为线图的特定段选择颜色的优雅方法? [英] Elegant way to select the color for a particular segment of a line plot?

问题描述

对于 n 对坐标 x,y 的列表，有没有办法在特定颜色上绘制不同点之间的线?

For a list of n pairs of coordinates x,y is there a way of plotting the line between different points on a specific color?

到目前为止我实现的解决方案不是使用 plot 函数，而是 lines 选择我想要颜色的范围.举个例子:

The solution I've implemented so far is not to use the plot function but lines selecting the range for which I want the color. Here an example:

x <- 1:100
y <- rnorm(100,1,100)
plot(x,y ,type='n')
lines(x[1:50],y[1:50], col='red')
lines(x[50:60],y[50:60], col='black')
lines(x[60:100],y[60:100], col='red')

有没有更简单的方法来做到这一点?

Is there an easier way of doing this?

推荐答案

是的，一种方法是使用 ggplot.

Yes, one way of doing this is to use ggplot.

ggplot 要求您的数据采用 data.frame 格式.在这个 data.frame 中，我添加了一列 col 来指示您想要的颜色.然后使用 ggplot、geom_line 和 scale_colour_identity 构建图，因为 col 变量已经是一种颜色:

ggplot requires your data to be in data.frame format. In this data.frame I add a column col that indicates your desired colour. The plot is then constructed with ggplot, geom_line, and scale_colour_identity since the col variable is already a colour:

library(ggplot2)

df <- data.frame(
  x = 1:100,
  y = rnorm(100,1,100),
  col = c(rep("red", 50), rep("black", 10), rep("red", 40))
)

ggplot(df, aes(x=x, y=y)) + 
  geom_line(aes(colour=col, group=1)) + 
  scale_colour_identity()

更一般地，每个线段可以是不同的颜色.在下一个示例中，我将颜色映射到 x 值，给出了一个从蓝色平滑地将颜色变为红色的图:

More generally, each line segment can be a different colour. In the next example I map colour to the x value, giving a plot that smoothly changes colour from blue to red:

df <- data.frame(
  x = 1:100,
  y = rnorm(100,1,100)
)

ggplot(df, aes(x=x, y=y)) + geom_line(aes(colour=x))

如果你坚持使用基础图形，那么使用 segments 如下:

And if you insist on using base graphics, then use segments as follows:

df <- data.frame(
  x = 1:100,
  y = rnorm(100,1,100),
  col = c(rep("red", 50), rep("black", 10), rep("red", 40))
)

plot(df$x, df$y, type="n")
for(i in 1:(length(df$x)-1)){
  segments(df$x[i], df$y[i], df$x[i+1], df$y[i+1], col=df$col[i])
}

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