使用拟合模型从Y值预测X值 [英] Predict X value from Y value with a fitted model

问题描述

我需要使用拟合模型预测新的y值对应的x值.

I need to predict the corresponding x value of a new y value using a fitted model.

通过使用predict函数从新的x值预测y值的通常情况很简单，但是我不知道该怎么做.

The usual case of predicting the y value from a new x value is straightforward by using the predict function, but I cannot figure out how to do the reverse.

对于具有多个x解决方案的情况，我希望获得x值范围内的所有解决方案，即1-10.并且新的y将始终在用于拟合模型的y值的范围内.

For cases with multiple x solutions, I wish to obtain all solutions within the range of x values, i.e. 1-10. And the new y will always be within the range of y values used for fitting the model.

请参见下面的示例代码，我需要在其中找到新的x值(new_x).

See below for an example code, where I need to find new x value (new_x).

x = seq(1:10)
y = c(60,30,40,45,35,20,10,15,25,10)

fit = lm(y ~ poly(x, 3, raw=T))

plot(x, y)
lines(sort(x), predict(fit)[order(x)], col='red') 

new_y = 30
new_x = predict(fit, data.frame(y=new_y)) #This line does not work as intended.

反拟合

由于我们得到了不同的模型/拟合线，因此拟合反向关系不会得到相同的模型.

Fitting the inversed relationship will not give the same model, since we get a different model/fitted line.

rev_fit = lm(x ~ poly(y, 3, raw=T))

plot(x, y)
lines(sort(x), predict(fit)[order(x)], col='red') 
lines(predict(rev_fit)[order(y)], sort(y), col='blue', lty=2) 

推荐答案

如

As hinted at in this answer you should be able to use approx() for your task. E.g. like this:

xval <- approx(x = fit$fitted.values, y = x, xout = 30)$y

points(xval, 30, col = "blue", lwd = 5)

给你:

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