Develop Reference

How can I crop an image in p5.js without drawing image on the screen

I've researched and found that you can "crop" an image by using get(), but the image has to be drawn on the screen where you then take a section of the canvas. Is it possible to load an image then save a cropped version of it in a variable? So maybe something like this:
var img;
var cropped;
function preload(){
img = imageLoad('dog.png', crop)
}
function crop(image){
cropped = crop(img, 0, 0, img.w/2, img.h) // Getting left half of image
}
Thank you.
EDIT:
This is the function I made using copy(), but I don't know if there's an easier way that I'm missing.
function crop(image, x, y, w, h) {
var cropped = createImage(w, h);
cropped.copy(image, x, y, x + w, y + h, 0, 0, x + w, y + h)
return cropped;
}

You can access the pixels of an image directly. You don't have to draw it to the canvas first.
Start by reading through the P5.Image reference.
At a high level, what you want to do is create a new graphics (the createGraphics() function is your friend, and then draw the section of the image you want to that graphics. Whether you draw the image or the graphics to the canvas is up to you.

Zoom my drawing on the background [duplicate]

This question already has an answer here:
HTML5 canvas zoom where mouse coordinates
(1 answer)
Closed 8 years ago.
I make program like a paint with HTML5 canvas and javascript. Drawing takes place on the background image. How to zoom my drawing on the background together.
Before zoom it:
After zoom it (need this result):
Note: zoom should be where clicked with the mouse on the background image

I've done this before!
First of all, I set a zoom level attribute on my canvas.
Main.canvas.zoomX = 1;
Main.canvas.zoomY = 1;
I also retain the original size of the canvas for reference.
Main.canvas.originW = Main.canvas.width;
Main.canvas.originH = Main.canvas.height;
I also retain the original left and top of the canvas for reference.
Main.canvas.gLeftStart = 0;
Main.canvas.gTopStart = 0;
I then set a zoom percentage. The zoom level will be adjusted by this amount every time that the zoom event occurs.
Main.canvas.zoomPerc = 0.05;
Next, I set an event listener on my canvas to watch for mousewheel.
Main.canvas.addEventListener('wheel', zoom, true);
Now, I'm going to write a quick function to retrieve the zoom, then I'll explain it.
function zoom(evt)
{
var x;
var y;
Main.canvas.xLayerS = (evt.layerX + (Main.canvas.gLeftStart * -1)) / (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.yLayerS = (evt.layerY + (Main.canvas.gTopStart * -1)) / (Main.canvas.originH * Main.canvas.zoomY);
Main.canvas.leftPerc = Main.canvas.gLeftStart / (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.topPerc = Main.canvas.gTopStart / (Main.canvas.originH * Main.canvas.zoomY);
if(evt.deltaY > 1)
{
Main.canvas.zoomX *= 1 + Main.canvas.zoomPerc;
Main.canvas.zoomY *= 1 + Main.canvas.zoomPerc;
}
else
{
Main.canvas.zoomX *= 1 - Main.canvas.zoomPerc;
Main.canvas.zoomY *= 1 - Main.canvas.zoomPerc;
}
var iiDS;
var cmd;
Main.canvas.xLayer = Main.canvas.xLayerS * (Main.canvas.originW * Main.canvas.zoomX);
Main.canvas.yLayer = Main.canvas.yLayerS * (Main.canvas.originH * Main.canvas.zoomY);
Main.context.clearRect(0, 0, Main.canvas.width, Main.canvas.height);
Main.context.beginPath();
Main.canvas.gLeftStart = (evt.layerX - Main.canvas.xLayer);
Main.canvas.gTopStart = (evt.layerY - Main.canvas.yLayer);
for(iiDS = 0; iiDS < Main.dataPoints.length; iiDS++)
{
if(iiDS === 0)
{
cmd = 'moveTo';
}
else
{
cmd = 'lineTo';
}
Main.dataPoints[iiDS].xPerc = Main.dataPoints[iiDS].x / Main.range.x;
Main.dataPoints[iiDS].yPerc = Main.dataPoints[iiDS].y / Main.range.y;
x = Main.canvas.gLeftStart + (Main.dataPoints[iiDS].xPerc * (Main.canvas.originW * Main.canvas.zoomX));
y = Main.canvas.gTopStart + (Main.dataPoints[iiDS].yPerc * (Main.canvas.originH * Main.canvas.zoomY));
Main.context[cmd](x, y);
}
Main.context.stroke();
}
Now that your canvas has been re-sized, you will need to redraw whatever was in it. Remember, any time that you re-size a canvas, you clear the canvas. If your canvas was holding an image, then that's simple, redraw that image at that size. If you canvas was holding data points (like a chart) then I would suggest that you make your data points have percentage like (probably a word for that) positions along your chart, not pixel positions.
More importantly though, I do not suggest that you ever re-size and re-position your canvas on zoom. Your page can get jumbled up and sloppy that way. Instead, use the percentages for size (like I showed you) and use the values for left and top positioning as starting points in your drawing. If a data point was a certain percentage of a way across a chart, it can be drawn at any size. Plus, you can draw outside of your canvas, it just won't be visible. Your canvas would then be more like a view-port.
You can do some really impressive charting this way, which a lot of companies pay a lot of money for. Have fun!

