Mickaël Perrier Mickaël Perrier - 1 year ago 61
R Question

How to 3D plot matrices that are not squared in R?

I would like to plot a heatmap using my own mathematical function instead of the Kernel density estimation. But for the moment my problem comes from the fact that I cannot 3D plot this function using

if my
axes are not squared. Indeed the heatmap is sized 855 x 670.

1) Is there a way to solve this problem?

2) Plus, does anyone know how to turn this into a heatmap?

Thanks in advance. Please, find a part of my script below.


Here are two functions we will need:

rep.row <- function(x, n){
matrix(rep(x, each = n), nrow = n)
rep.col <- function(x, n){
matrix(rep(x, each = n), ncol = n, byrow = TRUE)

This reads an image and extracts its dimensions (i.e., width and length):

png <- readPNG("myImage.png")
res <- dim(png)[2:1]

For information:

> dim(png)[2:1]
[1] 855 670

These are fixed parameters:

alphaW <- 53
alphaH <- 31
a <- 2.3

I create two vectors (i.e.,
) based on the image dimensions. Therefore,
is 855-cells long and
is 670-cells long. Then I use the functions above in order to create two matrices (i.e.,
) the same size of the image (i.e., 855 x 670).

e1 <- seq(-alphaW, alphaW, length = res[1])
e2 <- seq(-alphaH, alphaH, length = res[2])

E1 <- rep.row(e1, res[2])
E2 <- rep.col(e2, res[1])

A computation of these two matrices is used to create a 3rd matrix,

SV <- sqrt((a / (a + ((E1^2) + (E2^2)))))

Finally I want to plot a 3D representation of this matrix:

persp(x = e1, y = e2, z = SV,
col = "lightgoldenrod",
border = NA,
theta = 30,
phi = 15,
ticktype = "detailed",
ltheta = -120,
shade = 0.25)

This should output something like {this}, however, I receive:

Error in persp.default(e1, e2, SV, col = "lightgoldenrod", border = NA, :
argument 'z' incorrect

Answer Source

You switched x and y. If you look at the help page for persp(), you will notice that x should have length nrow(z) and y length ncol(z). So although intuitively you might expect rows to be on the vertical axis (how you visualise the matrix), it seems to be the other way around.

This works:

persp(y = e1, x = e2, z = SV,
      col = "lightgoldenrod",
      border = NA,
      theta = 30,
      phi = 15,
      ticktype = "detailed",
      ltheta = -120,
      shade = 0.25)
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