For example, I need to calculate a Math.sqrt of my data for each attr, how can I calculate only one time the Math.sqrt(d)?
var circle = svgContainer.data(dat
Probably, the most idiomatic way for doing this in D3 is using selection.each, which:
Invokes the specified function for each selected element, in order, being passed the current datum (d), the current index (i), and the current group (nodes), with this as the current DOM element (nodes[i]).
So, in your case:
circle.each(function(d){
//calculates the value just once for each datum:
var squareRoot = Math.sqrt(d)
//now use that value in the DOM element, which is 'this':
d3.select(this).attr("cx", squareRoot)
.attr("cy", squareRoot)
//etc...
});
Here is a demo:
var svg = d3.select("svg");
var data = d3.range(100, 1000, 100);
var ellipses = svg.selectAll(null)
.data(data)
.enter()
.append("ellipse")
.attr("fill", "gainsboro")
.attr("stroke", "darkslateblue")
.each(function(d) {
var squareRoot = Math.sqrt(d);
d3.select(this)
.attr("cx", function(d) {
return squareRoot * 3
})
.attr("cy", function(d) {
return squareRoot * 3
})
.attr("rx", function(d) {
return squareRoot + 3
})
.attr("ry", function(d) {
return squareRoot + 4
});
})
Another common approach in D3 codes is setting a new data property in the first attr
method, and retrieving it latter:
.attr("cx", function(d) {
//set a new property here
d.squareRoot = Math.sqrt(d.value);
return d.squareRoot * 3
})
.attr("cy", function(d) {
//retrieve it here
return d.squareRoot * 3
})
//etc...
That way you also perform the calculation only once per element.
Here is the demo:
var svg = d3.select("svg");
var data = d3.range(100, 1000, 100).map(function(d) {
return {
value: d
}
});
var ellipses = svg.selectAll(null)
.data(data)
.enter()
.append("ellipse")
.attr("fill", "gainsboro")
.attr("stroke", "darkslateblue")
.attr("cx", function(d) {
d.squareRoot = Math.sqrt(d.value);
return d.squareRoot * 3
})
.attr("cy", function(d) {
return d.squareRoot * 3
})
.attr("rx", function(d) {
return d.squareRoot + 3
})
.attr("ry", function(d) {
return d.squareRoot + 4
});
PS: by the way, your solution with var aux
will not work. Try it and you'll see.