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R 将nls应用于列表中的元素_R_List_Lapply_Nls - Fatal编程技术网

R 将nls应用于列表中的元素

R 将nls应用于列表中的元素,r,list,lapply,nls,R,List,Lapply,Nls,设想以下两个列表: seq<-seq(1,16) L1<-list(A=seq, B=seq, C=seq, D=seq) v1=c(11,15,17,19,21,22,24,25,26,27,28,28,29,29,30,30) v2=c(14,19,23,24,26,27,28,29,30,30,31,32,32,32,32,33) v3=c(2,4,5,6,6,7,8,8,9,9,9,10,10,10,11,11) v4=c(8,13,17,20,22,24,26,27,28

设想以下两个列表:

seq<-seq(1,16)
L1<-list(A=seq, B=seq, C=seq, D=seq)

v1=c(11,15,17,19,21,22,24,25,26,27,28,28,29,29,30,30)
v2=c(14,19,23,24,26,27,28,29,30,30,31,32,32,32,32,33)
v3=c(2,4,5,6,6,7,8,8,9,9,9,10,10,10,11,11)
v4=c(8,13,17,20,22,24,26,27,28,29,30,31,32,33,33,34)
L2<-list(A=v1,B=v2,C=v3,D=v4)

但这也不能回答我的问题(我真的需要这个该死的列表),我假设使用lappy的东西是最理想的,但是不知道如何使用lappy来做这样的事情。有什么想法吗?

mapply
与annonymus函数一起使用

res <- mapply(function(x,y){
  nls(y~(a/b)*(1-exp(-b*x)),
      start=list(a=1, b=0.1),
      trace= TRUE, data=data.frame(x, y))
},L1,L2, SIMPLIFY=FALSE)
res
for (i in L1) {
  nls(L2.i~(a/b)*(1-exp(-b*L1.i)),
      start=list(a=1, b=0.1),trace= TRUE,
  data=data.frame(L1$i, L2$i))
}
res <- mapply(function(x,y){
  nls(y~(a/b)*(1-exp(-b*x)),
      start=list(a=1, b=0.1),
      trace= TRUE, data=data.frame(x, y))
},L1,L2, SIMPLIFY=FALSE)