Did you try Context2d.scale(x, y)? You could do the following
var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');
context.scale(2, 2);
paintBackGround(context);
paintForeGround(context);
scale(factorWidth, factorHeight) Scales all coordinates in the canvas by the factors, so it will scale the background and the drawing. The example would double the size. You don't have to scale your coordinates by yourself, just let canvas do that for you.
Here is an example :
http://www.html5canvastutorials.com/advanced/html5-canvas-transform-scale-tutorial/
The only problem here: you need to scale before you draw, so you need a model that contains the original drawing in original unscaled coordinates, that can be drawn after scaling (paintForeGround() in my example)
Scale() is part of so called Transformations. You can Translate (move along a vector) rotate and scale the content of a canvas by using buildin functions of canvas. Just take a look at the html5canvastutorials. This works with matrix-mutliplications in the background, but it is really simple to use.

mouse position to isometric tile including height

Struggeling translating the position of the mouse to the location of the tiles in my grid. When it's all flat, the math looks like this:
this.position.x = Math.floor(((pos.y - 240) / 24) + ((pos.x - 320) / 48));
this.position.y = Math.floor(((pos.y - 240) / 24) - ((pos.x - 320) / 48));
where pos.x and pos.y are the position of the mouse, 240 and 320 are the offset, 24 and 48 the size of the tile. Position then contains the grid coordinate of the tile I'm hovering over. This works reasonably well on a flat surface.
Now I'm adding height, which the math does not take into account.
This grid is a 2D grid containing noise, that's being translated to height and tile type. Height is really just an adjustment to the 'Y' position of the tile, so it's possible for two tiles to be drawn in the same spot.
I don't know how to determine which tile I'm hovering over.
edit:
Made some headway... Before, I was depending on the mouseover event to calculate grid position. I just changed this to do the calculation in the draw loop itself, and check if the coordinates are within the limits of the tile currently being drawn. creates some overhead tho, not sure if I'm super happy with it but I'll confirm if it works.
edit 2018:
I have no answer, but since this ha[sd] an open bounty, help yourself to some code and a demo
The grid itself is, simplified;
let grid = [[10,15],[12,23]];
which leads to a drawing like:
for (var i = 0; i < grid.length; i++) {
for (var j = 0; j < grid[0].length; j++) {
let x = (j - i) * resourceWidth;
let y = ((i + j) * resourceHeight) + (grid[i][j] * -resourceHeight);
// the "+" bit is the adjustment for height according to perlin noise values
}
}
edit post-bounty:
See GIF. The accepted answer works. The delay is my fault, the screen doesn't update on mousemove (yet) and the frame rate is low-ish. It's clearly bringing back the right tile.
Source

