如何从线性模型(lm)预测x值 [英] How to predict x values from a linear model (lm)

问题描述

我有以下数据集:

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

我使用lm()计算了线性模型:

I calculated a linear model using lm():

model <- lm(y ~ x)

如果我有新的y值，例如，我想知道x的预测值. ynew <- c(5.5, 4.5, 3.5)，但是如果使用predict()函数，它将仅计算新的y值.

I want know the predicted values of x if I have new y values, e.g. ynew <- c(5.5, 4.5, 3.5), but if I use the predict() function, it calculates only new y values.

如果我有新的y值，如何预测新的x值?

How can I predict new x values if I have new y values?

推荐答案

由于这是化学中的典型问题(根据校准预测值)，因此包chemCal提供了inverse.predict.但是，此功能仅限于具有模型公式y〜x或y〜x-1的lm或rlm类的单变量模型对象".

Since this is a typical problem in chemistry (predict values from a calibration), package chemCal provides inverse.predict. However, this function is limited to "univariate model object[s] of class lm or rlm with model formula y ~ x or y ~ x - 1."

x <- c(0, 40, 80, 120, 160, 200)
y <- c(6.52, 5.10, 4.43, 3.99, 3.75, 3.60)
plot(x,y)
model <- lm(y ~ x)
abline(model)
require(chemCal)
ynew <- c(5.5, 4.5, 3.5)
xpred<-t(sapply(ynew,function(y) inverse.predict(model,y)[1:2]))
#  Prediction Standard Error
#[1,] 31.43007   -38.97289     
#[2,] 104.7669   -36.45131     
#[3,] 178.1037   -39.69539
points(xpred[,1],ynew,col="red")

警告:如果需要逆运算，此函数非常慢并且不适合.预测大量值.

Warning: This function is quite slow and not suitable, if you need to inverse.predict a large number of values.

如果我没记错的话，那就是负数.出现SE是因为函数期望斜率始终为正. SE的绝对值仍然应该正确.

If I remember correctly, the neg. SEs occur because the function expects the slope to be always positive. Absolute values of SE should still be correct.

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