一个图例行中有多个希腊字母

Question

我有以下一段 R 代码：

y1=rnorm(10,0,1)
y2=rnorm(10,0,1)
y3=rnorm(10,0,1)
x=seq(1,10,1)
mat=cbind(y1,y2,y3)
a=c(1,4,7)
b=c(2,5,8)
g=c(3,6,9)
v=as.factor(paste(expression(alpha)," = ",a,", ",
                  expression(beta)," = ",b,", ",
                  expression(gamma)," = ",g))
matplot(x,mat,type="l",col=1:3,lty=1,xlab="x",ylab="mat",main="Legend Practice")
legend("bottomright",col=1:3,lty=1,legend=levels(v))

在图例行中，结果显示为“alpha = 1,beta = 2, gamma = 3”，依此类推，对应于 3 个y 值的 3 行中的每一行。如何将其更改为 α = 1, β = 2, γ = 3 等等？

有什么想法吗？

Answer 1

我们可以为此使用bquote

matplot(x,mat,type="l",col=1:3,lty=1,xlab="x",ylab="mat",main="Legend Practice")
legend("bottomright",col=1:3,lty=1, legend=as.expression(c(
        bquote(alpha~'='~.(a[1])~', '~beta~'='~.(b[1])~', '~gamma~'='~.(g[1])),
        bquote(alpha~'='~.(a[2])~', '~beta~'='~.(b[2])~', '~gamma~'='~.(g[2])),
        bquote(alpha~'='~.(a[3])~', '~beta~'='~.(b[3])~', '~gamma~'='~.(g[3])) )))

这可以通过循环遍历@d.b 提到的向量序列来变得紧凑

legend("bottomright",col=1:3,lty=1, legend = as.expression(sapply(1:length(a),
  function(i) bquote(alpha~'='~.(a[i])~', '~beta~'='~.(b[i])~', '~gamma~'='~.(g[i])))))

或者mapply

legend("bottomright",col=1:3,lty=1, legend = as.expression(mapply(function(x,y, z) 
       bquote(alpha~'='~.(x)~', '~beta~'='~.(y)~', '~gamma~'='~.(z)),
               a, b, g)))