Intresting task.
Lets try to simplify it - lets resolve this concrete case
Solution
Working version is here: https://github.com/amuzalevskiy/perlin-landscape (changes https://github.com/jorgt/perlin-landscape/pull/1 )
Explanation
First what came into mind is:
Just two steps:
find an vertical column, which matches some set of tiles
iterate tiles in set from bottom to top, checking if cursor is placed lower than top line
Step 1
We need two functions here:
Detects column:
function getColumn(mouseX, firstTileXShiftAtScreen, columnWidth) {
return (mouseX - firstTileXShiftAtScreen) / columnWidth;
}
Function which extracts an array of tiles which correspond to this column.
Rotate image 45 deg in mind. The red numbers are columnNo. 3 column is highlighted. X axis is horizontal
function tileExists(x, y, width, height) {
return x >= 0 & y >= 0 & x < width & y < height;
}
function getTilesInColumn(columnNo, width, height) {
let startTileX = 0, startTileY = 0;
let xShift = true;
for (let i = 0; i < columnNo; i++) {
if (tileExists(startTileX + 1, startTileY, width, height)) {
startTileX++;
} else {
if (xShift) {
xShift = false;
} else {
startTileY++;
}
}
}
let tilesInColumn = [];
while(tileExists(startTileX, startTileY, width, height)) {
tilesInColumn.push({x: startTileX, y: startTileY, isLeft: xShift});
if (xShift) {
startTileX--;
} else {
startTileY++;
}
xShift = !xShift;
}
return tilesInColumn;
}
Step 2
A list of tiles to check is ready. Now for each tile we need to find a top line. Also we have two types of tiles: left and right. We already stored this info during building matching tiles set.
function getTileYIncrementByTileZ(tileZ) {
// implement here
return 0;
}
function findExactTile(mouseX, mouseY, tilesInColumn, tiles2d,
firstTileXShiftAtScreen, firstTileYShiftAtScreenAt0Height,
tileWidth, tileHeight) {
// we built a set of tiles where bottom ones come first
// iterate tiles from bottom to top
for(var i = 0; i < tilesInColumn; i++) {
let tileInfo = tilesInColumn[i];
let lineAB = findABForTopLineOfTile(tileInfo.x, tileInfo.y, tiles2d[tileInfo.x][tileInfo.y],
tileInfo.isLeft, tileWidth, tileHeight);
if ((mouseY - firstTileYShiftAtScreenAt0Height) >
(mouseX - firstTileXShiftAtScreen)*lineAB.a + lineAB.b) {
// WOHOO !!!
return tileInfo;
}
}
}
function findABForTopLineOfTile(tileX, tileY, tileZ, isLeftTopLine, tileWidth, tileHeight) {
// find a top line ~~~ a,b
// y = a * x + b;
let a = tileWidth / tileHeight;
if (isLeftTopLine) {
a = -a;
}
let b = isLeftTopLine ?
tileY * 2 * tileHeight :
- (tileX + 1) * 2 * tileHeight;
b -= getTileYIncrementByTileZ(tileZ);
return {a: a, b: b};
}

Please don't judge me as I am not posting any code. I am just suggesting an algorithm that can solve it without high memory usage.
The Algorithm:
Actually to determine which tile is on mouse hover we don't need to check all the tiles. At first we think the surface is 2D and find which tile the mouse pointer goes over with the formula OP posted. This is the farthest probable tile mouse cursor can point at this cursor position.
This tile can receive mouse pointer if it's at 0 height, by checking it's current height we can verify if this is really at the height to receive pointer, we mark it and move forward.
Then we find the next probable tile which is closer to the screen by incrementing or decrementing x,y grid values depending on the cursor position.
Then we keep on moving forward in a zigzag fashion until we reach a tile which cannot receive pointer even if it is at it's maximum height.
When we reach this point the last tile found that were at a height to receive pointer is the tile that we are looking for.
In this case we only checked 8 tiles to determine which tile is currently receiving pointer. This is very memory efficient in comparison to checking all the tiles present in the grid and yields faster result.

