如何在ggplot2`geom_curve`函数中传递各个曲率参数? [英] How can I pass individual `curvature` arguments in `ggplot2` `geom_curve` function?

问题描述

我有一个带有两个曲线定义的df，每个曲线定义由两个点和一个曲率值组成.目标是使用ggplot2 geom_curve(或替代方法)绘制两条单独的曲线.

I have a df with two curve definitions, each consists of two points and a curvature value. The goal is to plot two individual curves using ggplot2 geom_curve (or an alternative).

我可以使用以下命令生成期望的输出:

I can generate my expected output using:

df <- data.frame(x = c(0,.2), y = c(0,.3), xend = c(1,.4), yend = c(1,.6), curvature = c(-.2,.4))
ggplot(df) + geom_curve(data = df[1, ], aes(x = x, y = y, xend = xend, yend = yend), curvature = df$curvature[1]) + geom_curve(data = df[2, ], aes(x = x, y = y, xend = xend, yend = yend), curvature = df$curvature[2])

但这并不是真正的解决方案，因为在我的实际情况下，我有更多的曲线(而且我不知道事先有多少条曲线).

如何将单个curvature参数传递给geom_curve调用?

How can I pass an individual curvature argument to the geom_curve call?

我尝试过:

df <- data.frame(x = c(0,0), y = c(0,0), xend = c(1,1), yend = c(1,1), curvature = c(-.2,.8))
library(ggplot2)
ggplot(df) + geom_curve(aes(x = x, y = y, xend = xend, yend = yend, curvature = curvature))

这会使两条曲线相互重叠，并引发其他警告:

This plots both curves on top of each other and throws an additional warning:

警告:忽略未知的美学因素:曲率

Warning: Ignoring unknown aesthetics: curvature

---

所以我尝试了:

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So I tried:

ggplot(df) + geom_curve(aes(x = x, y = y, xend = xend, yend = yend), curvature = curvature)

这会引发错误:

图层错误(数据=数据，映射=映射，统计=统计，geom = GeomCurve 、: 找不到对象曲率"

Error in layer(data = data, mapping = mapping, stat = stat, geom = GeomCurve, : object 'curvature' not found

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所以我试图显式传递curvature冒号:

---

So I tried to explicitly pass the curvature colon:

ggplot(df) + geom_curve(aes(x = x, y = y, xend = xend, yend = yend), curvature = df$curvature)

这也会引发错误:

seq.default(0，dir * maxtheta，dir * maxtheta/(ncp + 1))中的错误:
'to'mussLänge1 haben另外:警告消息:1:如果发生 (曲率== 0){:条件的长度> 1，并且只有第一个 元素将被使用2:如果if(曲率> 0)手<-右"则else hand<-"left":条件的长度> 1，并且只有第一个 元素将被使用

Error in seq.default(0, dir * maxtheta, dir * maxtheta/(ncp + 1)) :
'to' muss Länge 1 haben In addition: Warning messages: 1: In if (curvature == 0) { : the condition has length > 1 and only the first element will be used 2: In if (curvature > 0) hand <- "right" else hand <- "left" : the condition has length > 1 and only the first element will be used

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我从@markus的解决方案中学到，我们可以将lists传递给ggplot对象，所以我尝试了:

---

From @markus' solution I learned, that we can pass lists to a ggplot object, so I tried:

ggplot(df) + 
  lapply(df$curvature, function(i) {
    geom_curve(aes(x = x, y = y, xend = xend, yend = yend), curvature = i) }
  )

但这会用两个curvature参数绘制每条曲线:

But this plots each curve with both curvature argument:

如何为每一行分别传递curvature自变量?

How can I pass that curvature argument individually for each row?

推荐答案

更新

您可以先拆分数据，然后使用lapply遍历结果列表，然后我们将其馈送到geom_curve()

You might split your data first and then use lapply to iterate over the resulting list which we'll feed to the data argument of geom_curve()

df2 <- data.frame(x = c(0,.2), y = c(0,.3), xend = c(1,.4), yend = c(1,.6), curvature = c(-.2,.4))
ggplot() + 
  lapply(split(df2, 1:nrow(df)), function(dat) {
    geom_curve(data = dat, aes(x = x, y = y, xend = xend, yend = yend), curvature = dat["curvature"]) }
  )

原始ansewr

curvature并不是一种美学.您可以将列表添加到ggplot()，以使其正常工作

curvature is not an aesthetic, as you have noted. You can add a list to ggplot(), to get it work

df <- data.frame(x = c(0,0), y = c(0,0), xend = c(1,1), yend = c(1,1), curvature = c(-.2,.8))
ggplot(df) + 
  lapply(df$curvature, function(i) {
    geom_curve(aes(x = x, y = y, xend = xend, yend = yend), curvature = i) }
    )

来自help("+.gg")

您可以添加什么?

...

您还可以提供一个列表，在这种情况下，列表中的每个元素都会依次添加.

You can also supply a list, in which case each element of the list will be added in turn.

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如果要在绘图中显示其他参数-每条线的颜色可能不同，大小不同等-使用Map

修改后的数据

df1 <- data.frame(x = c(0,0), y = c(0,0), xend = c(1,1), yend = c(1,1), curvature = c(-.2,.8),
                  colour = c("red", "blue"))

情节

ggplot(df1) + 
  Map(function(i, col) {
    geom_curve(aes(x = x, y = y, xend = xend, yend = yend), curvature = i, colour = col) },
    i = df1$curvature, col = df1$colour
  )

结果

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