改进ggplot2中概率密度函数的绘图

Question

我正在使用 ggplot 绘制多个已知的密度函数，例如 gamma 密度函数：

library(tidyverse)
apar<-c(1,2,7.5,9)
bpar<-c(2,2,1.3,0.5)
gmaxlim<-c(0, 25)
pgma1<-ggplot(data = data.frame(x = gmaxlim), aes(gmaxlim)) +
  stat_function(fun = dgamma, n = 101, args = list(shape = apar[1], scale = bpar[1]),aes(color="black")) +
  stat_function(fun = dgamma, n = 101, args = list(shape = apar[2], scale = bpar[2]),aes(color="red")) +
  stat_function(fun = dgamma, n = 101, args = list(shape = apar[3], scale = bpar[3]),aes(color="blue")) +
  stat_function(fun = dgamma, n = 101, args = list(shape = apar[4], scale = bpar[4]),aes(color="green")) +
  ylab(expression(paste("f(x|",alpha,",",beta,")"))) +xlab("x") + scale_x_continuous(breaks=seq(gmaxlim[1],gmaxlim[2], by =5)) + 
  scale_color_identity(name = "",
                       breaks = c("black", "red", "blue","green"),
                       labels = c(substitute(paste(alpha,"= ", v," ,",beta,"= ",s),list(v=apar[1],s=bpar[1])),
                                  substitute(paste(alpha,"= ", v," ,",beta,"= ",s),list(v=apar[2],s=bpar[2])), 
                                  substitute(paste(alpha,"= ", v," ,",beta,"= ",s),list(v=apar[3],s=bpar[3])),
                                  substitute(paste(alpha,"= ", v," ,",beta,"= ",s),list(v=apar[4],s=bpar[4]))),
                       guide = "legend")+
  theme_bw()
pgma1

^{由reprex package (v0.3.0) 于 2020 年 7 月 31 日创建}

然而，这段代码远非高效，而且违背了 ggplot 理念（也许是因为我们没有绘制任何“真实”数据集？）。有没有办法更有效地编写这个并且可以扩展到不同数量的参数对？我想只写一行stat_function 并在可能的情况下简化scale_color_identity。在颜色标签中保留数学表达式是强制性的

Answer 1

也许使用一些 lapply？

library(tidyverse)
apar <- c(1,2,7.5,9)
bpar <- c(2,2,1.3,0.5)
gmaxlim <- c(0, 25)
mycols <- c("black", "red", "blue", "green")

ggplot(data = data.frame(x = gmaxlim), aes(gmaxlim)) +
lapply(seq_along(apar), function(i){
    stat_function(fun = dgamma, n = 101, 
    args = list(shape = apar[i], scale = bpar[i]), aes( color=mycols[i]))
}) +
    scale_color_identity(name="", breaks = mycols,
    labels = lapply(seq_along(apar), function(i) 
        substitute(paste(alpha,"= ", v," ,",beta,"= ",s),
            list(v=apar[i], s=bpar[i]))), guide = "legend") +
    theme_bw()

^{由reprex package (v0.3.0) 于 2020 年 7 月 31 日创建}

Answer 2

我有点困惑，为什么这么多人尝试使用 ggplot 中的 stat 函数做这么多事情，而不是传递他们实际想要绘制的数据。使用stat_function 可以直接绘制奇数线，但是试图强迫它做复杂的事情，比如通过引用外部向量来绘制分布族，这似乎很难做到。

它更容易推理，并且需要更少的代码，只需计算出你想要绘制的内容并绘制它：

apar <- c(1, 2, 7.5, 9)
bpar <- c(2, 2, 1.3, 0.5)
x    <- seq(0, 25, 0.25)
y    <- as.vector(sapply(1:4, function(i) dgamma(x, apar[i], scale = bpar[i])))
df   <- data.frame(x = rep(x, 4), y, group = rep(letters[1:4], each = length(x)))
labs <- sapply(1:4, function(i) {
               substitute(paste(alpha,"= ", v," ,",beta,"= ",s), 
               list(v = apar[i], s = bpar[i]))})

ggplot(data = df, aes(x, y)) + geom_line(aes(color = group)) +
  ylab(expression(paste("f(x|", alpha, ",", beta,")"))) +
  scale_color_manual(values = c(1, 2, 4, 3), labels = labs) +
  theme_bw()