One way to solve this would be to follow the ray that goes from the clicked pixel on the screen into the map. For that, just determine the camera position in relation to the map and the direction it is looking at:
const camPos = {x: -5, y: -5, z: -5}
const camDirection = { x: 1, y:1, z:1}
The next step is to get the touch Position in the 3D world. In this certain perspective that is quite simple:
const touchPos = {
x: camPos.x + touch.x / Math.sqrt(2),
y: camPos.y - touch.x / Math.sqrt(2),
z: camPos.z - touch.y / Math.sqrt(2)
};
Now you just need to follow the ray into the layer (scale the directions so that they are smaller than one of your tiles dimensions):
for(let delta = 0; delta < 100; delta++){
const x = touchPos.x + camDirection.x * delta;
const y = touchPos.y + camDirection.y * delta;
const z = touchPos.z + camDirection.z * delta;
Now just take the tile at xz and check if y is smaller than its height;
const absX = ~~( x / 24 );
const absZ = ~~( z / 24 );
if(tiles[absX][absZ].height >= y){
// hanfle the over event
}

I had same situation on a game. first I tried with mathematics, but when I found that the clients wants to change the map type every day, I changed the solution with some graphical solution and pass it to the designer of the team. I captured the mouse position by listening the SVG elements click.
the main graphic directly used to capture and translate the mouse position to my required pixel.
https://blog.lavrton.com/hit-region-detection-for-html5-canvas-and-how-to-listen-to-click-events-on-canvas-shapes-815034d7e9f8
https://code.sololearn.com/Wq2bwzSxSnjl/#html

Here is the grid input I would define for the sake of this discussion. The output should be some tile (coordinate_1, coordinate_2) based on visibility on the users screen of the mouse:
I can offer two solutions from different perspectives, but you will need to convert this back into your problem domain. The first methodology is based on coloring tiles and can be more useful if the map is changing dynamically. The second solution is based on drawing coordinate bounding boxes based on the fact that tiles closer to the viewer like (0, 0) can never be occluded by tiles behind it (1,1).
Approach 1: Transparently Colored Tiles
The first approach is based on drawing and elaborated on here. I must give the credit to #haldagan for a particularly beautiful solution. In summary it relies on drawing a perfectly opaque layer on top of the original canvas and coloring every tile with a different color. This top layer should be subject to the same height transformations as the underlying layer. When the mouse hovers over a particular layer you can detect the color through canvas and thus the tile itself. This is the solution I would probably go with and this seems to be a not so rare issue in computer visualization and graphics (finding positions in a 3d isometric world).
Approach 2: Finding the Bounding Tile
This is based on the conjecture that the "front" row can never be occluded by "back" rows behind it. Furthermore, "closer to the screen" tiles cannot be occluded by tiles "farther from the screen". To make precise the meaning of "front", "back", "closer to the screen" and "farther from the screen", take a look at the following:
.
Based on this principle the approach is to build a set of polygons for each tile. So firstly we determine the coordinates on the canvas of just box (0, 0) after height scaling. Note that the height scale operation is simply a trapezoid stretched vertically based on height.
Then we determine the coordinates on the canvas of boxes (1, 0), (0, 1), (1, 1) after height scaling (we would need to subtract anything from those polygons which overlap with the polygon (0, 0)).
Proceed to build each boxes bounding coordinates by subtracting any occlusions from polygons closer to the screen, to eventually get coordinates of polygons for all boxes.
With these coordinates and some care you can ultimately determine which tile is pointed to by a binary search style through overlapping polygons by searching through bottom rows up.

