我试图绘制一个图以显示逻辑(或概率)回归背后的直觉。我将如何在ggplot中绘制出类似以下内容的图?
(Wolf&Best,《贤者回归分析和因果推断手册》,2015年,第155页)
实际上,我更愿意做的是沿y轴显示一个均值= 0的正态分布,并具有特定的方差,以便我可以绘制从线性预测器到y轴并横向的水平线正态分布。像这样:
应该显示的内容是(假设我没有误解)。到目前为止,我还没有取得太大的成就...
library(ggplot2)
x <- seq(1, 11, 1)
y <- x*0.5
x <- x - mean(x)
y <- y - mean(y)
df <- data.frame(x, y)
# Probability density function of a normal logistic distribution
pdfDeltaFun <- function(x) {
prob = (exp(x)/(1 + exp(x))^2)
return(prob)
}
# Tried switching the x and y to be able to turn the
# distribution overlay 90 degrees with coord_flip()
ggplot(df, aes(x = y, y = x)) +
geom_point() +
geom_line() +
stat_function(fun = pdfDeltaFun)+
coord_flip()
答案 0 :(得分:1)
我认为这与您提供的第一个插图非常接近。如果您不需要重复很多次,最好在绘制图形之前先计算密度曲线,然后使用单独的数据框进行绘制。
library(ggplot2)
x <- seq(1, 11, 1)
y <- x*0.5
x <- x - mean(x)
y <- y - mean(y)
df <- data.frame(x, y)
# For every row in `df`, compute a rotated normal density centered at `y` and shifted by `x`
curves <- lapply(seq_len(NROW(df)), function(i) {
mu <- df$y[i]
range <- mu + c(-3, 3)
seq <- seq(range[1], range[2], length.out = 100)
data.frame(
x = -1 * dnorm(seq, mean = mu) + df$x[i],
y = seq,
grp = i
)
})
# Combine above densities in one data.frame
curves <- do.call(rbind, curves)
ggplot(df, aes(x, y)) +
geom_point() +
geom_line() +
# The path draws the curve
geom_path(data = curves, aes(group = grp)) +
# The polygon does the shading. We can use `oob_squish()` to set a range.
geom_polygon(data = curves, aes(y = scales::oob_squish(y, c(0, Inf)),group = grp))
第二个插图非常接近您的代码。我通过标准的普通密度函数简化了密度函数,并向stat函数添加了一些额外的参数:
library(ggplot2)
x <- seq(1, 11, 1)
y <- x*0.5
x <- x - mean(x)
y <- y - mean(y)
df <- data.frame(x, y)
ggplot(df, aes(x, y)) +
geom_point() +
geom_line() +
stat_function(fun = dnorm,
aes(x = after_stat(-y * 4 - 5), y = after_stat(x)),
xlim = range(df$y)) +
# We fill with a polygon, squishing the y-range
stat_function(fun = dnorm, geom = "polygon",
aes(x = after_stat(-y * 4 - 5),
y = after_stat(scales::oob_squish(x, c(-Inf, -1)))),
xlim = range(df$y))