Mickaël Perrier - 9 months ago 34

R Question

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

`persp()`

`x`

`y`

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.

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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):

`require('png')`

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.,

`e1`

`e2`

`e1`

`e2`

`E1`

`E2`

`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`

`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

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)
```

Source (Stackoverflow)