It also matters what else is on the screen. Maths attempts work if your tiles are pretty much uniform. However if you are displaying various objects and want the user to pick them, it is far easier to have a canvas-sized map of identifiers.
function poly(ctx){var a=arguments;ctx.beginPath();ctx.moveTo(a[1],a[2]);
for(var i=3;i<a.length;i+=2)ctx.lineTo(a[i],a[i+1]);ctx.closePath();ctx.fill();ctx.stroke();}
function circle(ctx,x,y,r){ctx.beginPath();ctx.arc(x,y,r,0,2*Math.PI);ctx.fill();ctx.stroke();}
function Tile(h,c,f){
var cnv=document.createElement("canvas");cnv.width=100;cnv.height=h;
var ctx=cnv.getContext("2d");ctx.lineWidth=3;ctx.lineStyle="black";
ctx.fillStyle=c;poly(ctx,2,h-50,50,h-75,98,h-50,50,h-25);
poly(ctx,50,h-25,2,h-50,2,h-25,50,h-2);
poly(ctx,50,h-25,98,h-50,98,h-25,50,h-2);
f(ctx);return ctx.getImageData(0,0,100,h);
}
function put(x,y,tile,image,id,map){
var iw=image.width,tw=tile.width,th=tile.height,bdat=image.data,fdat=tile.data;
for(var i=0;i<tw;i++)
for(var j=0;j<th;j++){
var ijtw4=(i+j*tw)*4,a=fdat[ijtw4+3];
if(a!==0){
var xiyjiw=x+i+(y+j)*iw;
for(var k=0;k<3;k++)bdat[xiyjiw*4+k]=(bdat[xiyjiw*4+k]*(255-a)+fdat[ijtw4+k]*a)/255;
bdat[xiyjiw*4+3]=255;
map[xiyjiw]=id;
}
}
}
var cleanimage;
var pickmap;
function startup(){
var water=Tile(77,"blue",function(){});
var field=Tile(77,"lime",function(){});
var tree=Tile(200,"lime",function(ctx){
ctx.fillStyle="brown";poly(ctx,50,50,70,150,30,150);
ctx.fillStyle="forestgreen";circle(ctx,60,40,30);circle(ctx,68,70,30);circle(ctx,32,60,30);
});
var sheep=Tile(200,"lime",function(ctx){
ctx.fillStyle="white";poly(ctx,25,155,25,100);poly(ctx,75,155,75,100);
circle(ctx,50,100,45);circle(ctx,50,80,30);
poly(ctx,40,70,35,80);poly(ctx,60,70,65,80);
});
var cnv=document.getElementById("scape");
cnv.width=500;cnv.height=400;
var ctx=cnv.getContext("2d");
cleanimage=ctx.getImageData(0,0,500,400);
pickmap=new Uint8Array(500*400);
var tiles=[water,field,tree,sheep];
var map=[[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[1,1],[1,2],[3,2],[1,1]],
[[0,0],[1,1],[2,2],[3,2],[1,1]],
[[0,0],[1,1],[1,1],[1,1],[1,1]],
[[0,0],[0,0],[0,0],[0,0],[0,0]]];
for(var x=0;x<5;x++)
for(var y=0;y<5;y++){
var desc=map[y][x],tile=tiles[desc[0]];
put(200+x*50-y*50,200+x*25+y*25-tile.height-desc[1]*20,
tile,cleanimage,x+1+(y+1)*10,pickmap);
}
ctx.putImageData(cleanimage,0,0);
}
var mx,my,pick;
function mmove(event){
mx=Math.round(event.offsetX);
my=Math.round(event.offsetY);
if(mx>=0 && my>=0 && mx<cleanimage.width && my<cleanimage.height && pick!==pickmap[mx+my*cleanimage.width])
requestAnimationFrame(redraw);
}
function redraw(){
pick=pickmap[mx+my*cleanimage.width];
document.getElementById("pick").innerHTML=pick;
var ctx=document.getElementById("scape").getContext("2d");
ctx.putImageData(cleanimage,0,0);
if(pick!==0){
var temp=ctx.getImageData(0,0,cleanimage.width,cleanimage.height);
for(var i=0;i<pickmap.length;i++)
if(pickmap[i]===pick)
temp.data[i*4]=255;
ctx.putImageData(temp,0,0);
}
}
startup(); // in place of body.onload
<div id="pick">Move around</div>
<canvas id="scape" onmousemove="mmove(event)"></canvas>
Here the "id" is a simple x+1+(y+1)*10 (so it is nice when displayed) and fits into a byte (Uint8Array), which could go up to 15x15 display grid already, and there are wider types available too.
(Tried to draw it small, and it looked ok on the snippet editor screen but apparently it is still too large here)

