score:10

Accepted answer

Edited 19-Sep-2012 per comments with many thanks to nrabinowitz!

You will need to do some sort of search of the data returned by getPointAtLength. (See https://developer.mozilla.org/en-US/docs/DOM/SVGPathElement.)

// Line
var line = d3.svg.line()
     .interpolate("basis")
     .x(function (d) { return i; })
     .y(function(d, i) { return 100*Math.sin(i) + 100; });

// Append the path to the DOM
d3.select("svg#chart") //or whatever your SVG container is
     .append("svg:path")
     .attr("d", line([0,10,20,30,40,50,60,70,80,90,100]))
     .attr("id", "myline");

// Get the coordinates
function findYatX(x, linePath) {
     function getXY(len) {
          var point = linePath.getPointAtLength(len);
          return [point.x, point.y];
     }
     var curlen = 0;
     while (getXY(curlen)[0] < x) { curlen += 0.01; }
     return getXY(curlen);
}

console.log(findYatX(5, document.getElementById("myline")));

For me this returns [5.000403881072998, 140.6229248046875].

This search function, findYatX, is far from efficient (runs in O(n) time), but illustrates the point.

score:0

I have tried implementing findYatXbisection (as nicely suggested by bumbu), and I could not get it to work AS IS.

Instead of modifying the length as a function of length_end and length_start, I just decreased the length by 50% (if x < point.x) or increased by 50% (if x> point.x) but always relative to start length of zero. I have also incorporated revXscale/revYscale to convert pixels to x/y values as set by my d3.scale functions.

function findYatX(x,path,error){
    var length = apath.getTotalLength()
        , point = path.getPointAtLength(length)
        , bisection_iterations_max=50
        , bisection_iterations = 0
    error = error || 0.1
    while (x < revXscale(point.x) -error || x> revXscale(point.x + error) {
        point = path.getPointAtlength(length)
        if (x < revXscale(point.x)) {
             length = length/2
        } else {
             length = 3/2*length
        }
        if (bisection_iterations_max < ++ bisection_iterations) {
              break;
        }
    }
return revYscale(point.y)
}

score:20

This solution is much more efficient than the accepted answer. It's execution time is logarithmic (while accepted answer has linear complexity).

var findYatXbyBisection = function(x, path, error){
  var length_end = path.getTotalLength()
    , length_start = 0
    , point = path.getPointAtLength((length_end + length_start) / 2) // get the middle point
    , bisection_iterations_max = 50
    , bisection_iterations = 0

  error = error || 0.01

  while (x < point.x - error || x > point.x + error) {
    // get the middle point
    point = path.getPointAtLength((length_end + length_start) / 2)

    if (x < point.x) {
      length_end = (length_start + length_end)/2
    } else {
      length_start = (length_start + length_end)/2
    }

    // Increase iteration
    if(bisection_iterations_max < ++ bisection_iterations)
      break;
  }
  return point.y
}

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