Find r-squared of points around an x=y line (R)

I have some data I obtained from a lab experiment that shows the theoretical value of variable A, against its actual, experimentally derived value, for a range of variable A.

I've plotted the data as scatter plots and have an x=y line to demonstrate what would have been 100% accurate. I'd like to quantify this accuracy using R-squared.

minimal dataset:

library(ggplot2)

df <- data.frame(pred = c(-20, -10, 0, 10, 20),
real = c(-15, -7, 2, 8, 22))

plot <- ggplot(df, aes(real, pred))+
geom_point(color = "Red", size =4, shape = 4)+
geom_abline(color= "black", size = 0.6)

plot


So usually when I'm assessing the accuracy of a model I look at the R-squared in the summary but as both the real and predicted value were found experimentally I can't do that.

I feel like my best bet is to find the R-squared of these points against the line x=y, but I can't find anywhere how to do that without going through an lm, but surely any coefficients would meddle with the accuracy that I'm looking for?

I also tried defining R-squared manually as follows:

rsq <- function (x, y) {
rsq_scors <- cor(x, y) ^ 2
return(rsq_scors)
}

score <- rsq(df$$pred,df$$real)

# which returns:
[1] 0.9985488


Which looks okay except when you consider that if I multiple the real values by 10, I get the same outcome, which is why I really wan't to track it to x=y:

df <- data.frame(pred = c(-20, -10, 0, 10, 20),
real = c(-150, -70, 20, 100, 200))

score <- rsq(df$$pred,df$$real)

# which also returns:
[1] 0.9985488

• Be careful with the interpretation when you aren’t finding the line through a least squares fit, however. // You might be more interested in a measure like mean absolute deviation that measures the average amount by which predictions differ from observed values (slightly different from root mean squared deviation/error).
– Dave
Oct 12 at 11:00
• Because the squared distance from a point $(x,y)$ to the diagonal line is simply $(x-y)^2/2,$ just compute these values and average them to obtain a mean squared difference. You do not want to divide by some "variance" to create an $R^2$ value because the result will be deceptively high and will depend somewhat arbitrarily on the range of experimental values, making it impossible to make valid comparisons between any two $R^2$ values.
– whuber
Oct 12 at 14:54

$$R^2$$ aka Coefficient of determination has the following meaning:
$$R^2=1 - \frac{SS_{residual}}{SS_{total}}$$, where $$SS$$ denotes sum of squares and we have sum of squares of the residual divided by total sum of squares. A more meaningful way is to put the number of observations $$n$$ in denominator and numerator and get sample estimates for variance of residuals divided by total variance.
Since you model is fixed to $$y = 0 + 1*x$$. Than you have $$SS_{residual} = \sum_{i=1}^n (y_i - x_i)^2$$ and the initial sum of squares is $$SS_{total} = \sum_{i=1}^n (y_i - \bar{y})^2$$.
• This is not the $R^2$ described in the question.