从数据绘制 3D 表面

Question

I have written C++ code to numerically solve a PDE. I would like to plot the result. I have outputted the data to an ascii file, as 3 columns of numbers. The x-coordinate, the y-coordinate and the z-coordinate. This might look like

0.01 7 -3
-12 1.2 -0.24
...

I often have in excess of 1000 data points. I want to plot a surface. I was able to load the data in both R and octave. In R scatterplot3D worked, and in octave plot3 worked. However, I wish to produce a surface, and not distinct points (scatterplot3d), or a curve (plot3). I am struggling to get mesh or surf to work from data in octave. I am looking for a simple way to plot a surface in 3D space with octave, R, C++ or any other program.

Answer 1

You could coerce the data into the correct format for plotting with the base R function persp . This requires a vector of unique x values, a vector of unique y values, and a matrix of z values which is a length(unique(x)) by length(unique(y)) matrix.

Suppose your data looks like this:

x    <- y <- seq(-pi, pi, length = 20)
df   <- expand.grid(x = x, y = y)
df$z <- cos(df$x) + sin(df$y)

head(df)
#>           x         y           z
#> 1 -3.141593 -3.141593 -1.00000000
#> 2 -2.810899 -3.141593 -0.94581724
#> 3 -2.480205 -3.141593 -0.78914051
#> 4 -2.149511 -3.141593 -0.54694816
#> 5 -1.818817 -3.141593 -0.24548549
#> 6 -1.488123 -3.141593  0.08257935

Then you can create a matrix like this:

z <- tapply(df$z, list(df$x, df$y), mean)

So your plot would look like this:

persp(unique(df$x), unique(df$y), z, 
      col = "gold", theta = 45, shade = 0.75, ltheta = 90)

If your x and y co-ordinates are not nicely aligned, then a more general approach would be:

z <- tapply(df$z, list(cut(df$x, 20), cut(df$y, 20)), mean, na.rm = TRUE)

persp(as.numeric(factor(levels(cut(df$x, 20)), levels(cut(df$x, 20)))), 
      as.numeric(factor(levels(cut(df$y, 20)), levels(cut(df$y, 20)))), 
      z, col = "gold", theta = 45, shade = 0.75, ltheta = 90, xlab = "x", 
      ylab = "y")