Question

首先，我有你的载体，

x.seq

对于x，Fm和x的平滑向量是确定向量x.seq <- seq(qjx[1],qjx[m+1],by=0.01) Fm <- function(data,x){ #data = smooth data, x = intial data x.smooth <- NULL; xsmooth <- matrix(,nrow=(length(x)),ncol=length(data)) xsmooth. <- matrix(,nrow=(length(x)),ncol=length(data)) for(i in 1:m){ for(j in 1:length(data)){ xsmooth[i,j] <- (data[j] - qjx[i])/rjx[i] if(xsmooth[i,j] < 0){ xsmooth.[i,j] <- 0 } else if(xsmooth[i,j] >= 0 && xsmooth[i,j] <= 1){ xsmooth.[i,j] = xsmooth[i,j] } else if(xsmooth[i,j] > 1){ xsmooth.[i,j] <- 1 } } } xsmoothing <- NULL for(k in 1:ncol(xsmooth.)){ xsmoothing[k] <- mean(xsmooth.[,k]) } print(as.numeric(xsmoothing)) }的c.d.f值的函数

接下来我想用函数Fm计算m <- length(x) Finv <- function(t){ if(t > 0){ f.inv <- matrix(,ncol=2,nrow=m) k <- 1;f.invers <- NULL while(k <= m){ f.inv[k,1] <- (k-1)/m f.inv[k,2] <- k/m if(t >= f.inv[k,1] && t <= f.inv[k,2]){ f.invers[k] <- (((m*t)-(k-1))*rjx[k])+qjx[k] } else f.invers[k] <- NA k <- k+1 } hasil <- as.numeric(f.invers[!is.na(f.invers)]) print(hasil) } else hasil <- L # L = the initial value,for example L is 2.06 print(as.numeric(hasil)) }的逆值，我的函数在这里

原因是为什么我做了这个功能，因为q有许多序列。所以，我从1开始编译每个t，直到t <- Fm(x.seq,x) invers <- NULL for(q in 1:length(t)){ print(invers[q] <- Finv(1-t[q])) } invers

的向量的最后一个顺序

Warning messages:
1: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length
2: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length
3: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length
4: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length
5: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length
6: In invers[q] <- Finv(1 - t[q]) :
  number of items to replace is not a multiple of replacement length

它工作正常，但在处理反转的值时，会有警告

foo

我只想知道反函数的错误在哪里。感谢

Answer 1

在t之后尝试这个，

invers<-vector()
invers <- NULL
for(q in 1:(length(t)-1)){
  invers<-c(invers,Finv(1-t[q]))
}
invers

我将测试这个：

 invers <- NULL
 for(q in 1:(length(t))){
   invers<-c(invers,Finv(1-t[q]))
 }
 invers
 invers1<-vector()
 invers1 <- NULL
 for(q in 1:(length(t)-1)){
   invers1<-c(invers,Finv(1-t[q]))
 }
 invers1
 which(invers[1:1134]!=invers1[1:1134])

这是控制台的结果。

 > which(invers[1:1134]!=invers1[1:1134])
 integer(0)

要更换的项目数量不是更换长度的倍数警告是我的功能

1 个答案: