问题
http://www.texample.net/tikz/examples/lindenmayer-systems/
My sample code shown below, I don't know how to colored with hue color.
plot.koch <- function(k,col="blue"){
plot(0,0,xlim=c(0,1), ylim=c(-sqrt(3)/6,sqrt(3)/2), asp = 1,type="n",xlab="", ylab="")
plotkoch <- function(x1,y1,x2,y2,n){
if (n > 1){
plotkoch(x1,y1,(2*x1+x2)/3,(2*y1+y2)/3,n-1);
plotkoch((2*x1+x2)/3,(2*y1+y2)/3,(x1+x2)/2-(y1-y2)*sqrt(3)/6,(y1+y2)/2-(x2-x1) *sqrt(3)/6,n-1);
plotkoch((x1+x2)/2-(y1-y2)*sqrt(3)/6,(y1+y2)/2-(x2-x1)*sqrt(3)/6,(2*x2+x1)/3,(2 *y2+y1)/3,n-1);
plotkoch((2*x2+x1)/3,(2*y2+y1)/3,x2,y2,n-1)
}
else {
x=c(x1,(2*x1+x2)/3,(x1+x2)/2-(y1-y2)*sqrt(3)/6,(2*x2+x1)/3,x2);
y=c(y1,(2*y1+y2)/3,(y1+y2)/2-(x2-x1)*sqrt(3)/6,(2*y2+y1)/3,y2);
polygon(x,y,type="l",col=col)
}
}
plotkoch(0,0,1,0,k)
plotkoch(0.5,sqrt(3)/2,0,0,k)
plotkoch(1,0,0.5,sqrt(3)/2,k)
}
plot.koch(3, col=3)
回答1:
Here's a method using spatial objects in R, with sp, rgeos and raster packages in the mix.
Slight modifications to the function to return the
x,ycoordinates to the user (and in the correct order):koch <- function(k) { yy <- xx <- numeric(0) Koch <- function(x1, y1, x2, y2, n) { if (n > 1){ Koch(x1, y1, (2*x1+x2)/3, (2*y1+y2)/3, n-1); Koch((2*x1+x2)/3, (2*y1+y2)/3, (x1+x2)/2-(y1-y2)*sqrt(3)/6, (y1+y2)/2-(x2-x1) *sqrt(3)/6, n-1); Koch((x1+x2)/2-(y1-y2)*sqrt(3)/6, (y1+y2)/2-(x2-x1)*sqrt(3)/6, (2*x2+x1)/3, (2 *y2+y1)/3, n-1); Koch((2*x2+x1)/3, (2*y2+y1)/3, x2, y2, n-1) } else { x <- c(x1, (2*x1+x2)/3, (x1+x2)/2-(y1-y2)*sqrt(3)/6, (2*x2+x1)/3, x2); xx <<- c(xx, x) y <- c(y1, (2*y1+y2)/3, (y1+y2)/2-(x2-x1)*sqrt(3)/6, (2*y2+y1)/3, y2); yy <<- c(yy, y) } } Koch(0, 0, 1, 0, k) Koch(1, 0, 0.5, sqrt(3)/2, k) Koch(0.5, sqrt(3)/2, 0, 0, k) xy <- data.frame(x=xx, y=yy) rbind(unique(xy), xy[1, ]) }Create a colour ramp:
colr <- colorRampPalette(hcl(h=seq(0, 360, len=100), c=100))Use
kochfunction to get vertices:xy <- koch(4)Load spatial packages and create
SpatialPolygonsobject from fractal and plot it once to set up the plot area.library(sp) library(rgeos) library(raster) poly <- SpatialPolygons(list(Polygons(list(Polygon(xy)), 1))) plot(poly)Plot a series of segments with desired origin and large enough radius to cover the fractal polygon (here we use radius
r <- 1).r <- 1 mapply(function(theta, col) { segments(0.5, 0.3, 0.5 + r*cos(theta), 0.3 + r*sin(theta), lwd=3, col=col) }, seq(0, 360, length=1000)*pi/180, colr(1000))Create a second polygon of the difference between the plot area and the fractal polygon, and plot this (with
col='white') to mask out the unwanted gradient area.plot(gDifference(as(extent(par('usr')), 'SpatialPolygons'), poly), col='white', border='white', add=TRUE)Plot the polygon once more.
plot(poly, add=TRUE)
回答2:
Here's my attempt at solving your question. Currently it draws the color also outside of the snowflake. If you can figure out if points are inside or outside the snowflake, you should be able to just remove outside points in the df_fill.
Here I'm first creating the data.frame used for plotting the polygon. Then I'm creating the data.frame for the background color. And finally I'm using ggplot2 to plot the data.
