I'm trying to place the results of a regression (i.e., R2) in a graph, but can't seem to figure out how to call a variable from within an expression (it pastes the variable name).
Here is my code.
R2Val <- signif(summary(sMod_pre90)$r.squared[1], 2)
text(92, 4, expression(paste(R^2, " = ", R2Val)), adj = 0, cex = 0.85)
Use bquote(). Here is an example with dummy data:
set.seed(1)
DF <- data.frame(A = rnorm(100), B = rnorm(100))
mod <- lm(B ~ A, data = DF)
R2Val<-signif(summary(mod)$r.squared[1], 2)
The parts of the expression wrapped in .() get evaluated in the environment, i.e. the value of R2Val is substituted.
plot(B ~ A, data = DF)
text(1.5, 2, labels = bquote(R^2 == .(R2Val)), adj = 0, cex = 0.85)
Another potential solution is substitute, which works similarly:
plot(B ~ A, data = DF)
text(1.5, 2, labels = substitute(R^2 == A, list(A = R2Val)), adj = 0, cex = 0.85)
To combine substitute() and paste(), I use the following code;
waic <-1;
chisquare <-2;
plot(x=1:2,y=1:2,
main = substitute(paste(
"chi^2 goodness of fit ",
chi^2*(D/theta) , "=", chisquare ,
". WAIC =",waic),
list(chisquare=chisquare,waic=waic)
)
)
The result is the following;
I've managed to put it together using the substitute function.
R2Val <- signif(summary(sMod_pre90)$r.squared[1], 2)
text(92, 4, substitute(R^2 ~ "=" ~ R2Val), adj = 0, cex = 0.85)
All is good.
substitute which variables to substitute. See my answer.
R2Val is printed literally, rather than being interpreted, on the plot. I think ?substitute is quite clear that unless you provide something for the env argument substitution will not take place in this example.