vector specification to "col", "lwd" and "lty" not working when drawing lines?

Question

I have matrix dsts with 3 columns; third is a factor. I want my linear plot to be colored by the factor but this command is not working:

plot(dsts[ ,'x'],dsts[,'dist'],col=dsts[,'i'],type='l')

and,

plot(dsts[ ,'x'],dsts[,'dist'],col=dsts[,'i'],type='n')
lines(dsts[ ,'x'],dsts[,'dist'],col=dsts[,'i']) 

is not working either!!!

I want to avoid using matplot which accepts matrices.

Answer 1

On top of @Zheyuan Li valuable insight, I wrote a simple function to solve the problem:

plot_line_color <- function(x,y,fact,lwd=2,...) 
{
  plot(x,y,type='n')
  xy <- cbind(x,y)
  invisible(
   lapply(1:length(unique(fact)), function(j) { 
     xy2 <- subset(xy,fact==j)
    lines(xy2[ ,1],xy2[,2],col=j,lwd=lwd,...)
    })
  )
}

A simple simulation:

k <- 1:5
x <- seq(0,10,length.out =  100)
dsts <- lapply(1:length(k), function(i) cbind(x=x, distri=dchisq(x,k[i]),fact=i) )
dsts <- do.call(rbind,dsts)

plot_line_color(x=dsts[,1],y=dsts[,2],fact=dsts[,3])

Answer 2

The col option, though able to take vector input, only effectively controls point colour instead of line colour, so type = "p" works but not pch = "l". For pch = "b", only points will have correct colours.

If you want to have several lines with different colours, you have to plot them with separate plot or lines calls. A better way to go is to reshape your data, then use matplot. It takes a matrix, and plot its columns one by one via a for loop.

Since you've already got a function to reshape data, you have the right way to go.

---

The reason that plot and lines depreciate vector values in col for line display, is that they have no idea of whether this vector has a reasonable, non-random pattern. They will do something safe, by using only col[1]. I will elaborate on this by two steps.

Firstly, consider this example to see that plot will always use col[1] when type = "l":

set.seed(0); mat1 <- round(cbind(rnorm(9),rnorm(9),rep(1:3, each = 3)), 1)
#      [,1] [,2] [,3]
# [1,]  1.3  2.4    1
# [2,] -0.3  0.8    1
# [3,]  1.3 -0.8    1
# [4,]  1.3 -1.1    2
# [5,]  0.4 -0.3    2
# [6,] -1.5 -0.3    2
# [7,] -0.9 -0.4    3
# [8,] -0.3  0.3    3
# [9,]  0.0 -0.9    3

Then we reorder the rows of mat1:

mat2 <- mat1[c(4:9,1:3), ]
#      [,1] [,2] [,3]
# [1,]  1.3 -1.1    2
# [2,]  0.4 -0.3    2
# [3,] -1.5 -0.3    2
# [4,] -0.9 -0.4    3
# [5,] -0.3  0.3    3
# [6,]  0.0 -0.9    3
# [7,]  1.3  2.4    1
# [8,] -0.3  0.8    1
# [9,]  1.3 -0.8    1

We use the 3rd column for col, now compare:

par(mfrow = c(1,2))
plot(mat1[,1], mat1[,2], col = mat1[,3], type = "l")
plot(mat2[,1], mat2[,2], col = mat2[,3], type = "l")

mat1[, 3] starts with 1, so the line colour is black; mat2[,3] starts with 2, so the line colour is red.

Now it is time to say why plot and lines depreciate vector col when type = "l". Consider a random row shuffle of mat1:

set.seed(0); mat3 <- mat1[sample(9), ]
#      [,1] [,2] [,3]
# [1,]  0.0 -0.9    3
# [2,]  1.3 -0.8    1
# [3,] -0.3  0.3    3
# [4,]  1.3 -1.1    2
# [5,]  0.4 -0.3    2
# [6,]  1.3  2.4    1
# [7,] -0.9 -0.4    3
# [8,] -0.3  0.8    1
# [9,] -1.5 -0.3    2

plot(..., type = "l") will line up points one by one. Be aware that a line of a single colour can only be drawn, if data points on this path have the same colour specification. Now, the 3rd column is completely random: there is no way to line points up with such colour specification.

The best & safest assumption plot and lines can take is that col vector is completely random. Thus, it will only retain col[1] to produce a single colour plot. The full vector will only be used, when type = "p".

Note, the same logic applies to lwd and lty, too. Any argument associated with line display will take only the first vector element. As I said earlier, if you do want to draw several different lines in different styles, do them one by one.