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Unable to resize Cubism graph D3.js

I got the cubism code from the Cubism Demo. The timeframe by default in the demo code is 4 hours. I'm trying to reduce it to 15 min. I successfully modified the option to make it 15 min but the graph got shrinked.
Here is the JavaScript code :
var context = cubism.context()
.step(10000)
.size(1440); // Modified this to 90 to make it 15 min
d3.select("body").selectAll(".axis")
.data(["top", "bottom"])
.enter().append("div")
.attr("class", function(d) { return d + " axis"; })
.each(function(d) { d3.select(this).call(context.axis().ticks(12).orient(d)); });
d3.select("body").append("div")
.attr("class", "rule")
.call(context.rule());
d3.select("body").selectAll(".horizon")
.data(d3.range(1, 3).map(random))
.enter().insert("div", ".bottom")
.attr("class", "horizon")
.call(context.horizon().extent([-10, 10]));
context.on("focus", function(i) {
d3.selectAll(".value").style("right", i == null ? null : context.size() - i + "px");
});
// Replace this with context.graphite and graphite.metric!
function random(x) {
var value = 0,
values = [],
i = 0,
last;
return context.metric(function(start, stop, step, callback) {
start = +start, stop = +stop;
if (isNaN(last)) last = start;
while (last < stop) {
last += step;
value = Math.max(-10, Math.min(10, value + .8 * Math.random() - .4 + .2 * Math.cos(i += x * .02)));
values.push(value);
}
callback(null, values = values.slice((start - stop) / step));
}, x);
}
Here is the demo.
How to display 15 min or lower time frame in Cubism graph?

Size option defines the number of values shown in the graph so 1440 values = 1440px width. According to the docs, step set the context step in milliseconds.
So, if you want to display 15min = 900000ms, in a graph that is 1440px wide you have to use a step that is 900000/1440 = 625.
var context = cubism.context()
.step(625)
.size(1440);
Edited your fiddle

Display all the leafs on the same level with D3.JS

I recently started to use D3.js in order to visualize a tree in the form of a radial tree, as presented in Colapsible Radial Tree , but I encountered some problems while modifying the code.
I used the code in 1 to display all the leafs on the same 'level'. My methodology is the following:
First, compute the maximum depth of the tree to know where to put all the leafs of the tree, by using the following code:
var maxDepth = getDepth(source);
function getDepth(obj) {
var depth = 0;
if (obj.children) {
obj.children.forEach(function (d) {
var tmpDepth = getDepth(d)
if (tmpDepth > depth) {
depth = tmpDepth
}
})
}
return 1 + depth}
Then, loop over all the nodes of the tree and whenever a node without children is encountered (= leaf), the position is set to:
nodes.forEach(function(d) {
if (d.children)
{
d.y = d.depth * 60;
}else
d.y = maxDepth * 60;})
It seems to work correctly, except that the positions of some nodes are not correct and are overlapping some times.
How can I adjust the code to spread the nodes on a more efficient way ? The full code is available here.

Use d3's cluster layout, it's designed to place leaves at the same level
https://github.com/mbostock/d3/wiki/Cluster-Layout
https://bl.ocks.org/mbostock/4063570
var tree = d3.layout.cluster()
.size([360, diameter / 2 - 80])
.separation(function(a, b) { return (a.parent == b.parent ? 1 : 5) / a.depth; });
...
// Normalize for fixed-depth.
//nodes.forEach(function(d) { d.y = d.depth * 80; }); // get rid of this line (or amend it) as it resets the leaves to one level below the parent

Creating variable number of elements with D3.js

I am pretty new to d3.js and maybe my question is very basic, but I haven't been able to find an answer...
I am trying to achieve the following, based on an array of data like this:
var data = [
{
"items": 10,
"selected": 8
},
{
"items": 12,
"selected": 4
}];
I would like to create a row of circles for every element of the array. The number of circles should be the equal to the items property and in every row, the circle in selected position should be special (like a different color). For the example above, it should display something similar to:
OOOOOOO*OO
OOO*OOOOOOOO
For the first step, any tips on how to create a variable number of SVG elements based on data values would be a great help.

Here's an example I made on codepen.
Check out the code below and/or fork the codepen and have a play.
Essentially, what is happening here is that I add a g element for each item in your data array. Normally we might be-able to create a circle for each data element, but since that is contained within a property and variable, I've used an each loop (d3 each). This creates a loop of the number of items, and creates a circle for each. If the element is selected, the fill color changes.
things to note:
The g element has a transform of 30 * i on the y axis. This means that each group will stack down the page.
In the for loop we get access to the g element using this, but we must use the d3 select function to reference it as a d3 object so we can append.
The code:
//add an svg canvas to the page
var svg = d3.select("body")
.append("svg")
.attr("transform", "translate(" + 20 + "," + 20 + ")"); //this will give it a bottom and left margin
//set up some basic data
var data = [
{
"items": 10,
"selected": 8
},
{
"items": 12,
"selected": 4
}];
var groups = svg.selectAll("g")
.data(data)
.enter()
.append("g").attr("transform", function(d, i) {
return "translate(0," + i * 30 + ")";
});
groups.each(function(d,i) {
for (var x = 1; x <= d.items; x++) {
d3.select(this).append('circle').attr({cx: function(d,i) { return (x + 1) * 22; } ,
cy: 10 ,
r: 10 ,
fill: function(d,i) { return x == d.selected ? "yellow" : "blue" }
});
}
});

d3.js spreading labels for pie charts

I'm using d3.js - I have a pie chart here. The problem though is when the slices are small - the labels overlap. What is the best way of spreading out the labels.
http://jsfiddle.net/BxLHd/16/
Here is the code for the labels. I am curious - is it possible to mock a 3d pie chart with d3?
//draw labels
valueLabels = label_group.selectAll("text.value").data(filteredData)
valueLabels.enter().append("svg:text")
.attr("class", "value")
.attr("transform", function(d) {
return "translate(" + Math.cos(((d.startAngle+d.endAngle - Math.PI)/2)) * (that.r + that.textOffset) + "," + Math.sin((d.startAngle+d.endAngle - Math.PI)/2) * (that.r + that.textOffset) + ")";
})
.attr("dy", function(d){
if ((d.startAngle+d.endAngle)/2 > Math.PI/2 && (d.startAngle+d.endAngle)/2 < Math.PI*1.5 ) {
return 5;
} else {
return -7;
}
})
.attr("text-anchor", function(d){
if ( (d.startAngle+d.endAngle)/2 < Math.PI ){
return "beginning";
} else {
return "end";
}
}).text(function(d){
//if value is greater than threshold show percentage
if(d.value > threshold){
var percentage = (d.value/that.totalOctets)*100;
return percentage.toFixed(2)+"%";
}
});
valueLabels.transition().duration(this.tweenDuration).attrTween("transform", this.textTween);
valueLabels.exit().remove();

As #The Old County discovered, the previous answer I posted fails in firefox because it relies on the SVG method .getIntersectionList() to find conflicts, and that method hasn't been implemented yet in Firefox.
That just means we have to keep track of label positions and test for conflicts ourselves. With d3, the most efficient way to check for layout conflicts involves using a quadtree data structure to store positions, that way you don't have to check every label for overlap, just those in a similar area of the visualization.
The second part of the code from the previous answer gets replaced with:
/* check whether the default position
overlaps any other labels*/
var conflicts = [];
labelLayout.visit(function(node, x1, y1, x2, y2){
//recurse down the tree, adding any overlapping labels
//to the conflicts array
//node is the node in the quadtree,
//node.point is the value that we added to the tree
//x1,y1,x2,y2 are the bounds of the rectangle that
//this node covers
if ( (x1 > d.r + maxLabelWidth/2)
//left edge of node is to the right of right edge of label
||(x2 < d.l - maxLabelWidth/2)
//right edge of node is to the left of left edge of label
||(y1 > d.b + maxLabelHeight/2)
//top (minY) edge of node is greater than the bottom of label
||(y2 < d.t - maxLabelHeight/2 ) )
//bottom (maxY) edge of node is less than the top of label
return true; //don't bother visiting children or checking this node
var p = node.point;
var v = false, h = false;
if ( p ) { //p is defined, i.e., there is a value stored in this node
h = ( ((p.l > d.l) && (p.l <= d.r))
|| ((p.r > d.l) && (p.r <= d.r))
|| ((p.l < d.l)&&(p.r >=d.r) ) ); //horizontal conflict
v = ( ((p.t > d.t) && (p.t <= d.b))
|| ((p.b > d.t) && (p.b <= d.b))
|| ((p.t < d.t)&&(p.b >=d.b) ) ); //vertical conflict
if (h&&v)
conflicts.push(p); //add to conflict list
}
});
if (conflicts.length) {
console.log(d, " conflicts with ", conflicts);
var rightEdge = d3.max(conflicts, function(d2) {
return d2.r;
});
d.l = rightEdge;
d.x = d.l + bbox.width / 2 + 5;
d.r = d.l + bbox.width + 10;
}
else console.log("no conflicts for ", d);
/* add this label to the quadtree, so it will show up as a conflict
for future labels. */
labelLayout.add( d );
var maxLabelWidth = Math.max(maxLabelWidth, bbox.width+10);
var maxLabelHeight = Math.max(maxLabelHeight, bbox.height+10);
Note that I've changed the parameter names for the edges of the label to l/r/b/t (left/right/bottom/top) to keep everything logical in my mind.
Live fiddle here: http://jsfiddle.net/Qh9X5/1249/
An added benefit of doing it this way is that you can check for conflicts based on the final position of the labels, before actually setting the position. Which means that you can use transitions for moving the labels into position after figuring out the positions for all the labels.

Should be possible to do. How exactly you want to do it will depend on what you want to do with spacing out the labels. There is not, however, a built in way of doing this.
The main problem with the labels is that, in your example, they rely on the same data for positioning that you are using for the slices of your pie chart. If you want them to space out more like excel does (i.e. give them room), you'll have to get creative. The information you have is their starting position, their height, and their width.
A really fun (my definition of fun) way to go about solving this would be to create a stochastic solver for an optimal arrangement of labels. You could do this with an energy-based method. Define an energy function where energy increases based on two criteria: distance from start point and overlap with nearby labels. You can do simple gradient descent based on that energy criteria to find a locally optimal solution with regards to your total energy, which would result in your labels being as close as possible to their original points without a significant amount of overlap, and without pushing more points away from their original points.
How much overlap is tolerable would depend on the energy function you specify, which should be tunable to give a good looking distribution of points. Similarly, how much you're willing to budge on point closeness would depend on the shape of your energy increase function for distance from the original point. (A linear energy increase will result in closer points, but greater outliers. A quadratic or a cubic will have greater average distance, but smaller outliers.)
There might also be an analytical way of solving for the minima, but that would be harder. You could probably develop a heuristic for positioning things, which is probably what excel does, but that would be less fun.

One way to check for conflicts is to use the <svg> element's getIntersectionList() method. That method requires you to pass in an SVGRect object (which is different from a <rect> element!), such as the object returned by a graphical element's .getBBox() method.
With those two methods, you can figure out where a label is within the screen and if it overlaps anything. However, one complication is that the rectangle coordinates passed to getIntersectionList are interpretted within the root SVG's coordinates, while the coordinates returned by getBBox are in the local coordinate system. So you also need the method getCTM() (get cumulative transformation matrix) to convert between the two.
I started with the example from Lars Khottof that #TheOldCounty had posted in a comment, as it already included lines between the arc segments and the labels. I did a little re-organization to put the labels, lines and arc segments in separate <g> elements. That avoids strange overlaps (arcs drawn on top of pointer lines) on update, and it also makes it easy to define which elements we're worried about overlapping -- other labels only, not the pointer lines or arcs -- by passing the parent <g> element as the second parameter to getIntersectionList.
The labels are positioned one at a time using an each function, and they have to be actually positioned (i.e., the attribute set to its final value, no transitions) at the time the position is calculated, so that they are in place when getIntersectionList is called for the next label's default position.
The decision of where to move a label if it overlaps a previous label is a complex one, as #ckersch's answer outlines. I keep it simple and just move it to the right of all the overlapped elements. This could cause a problem at the top of the pie, where labels from the last segments could be moved so that they overlap labels from the first segments, but that's unlikely if the pie chart is sorted by segment size.
Here's the key code:
labels.text(function (d) {
// Set the text *first*, so we can query the size
// of the label with .getBBox()
return d.value;
})
.each(function (d, i) {
// Move all calculations into the each function.
// Position values are stored in the data object
// so can be accessed later when drawing the line
/* calculate the position of the center marker */
var a = (d.startAngle + d.endAngle) / 2 ;
//trig functions adjusted to use the angle relative
//to the "12 o'clock" vector:
d.cx = Math.sin(a) * (that.radius - 75);
d.cy = -Math.cos(a) * (that.radius - 75);
/* calculate the default position for the label,
so that the middle of the label is centered in the arc*/
var bbox = this.getBBox();
//bbox.width and bbox.height will
//describe the size of the label text
var labelRadius = that.radius - 20;
d.x = Math.sin(a) * (labelRadius);
d.sx = d.x - bbox.width / 2 - 2;
d.ox = d.x + bbox.width / 2 + 2;
d.y = -Math.cos(a) * (that.radius - 20);
d.sy = d.oy = d.y + 5;
/* check whether the default position
overlaps any other labels*/
//adjust the bbox according to the default position
//AND the transform in effect
var matrix = this.getCTM();
bbox.x = d.x + matrix.e;
bbox.y = d.y + matrix.f;
var conflicts = this.ownerSVGElement
.getIntersectionList(bbox, this.parentNode);
/* clear conflicts */
if (conflicts.length) {
console.log("Conflict for ", d.data, conflicts);
var maxX = d3.max(conflicts, function(node) {
var bb = node.getBBox();
return bb.x + bb.width;
})
d.x = maxX + 13;
d.sx = d.x - bbox.width / 2 - 2;
d.ox = d.x + bbox.width / 2 + 2;
}
/* position this label, so it will show up as a conflict
for future labels. (Unfortunately, you can't use transitions.) */
d3.select(this)
.attr("x", function (d) {
return d.x;
})
.attr("y", function (d) {
return d.y;
});
});
And here's the working fiddle: http://jsfiddle.net/Qh9X5/1237/

D3 tree vertical separation

I am using the D3 tree layout, such as this one: http://mbostock.github.com/d3/talk/20111018/tree.html
I have modified it for my needs and am running into an issue. The example has the same issue too where if you have too many nodes open then they become compact and makes reading and interacting difficult. I am wanting to defined a minimum vertical space between nodes while re-sizing stage to allow for such spacing.
I tried modifying the separation algorithm to make it work:
.separation(function (a, b) {
return (a.parent == b.parent ? 1 : 2) / a.depth;
})
That didn't work. I also tried calculating which depth had the most children then telling the height of the stage to be children * spaceBetweenNodes. That got me closer, but still was not accurate.
depthCounts = [];
nodes.forEach(function(d, i) {
d.y = d.depth * 180;
if(!depthCounts[d.depth])
depthCounts[d.depth] = 0;
if(d.children)
{
depthCounts[d.depth] += d.children.length;
}
});
tree_resize(largest_depth(depthCounts) * spaceBetweenNodes);
I also tried to change the node's x value too in the method below where it calculates the y separation, but no cigar. I would post that change too but I removed it from my code.
nodes.forEach(function(d, i) {
d.y = d.depth * 180;
});
If you can suggest a way or know a way that I can accomplish a minimum spacing vertically between nodes please post. I will be very grateful. I am probably missing something very simple.

As of 2016, I was able to achieve this using just
tree.nodeSize([height, width])
https://github.com/mbostock/d3/wiki/Tree-Layout#nodeSize
The API Reference is a bit poor, but is works pretty straight forward. Be sure to use it after tree.size([height, width]) or else you will be overriding your values again.
For more reference: D3 Tree Layout Separation Between Nodes using NodeSize

I was able to figure this out with help from a user on Google Groups. I was not able to find the post. The solution requires you to modify D3.js in one spot, which is not recommended but it was the only to get around this issue that I could find.
Starting around line 5724 or this method: d3_layout_treeVisitAfter
change:
d3_layout_treeVisitAfter(root, function(node) {
node.x = (node.x - x0) / (x1 - x0) * size[0];
node.y = node.depth / y1 * size[1];
delete node._tree;
});
to:
d3_layout_treeVisitAfter(root, function(node) {
// make sure size is null, we will make it null when we create the tree
if(size === undefined || size == null)
{
node.x = (node.x - x0) * elementsize[0];
node.y = node.depth * elementsize[1];
}
else
{
node.x = (node.x - x0) / (x1 - x0) * size[0];
node.y = node.depth / y1 * size[1];
}
delete node._tree;
});
Below add a new variable called: elementsize and default it to [ 1, 1 ] to line 5731
var hierarchy = d3.layout.hierarchy().sort(null).value(null)
, separation = d3_layout_treeSeparation
, elementsize = [ 1, 1 ] // Right here
, size = [ 1, 1 ];
Below that there is a method called tree.size = function(x). Add the following below that definition:
tree.elementsize = function(x) {
if (!arguments.length) return elementsize;
elementsize = x;
return tree;
};
Finally when you create the tree you can change the elementsize like so
var tree = d3.layout.tree()
.size(null)
.elementsize(50, 240);

I know I'm not supposed to respond to other answers, but I don't have enough reputation to add a comment.
Anyway, I just wanted to update this for people using the latest d3.v3.js file. (I assume this is because of a new version, because the line references in the accepted answer were wrong for me.)
The d3.layout.tree function that you are editing is found between lines 6236 and 6345. d3_layout_treeVisitAfter starts on line 6318. The hierarchy variable is declared on line 6237. The bit about tree.elementsize still stands - I put it on line 6343.
Lastly (I assume this was an error): when you create the tree, put the dimensions inside square brackets, like you normally do with "size". So:
var tree = d3.layout.tree()
.size(null)
.elementsize([50, 240]);

The original fix you proposed will work, you just have to make sure you do it after you add everything to the canvas. d3 recalculates the layout each time you enter, exit, append, etc. Once you've done all that, then you can fiddle with the d.y to fix the depth.
nodes.forEach(function(d) { d.y = d.depth * fixdepth});