How to predict x values from a linear model (lm)

前端 未结 3 1690
Happy的楠姐
Happy的楠姐 2021-02-08 12:23

I have this data set:

x <- c(0, 40, 80, 120, 160, 200)
y <- c(6.52, 5.10, 4.43, 3.99, 3.75, 3.60)

I calculated a linear model using

3条回答
  •  伪装坚强ぢ
    2021-02-08 12:44

    If your relationship is nonmonotone or if you have multiple predictor values then there can be multiple x-values for a given y-value and you need to decide how to deal with that.

    One option that could be slow (and may be the method used in the other packages mentioned) is to use the uniroot function:

    x <- runif(100, min=-1,max=2)
    y <- exp(x) + rnorm(100,0,0.2)
    
    fit <- lm( y ~ poly(x,3), x=TRUE )
    (tmp <- uniroot( function(x) predict(fit, data.frame(x=x)) - 4, c(-1, 2) )$root)
    library(TeachingDemos)
    plot(x,y)
    Predict.Plot(fit, 'x', data=data.frame(x=x), add=TRUE, ref.val=tmp)
    

    You could use the TkPredict function from the TeachingDemos package to eyeball a solution.

    Or you could get a fairly quick approximation by generating a lot of predicted points, then feeding them to the approxfun or splinfun functions to produce the approximations:

    tmpx <- seq(min(x), max(x), length.out=250)
    tmpy <- predict(fit, data.frame(x=tmpx) )
    tmpfun <- splinefun( tmpy, tmpx )
    tmpfun(4)
    

提交回复
热议问题