR模型
3.每个PLSR模型用于预测9个测试集
4.对于每个预测,计算预测偏差
far5<- function(data, nSubsets, nSkip){
sub <- vector("list", length = nSubsets)
names(sub) <- c( paste("sub", 1:nSubsets, sep = ""))
totRow <- nrow(data)
for (i in seq_len(nSubsets)) {
rowsToGrab <- seq(i, totRow, nSkip)
sub[[i]] <- data[rowsToGrab ,]} #sub is the subsets created
mop<- lapply(sub,cpr2) #assigning output from cpr to mop
names(mop)<-c(paste("mop", mop, sep=""))
return(names(mop))
}
call: far5(data,nSubsets, nSkip))
第一部分-选择子集是基于对我问题的回答Select several subsets by taking different row interval and appy function to all subsets
然后,我可以将函数cpr2应用于使用"lApply"创建的子集,而不是像前面那样使用"for"循环。
CP2是对CPO的修改,只提供数据,需要预测的新数据直接在函数中使用,如下图所示。
cpr2<-function(data){
data.pls<-plsr(protein~.,8,data=data,validation="LOO") #make plsr model
gag11p.pred<-predict(data.pls,8,newdata=gag11p) #predict each test set
gag12p.pred<-predict(data.pls,8,newdata=gag12p)
gag13p.pred<-predict(data.pls,8,newdata=gag13p)
gag21p.pred<-predict(data.pls,8,newdata=gag21p)
gag22p.pred<-predict(data.pls,8,newdata=gag22p)
gag23p.pred<-predict(data.pls,8,newdata=gag23p)
gag31p.pred<-predict(data.pls,8,newdata=gag31p)
gag32p.pred<-predict(data.pls,8,newdata=gag32p)
gag33p.pred<-predict(data.pls,8,newdata=gag33p)
pred.bias1<-mean(gag11p.pred-gag11p[742]) #calculate prediction bias
pred.bias2<-mean(gag12p.pred-gag12p[742])
pred.bias3<-mean(gag13p.pred-gag13p[742])
pred.bias4<-mean(gag21p.pred-gag21p[742])
pred.bias5<-mean(gag22p.pred-gag22p[742])
pred.bias6<-mean(gag23p.pred-gag23p[742])
pred.bias7<-mean(gag31p.pred-gag31p[742])
pred.bias8<-mean(gag32p.pred-gag32p[742])
pred.bias9<-mean(gag33p.pred-gag33p[742])
r<-signif(c(pred.bias1,pred.bias2,pred.bias3,pred.bias4,pred.bias5,
pred.bias6,pred.bias7,pred.bias8,pred.bias9),2)
out<-c(R2(data.pls,"train",ncomp=8),RMSEP(data.pls,"train",ncomp=8),r)
return(out)
} #signif use to return 2 decimal place for prediction bias
call:cpr2(data)
我能够使用它来解决我的问题,但是,因为要预测的新数据量只有9个,所以可以像我一样列出它们。如果有更通用的方法来做到这一点,我很有兴趣学习。
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