R help - Jan 2021 - How to predict/interpolate new Y given knwon Xs and Ys?

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

model2 <- lm( x~y )
predict(model2, data.frame(y=26))

model2 is however not the inverse of model... if you need that then you need to
handle that some other way than using predict, such as an invertible monotonic
spline (or in this case a little algebra).

On January 26, 2021 1:11:39 AM PST, Luigi Marongiu <marongiu.luigi at
gmail.com> wrote:
>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
>
>______________________________________________
>R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
-- 
Sent from my phone. Please excuse my brevity.

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

?s 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
> 
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>