data <- dput(data): structure(list(x = 1:16, y = c(-79.62962963, -84.72222222, -88.42592593, -74.07407407, -29.62962963, 51.38888889, 79.62962963, 96.2962963, 87.96296296, 88.42592593, 73.14814815, 12.96296296, -63.42592593, -87.03703704, -87.5, -87.96296296)), .Names = c("x", "y"), row.names = c(NA, 16L), class = "data.frame")
I calculated in R with ggplot2 a smooth line for my dataset:
p1 <- ggplot(data, aes(x=x(°), y=(%)))
library(splines)
library(MASS)
(p2 <- p1 + stat_smooth(method = "lm", formula = y ~ ns(x,3)) +
geom_point()
)
How can I calculate the x-value of the curve maximum of the smooth line?
You'll need to do some math on the resultant data.frame calculated by stat_smooth
:
library(ggplot2)
library(splines)
library(MASS)
data <- structure(list(x = 1:16,
y = c(-79.62962963, -84.72222222, -88.42592593,
-74.07407407, -29.62962963, 51.38888889,
79.62962963, 96.2962963, 87.96296296,
88.42592593, 73.14814815, 12.96296296,
-63.42592593, -87.03703704, -87.5,
-87.96296296)), .Names = c("x", "y"),
row.names = c(NA, 16L), class = "data.frame")
p1 <- ggplot(data, aes(x=x, y=y))
p1 <- p1 + stat_smooth(method = "lm", formula = y ~ ns(x,3))
p1 <- p1 + geom_point()
p1
gb <- ggplot_build(p1)
exact_x_value_of_the_curve_maximum <- gb$data[[1]]$x[which(diff(sign(diff(gb$data[[1]]$y)))==-2)+1]
p1 + geom_vline(xintercept=exact_x_value_of_the_curve_maximum)
exact_x_value_of_the_curve_maximum
[1] 9.164557
There are more robust ways as well, but you still need the ggplot_build
part to get to the data.