Computer graphics is fun, right?
This is a special case of the more standard computational geometry "point location problem". You could also express it as a nearest neighbour search.
To make this look like a point location problem you just need to express your tiles as non-overlapping polygons in a 2D plane. If you want to keep your shapes in a 3D space (e.g. with a z buffer) this becomes the related "ray casting problem".
One source of good geometry algorithms is W. Randolf Franklin's website and turf.js contains an implementation of his PNPOLY algorithm.
For this special case we can be even faster than the general algorithms by treating our prior knowledge about the shape of the tiles as a coarse R-tree (a type of spatial index).

Javascript only draws last of multiple drawings

Okay so i'm embarrassingly unknowledgeable of javascript. I know my ruby and rails pretty solid but have never used js as extensively as i am in my current project.
i have a map that's been drawn on a canvas. on that map, i want drawn multiple position markers (via the provided function). i've given it a list (via rails) of locations to mark. for some reason, it's only drawing the last coordinate.
the javascript is improvised from another source, not mine. this is just the problematic portion:
function plotPosition(long,lat) {
// Grab a handle to the canvas
var canvas = document.getElementById('map'),
ctx;
// Canvas supported?
if (canvas.getContext) {
// Grab the context
ctx = canvas.getContext('2d');
ctx.beginPath();
// Draw a arc that represent the geo-location of the request
ctx.arc(
degreesOfLongitudeToScreenX(long),
degreesOfLatitudeToScreenY(lat),
5,
0,
2 * Math.PI,
false
);
// Point style
ctx.fillStyle = 'rgb(0,0,0)';
ctx.fill();
ctx.stroke();
}
}
function draw() {
// Main entry point got the map canvas example
var canvas = document.getElementById('map'),
ctx;
// Canvas supported?
if (canvas.getContext) {
ctx = canvas.getContext('2d');
// Draw the background
drawBackground(ctx);
// Draw the map background
drawMapBackground(ctx);
// Draw the map background
// drawGraticule(ctx);
// Draw the land
drawLandMass(ctx);
<% #events.each do |e| %>
plotPosition('<%= e.longitude %>','<%= e.latitude %>');
<% end %>
} else {
alert("Canvas not supported!");
}
}
so my thinking is that the plotPosition function is drawing each coordinate, but every time it's given a new coordinate, the old one is erased/moved and replaced with the new one. probably a simple fix here, but i've been at this for hours and to no avail. banging my head.
any ideas as to the problem?

