绘制非连续函数ggplot2 [英] Plotting Noncontinuous Function ggplot2

问题描述

我正在尝试将此函数绘制在值0-1上:

I am attempting to plot this function over the values 0 - 1:

dweird <- function(x){
  if (x< 0){return(0)}
  if (x> 1){return(0)}
  if (x >= 0 & x < (1/3)) {return((1))}
  if (x >= (1/3) & x < (2/3)){return(3/2)}
  if (x >= (2/3) & x <= 1){return((1/2))}
}

这是我正在使用的代码

library(ggplot2)
ggplot(data.frame(x=c(0, 1)), aes(x)) + 
  stat_function(fun=function(x) dweird(x), linetype="dotted")

但这会返回错误消息

警告信息:如果if(x> = 0& x<(1/3)){:条件的长度> 1，并且只会使用第一个元素

Warning message: In if (x >= 0 & x < (1/3)) { : the condition has length > 1 and only the first element will be used

为清楚起见，该函数应绘制一条直线，其中y = 1从0-1/3开始，另一条直线在y = 1.5从1/3-2/3开始，另一条直线在1/2从2/3开始到1.

To be clear, the function should plot one straight line at y= 1 from 0-1/3, another at y=1.5 from 1/3-2/3, and another line at 1/2 from 2/3 to 1.

我为什么收到该错误消息的任何想法?

Any ideas why I'm getting that error message?

推荐答案

您需要向量化您的函数.ggplot不能一次评估一个点.懒惰的方法是使用 vectorize

You need to vectorize your function. ggplot doesn't expect to evaluate it one point at a time. The lazy way is to use vectorize

dweird_v_lazy = Vectorize(dweird)

但是更好的方法是首先以这种方式进行编码:

but the better way is to just code it that way in the first place:

dweird_v = function(x) {
    ifelse(x < 0, 0,
           ifelse(x < 1/3, 1,
                  ifelse(x < 2/3, 3/2,
                         ifelse(x < 1, 1/2, 0))))
}

# or, more concisely with `cut`:
dweird_cut = function(x) {
  as.numeric(as.character(
    cut(x,
        breaks = c(-Inf, 0, 1/3, 2/3, 1, Inf),
        labels = c(0, 1, 1.5, .5, 0)
     )
  ))
}

然后这将很好地工作:

x = seq(-.2, 1.2, length.out = 15)
dweird_v(x)
 [1] 0.0 0.0 0.0 1.0 1.0 1.0 1.5 1.5 1.5 0.5 0.5 0.5 0.0 0.0 0.0

您的情节也会如此:

library(ggplot2)
ggplot(data.frame(x=c(0, 1)), aes(x)) + 
    stat_function(fun= dweird_v, linetype="dotted")

请注意，当您将单个函数传递给 stat_function 时，不必将其转换为匿名函数，只需告诉它函数的名称即可.

Note that when you're passing a single function to stat_function, you don't have to turn it into an anonymous function, you can just tell it the name of your function.

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