# creating relevant data
data.koch <- function(k){
df <- data.frame(x = 0,
y = 0,
grp = 0)
plotkoch <- function(x1, y1, x2, y2, n, data){
if (n==1) {
x=c(x1,(2*x1+x2)/3,(x1+x2)/2-(y1-y2)*sqrt(3)/6,(2*x2+x1)/3,x2)
y=c(y1,(2*y1+y2)/3,(y1+y2)/2-(x2-x1)*sqrt(3)/6,(2*y2+y1)/3,y2)
df <- rbind(data, data.frame(x, y, grp=max(data$grp)+1))
}
if (n > 1){
df <- plotkoch(x1,y1,(2*x1+x2)/3,(2*y1+y2)/3,n-1, data = data)
df <- plotkoch((2*x1+x2)/3,(2*y1+y2)/3,(x1+x2)/2-(y1-y2)*sqrt(3)/6,(y1+y2)/2-(x2-x1) *sqrt(3)/6,n-1, data=df)
df <- plotkoch((x1+x2)/2-(y1-y2)*sqrt(3)/6,(y1+y2)/2-(x2-x1)*sqrt(3)/6,(2*x2+x1)/3,(2 *y2+y1)/3,n-1, data=df)
df <- plotkoch((2*x2+x1)/3,(2*y2+y1)/3,x2,y2,n-1, data=df)
}
return(df)
}
df <- plotkoch(0,0,1,0,k, data = df)
df <- plotkoch(0.5,sqrt(3)/2,0,0,k, data = df)
df <- plotkoch(1,0,0.5,sqrt(3)/2,k, data = df)
return(df)
}
# plotting functon
plot.koch <- function(k){
stopifnot(require(ggplot2))
if (is.data.frame(k)) df <- k
else df <- data.koch(k)
# filling data (CHANGE HERE TO GET ONLY INSIDE POINTS)
l <- 500
df_fill <- expand.grid(x=seq(0, 1, length=l),
y=seq(-sqrt(3)/6, sqrt(3)/2, length=l))
df_fill[, "z"] <- atan2(-df_fill[, "y"] + sqrt(3)/6, df_fill[, "x"] - 0.5) + pi/2
df_fill[df_fill[, "z"] < 0, "z"] <- df_fill[df_fill[, "z"] < 0, "z"] + 2*pi
# plotting
ggplot(df, aes(x, y, group=grp)) +
geom_raster(data = df_fill,
aes(fill=z, group=NULL),
hjust = 0,
vjust = 0,
linetype='blank') +
geom_path(data=df, size=1) +
scale_fill_gradientn(colours = rainbow(30), guide = 'none') +
scale_x_continuous(name = '', limits = c(0, 1), expand=c(0, 0)) +
scale_y_continuous(name = '', limits = c(-sqrt(3)/6,sqrt(3)/2), expand=c(0, 0)) +
coord_fixed() +
theme_bw() +
theme(axis.line = element_blank(),
panel.grid = element_blank(),
axis.ticks = element_blank(),
axis.text = element_blank())
}
#
p <- plot.koch(4)
print(p)
回答3:
I would do it like this:
- for any drawed pixel obtain its position
x,y compute the
angle=atan2(y-y0,x-x0)where
x0,y0is the koch's snowflake mid positioncompute the color based on angle
if you use HSV then
hue=angleand compute the target color value (I assume RGB). If you want the visible spectra colors you can try mine:- RGB values of visible spectrum
Just convert the angle range
angle=<0,2*Pi> [rad]to wavelengthl=<400,700> [nm]so:l = 400.0 + (700.0-400.0)*angle/(2.0*M_PI)render the pixel
[Notes]
not using R nor Matlab so you need to code it yourself. The angle may need some shifting to match your coordinate system for example:
const angle0=???; // some shift constant [rad]
angle+=angle0; // or angle=angle0-angle; if the direction is oposite
if (angle>=2.0*M_PI) angle-=2.0*M_PI;
if (angle< 0.0) angle+=2.0*M_PI;
If you drawing this as polygon then you need to compute color per vertex not per pixel but then you can get to problems because this is not convex polygon. So how to ensure the mid point color ??? I am afraid you will need to use some sort of triangulation because simple triangle fan will fail ...
The only thing that is obvious is to fill the color for whole space and then draw the outline with black color then flood fill all non black pixels from outside with white color
回答4:
It's my solution with grid package.
##data
koch <- function(k) {
yy <- xx <- numeric(0)
Koch <- function(x1, y1, x2, y2, n) {
if (n > 1) {
Koch(x1, y1, (2 * x1 + x2)/3, (2 * y1 + y2)/3, n - 1)
Koch((2 * x1 + x2)/3, (2 * y1 + y2)/3, (x1 + x2)/2 - (y1 -
y2) * sqrt(3)/6, (y1 + y2)/2 - (x2 - x1) * sqrt(3)/6,
n - 1)
Koch((x1 + x2)/2 - (y1 - y2) * sqrt(3)/6, (y1 + y2)/2 -
(x2 - x1) * sqrt(3)/6, (2 * x2 + x1)/3, (2 * y2 + y1)/3,
n - 1)
Koch((2 * x2 + x1)/3, (2 * y2 + y1)/3, x2, y2, n - 1)
} else {
x <- c(x1, (2 * x1 + x2)/3, (x1 + x2)/2 - (y1 - y2) * sqrt(3)/6,
(2 * x2 + x1)/3, x2)
xx <<- c(xx, x)
y <- c(y1, (2 * y1 + y2)/3, (y1 + y2)/2 - (x2 - x1) * sqrt(3)/6,
(2 * y2 + y1)/3, y2)
yy <<- c(yy, y)
}
}
Koch(0, 0, 1, 0, k)
Koch(1, 0, 0.5, sqrt(3)/2, k)
Koch(0.5, sqrt(3)/2, 0, 0, k)
xy <- data.frame(x = (xx - min(xx))/(max(xx) - min(xx)), y = (yy -
min(yy))/(max(yy) - min(yy)))
rbind(unique(xy), xy[1, ])
}
xy <- koch(5)
##Plot
library(grid)
grid.newpage()
pushViewport(dataViewport(xy$x, xy$y), plotViewport(c(1, 1, 1, 1)))
for (i in 1:nrow(xy)) {
grid.path(x = c(xy[i, 1], xy[i + 1, 1], mean(xy$x)),
y = c(xy[i, 2], xy[i + 1, 2], mean(xy$y)),
gp = gpar(col = rainbow(nrow(xy))[i],
fill = rainbow(nrow(xy))[i]))
}
来源：https://stackoverflow.com/questions/28422985/draw-a-polygon-colored-like-this-in-r-or-matlab