From the code example it seems as you are clearing the background too - you're not showing the calls to these methods but I will assume you do this before drawing the new point(s) (?) somewhere else in the code.
If so -
You need to redraw all the points you have drawn before you clear the canvas.
One way to do this is to store the points you already have in an array and then call a redraw function to draw everything in there. Or store the points from server directly to the array with converted values and when all points has been converted redraw everything.
Simplified example based on latter approach (which will require some re-factoring):
/// GLOBALS (put ctx here instead as well..)
var points = [],
canvas = document.getElementById('map'),
ctx = canvas.getContext ? canvas.getContext('2d') : null;
if (ctx === null) {...sorry...};
...
/// somewhere comes the logic to get the points themselves.
...
<% #events.each do |e| %>
storePosition('<%= e.longitude %>','<%= e.latitude %>');
<% end %>
...
/// after new points are added, render them all
renderAll();
function storePosition(long, lat) {
points.push{
x: degreesOfLongitudeToScreenX(long)
y: degreesOfLatitudeToScreenY(lat),
long: long,
lat: lat
}
}
function renderAll();
/// clear and redraw background of canvas
drawBackground(ctx);
// Draw the map background
drawMapBackground(ctx);
// Draw the map background
// drawGraticule(ctx);
// Draw the land
drawLandMass(ctx);
/// now plot all the stored points
ctx.fillStyle = 'rgb(0,0,0)';
for(var i = 0, point; point = points[i]; i++) {
ctx.beginPath();
ctx.arc(
point.x,
point.y,
5,
0,
2 * Math.PI
);
ctx.closePath();
ctx.fill();
}
}
You can implement a check of x and y to see that they are inside canvas but this is strictly not necessary as canvas will clip them for you and the internal clipping do faster checking than JavaScript will. But for debugging purposes you could always do a check if you suspect this. But you write that the point is plotted but erased when a new point is drawn so I don't think this is the problem here.

Raphael-JS: Rect with one round corner

paper.rect(0, 0, settings.width, settings.height, settings.radius);
Creates a nice rectangle with rounded corners. Is it possible to create a rectangle with just one round corner?

If you use Raphael JS:
Raphael.fn.roundedRectangle = function (x, y, w, h, r1, r2, r3, r4){
var array = [];
array = array.concat(["M",x,r1+y, "Q",x,y, x+r1,y]); //A
array = array.concat(["L",x+w-r2,y, "Q",x+w,y, x+w,y+r2]); //B
array = array.concat(["L",x+w,y+h-r3, "Q",x+w,y+h, x+w-r3,y+h]); //C
array = array.concat(["L",x+r4,y+h, "Q",x,y+h, x,y+h-r4, "Z"]); //D
return this.path(array);
};
To have a rectangle with only the upper-right corner rounded
var paper = Raphael("canvas", 840, 480);
paper.roundedRectangle(10, 10, 80, 80, 0, 20, 0, 0);
Source and online example: http://www.remy-mellet.com/blog/179-draw-rectangle-with-123-or-4-rounded-corner/

The rounded corners feature maps directly on to the underlying SVG rx and ry attributes, they apply to whole rectangles and so there's no possibility of just setting it on a single corner.
This blog post discusses an approach in SVG of basically covering up the corners you don't want rounded. Although his examples appear to be offline now, the approach should be fairly easy to reverse engineer into SVG.
An alternative approach would be to use a path instead of a rectangle object and draw the whole outline yourself. The syntax is a little obscure but is easy enough once you understand what's going on. Try Jakob Jenkov's SVG Path tutorial for an introduction.

Very old question, here's a better path. I converted it to relative coordinates, which should be better at animations...
Raphael.fn.roundedRectangle = function (x, y, w, h, r1, r2, r3, r4){
var array = [];
array = array.concat(["M",x+r1,y]);
array = array.concat(['l',w-r1-r2,0]);//T
array = array.concat(["q",r2,0, r2,r2]); //TR
array = array.concat(['l',0,h-r3-r2]);//R
array = array.concat(["q",0,r3, -r3,r3]); //BR
array = array.concat(['l',-w+r4+r3,0]);//B
array = array.concat(["q",-r4,0, -r4,-r4]); //BL
array = array.concat(['l',0,-h+r4+r1]);//L
array = array.concat(["q",0,-r1, r1,-r1]); //TL
array = array.concat(["z"]); //end
return this.path(array);
};