Hello,
You can predict y on x, not the other way around, like you are doing in
the second call to predict.lm.
The 10 values you are getting are the predicted values on the original x
values, just see that x=7.5 gives ypred=30, right in the middle of x=7
and x=8 -> ypred=29 and ypred=31.
As for the inverse regression, how do you account for the errors? In
linear regression the only rv is the errors vector, the inverse of
y = a + b*x + e
is not
x = (y - a)/b
though you can write a function that computes this value:
pred_x <- function(model, newdata){
beta <- coef(model)
y <- newdata[[1]]
x <- (y - beta[1])/beta[2]
unname(x)
}
pred_x(model, data.frame(y = 26))
#[1] 5.5
There is a CRAN package, investr that computes the standard errors:
investr::calibrate(model, y0 = 26)
#estimate lower upper
# 5.5 5.5 5.5
See the decumentation in [1]
[1] https://CRAN.R-project.org/package=investr
Hope this helps,
Rui Barradas
Às 09:11 de 26/01/21, Luigi Marongiu escreveu:
Hello,
I have a series of x/y and a model. I can interpolate a new value of x
using this model, but I get funny results if I give the y and look for
the correspondent x:
```
x = 1:10
y = 2*x+15
model <- lm(y~x)
predict(model, data.frame(x=7.5))
1
30
predict(model, data.frame(y=26))
1 2 3 4 5 6 7 8 9 10
17 19 21 23 25 27 29 31 33 35
Warning message:
'newdata' had 1 row but variables found have 10 rows
data.frame(x=7.5)
x
1 7.5
data.frame(y=26)
y
1 26
```
what is the correct syntax?
Thank you
Luigi
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