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How to use the D3 diagonal function to draw curved lines?


D3 Sankey chart using circle node instead of rectangle nodeDraw curve between two points using diagonal function in d3 jsIs there an “exists” function for jQuery?How do JavaScript closures work?How do I check if an element is hidden in jQuery?How do I remove a property from a JavaScript object?var functionName = function() vs function functionName() How do I redirect to another webpage?How do I include a JavaScript file in another JavaScript file?How to check whether a string contains a substring in JavaScript?How do I remove a particular element from an array in JavaScript?How do I return the response from an asynchronous call?













19















I looked at sample http://bl.ocks.org/mbostock/raw/4063570/:



enter image description here



It produces nice merged lines from source target from left to right.



In my case I need to layout nodes manually and put x, y coordinates. In this case the lines are not merged at source nodes. Here is the test code that reproduce this problem:



var data = [ name: "p1", children: [name: "c1", name: "c2", name: "c3", name: "c4"]];
var width = 400, height = 200, radius = 10, gap = 50;

// test layout
var nodes = [];
var links = [];
data.forEach(function(d, i)
d.x = width/4;
d.y = height/2;
nodes.push(d);
d.children.forEach(function(c, i)
c.x = 3*width/4;
c.y = gap * (i +1) -2*radius;
nodes.push(c);
links.push(source: d, target: c);
)
)

var color = d3.scale.category20();

var svg = d3.select("#chart").append("svg")
.attr("width", width)
.attr("height", height)
.append("g");
var diagonal = d3.svg.diagonal()
.projection(function(d) return [d.x, d.y]; );

var link = svg.selectAll(".link")
.data(links)
.enter().append("path")
.attr("class", "link")
.attr("d", diagonal);

var circle = svg.selectAll(".circle")
.data(nodes)
.enter()
.append("g")
.attr("class", "circle");

var el = circle.append("circle")
.attr("cx", function(d) return d.x)
.attr("cy", function(d) return d.y)
.attr("r", radius)
.style("fill", function(d) return color(d.name))
.append("title").text(function(d) return d.name);


There is sample of this at http://jsfiddle.net/zmagdum/qsEbd/:



enter image description here



However, it looks like the behavior of curves close to nodes are opposite of desired. I would like them to start straight horizontally at the nodes and make a curve in the middle. Is there a trick to do this?










share|improve this question




























    19















    I looked at sample http://bl.ocks.org/mbostock/raw/4063570/:



    enter image description here



    It produces nice merged lines from source target from left to right.



    In my case I need to layout nodes manually and put x, y coordinates. In this case the lines are not merged at source nodes. Here is the test code that reproduce this problem:



    var data = [ name: "p1", children: [name: "c1", name: "c2", name: "c3", name: "c4"]];
    var width = 400, height = 200, radius = 10, gap = 50;

    // test layout
    var nodes = [];
    var links = [];
    data.forEach(function(d, i)
    d.x = width/4;
    d.y = height/2;
    nodes.push(d);
    d.children.forEach(function(c, i)
    c.x = 3*width/4;
    c.y = gap * (i +1) -2*radius;
    nodes.push(c);
    links.push(source: d, target: c);
    )
    )

    var color = d3.scale.category20();

    var svg = d3.select("#chart").append("svg")
    .attr("width", width)
    .attr("height", height)
    .append("g");
    var diagonal = d3.svg.diagonal()
    .projection(function(d) return [d.x, d.y]; );

    var link = svg.selectAll(".link")
    .data(links)
    .enter().append("path")
    .attr("class", "link")
    .attr("d", diagonal);

    var circle = svg.selectAll(".circle")
    .data(nodes)
    .enter()
    .append("g")
    .attr("class", "circle");

    var el = circle.append("circle")
    .attr("cx", function(d) return d.x)
    .attr("cy", function(d) return d.y)
    .attr("r", radius)
    .style("fill", function(d) return color(d.name))
    .append("title").text(function(d) return d.name);


    There is sample of this at http://jsfiddle.net/zmagdum/qsEbd/:



    enter image description here



    However, it looks like the behavior of curves close to nodes are opposite of desired. I would like them to start straight horizontally at the nodes and make a curve in the middle. Is there a trick to do this?










    share|improve this question


























      19












      19








      19


      7






      I looked at sample http://bl.ocks.org/mbostock/raw/4063570/:



      enter image description here



      It produces nice merged lines from source target from left to right.



      In my case I need to layout nodes manually and put x, y coordinates. In this case the lines are not merged at source nodes. Here is the test code that reproduce this problem:



      var data = [ name: "p1", children: [name: "c1", name: "c2", name: "c3", name: "c4"]];
      var width = 400, height = 200, radius = 10, gap = 50;

      // test layout
      var nodes = [];
      var links = [];
      data.forEach(function(d, i)
      d.x = width/4;
      d.y = height/2;
      nodes.push(d);
      d.children.forEach(function(c, i)
      c.x = 3*width/4;
      c.y = gap * (i +1) -2*radius;
      nodes.push(c);
      links.push(source: d, target: c);
      )
      )

      var color = d3.scale.category20();

      var svg = d3.select("#chart").append("svg")
      .attr("width", width)
      .attr("height", height)
      .append("g");
      var diagonal = d3.svg.diagonal()
      .projection(function(d) return [d.x, d.y]; );

      var link = svg.selectAll(".link")
      .data(links)
      .enter().append("path")
      .attr("class", "link")
      .attr("d", diagonal);

      var circle = svg.selectAll(".circle")
      .data(nodes)
      .enter()
      .append("g")
      .attr("class", "circle");

      var el = circle.append("circle")
      .attr("cx", function(d) return d.x)
      .attr("cy", function(d) return d.y)
      .attr("r", radius)
      .style("fill", function(d) return color(d.name))
      .append("title").text(function(d) return d.name);


      There is sample of this at http://jsfiddle.net/zmagdum/qsEbd/:



      enter image description here



      However, it looks like the behavior of curves close to nodes are opposite of desired. I would like them to start straight horizontally at the nodes and make a curve in the middle. Is there a trick to do this?










      share|improve this question
















      I looked at sample http://bl.ocks.org/mbostock/raw/4063570/:



      enter image description here



      It produces nice merged lines from source target from left to right.



      In my case I need to layout nodes manually and put x, y coordinates. In this case the lines are not merged at source nodes. Here is the test code that reproduce this problem:



      var data = [ name: "p1", children: [name: "c1", name: "c2", name: "c3", name: "c4"]];
      var width = 400, height = 200, radius = 10, gap = 50;

      // test layout
      var nodes = [];
      var links = [];
      data.forEach(function(d, i)
      d.x = width/4;
      d.y = height/2;
      nodes.push(d);
      d.children.forEach(function(c, i)
      c.x = 3*width/4;
      c.y = gap * (i +1) -2*radius;
      nodes.push(c);
      links.push(source: d, target: c);
      )
      )

      var color = d3.scale.category20();

      var svg = d3.select("#chart").append("svg")
      .attr("width", width)
      .attr("height", height)
      .append("g");
      var diagonal = d3.svg.diagonal()
      .projection(function(d) return [d.x, d.y]; );

      var link = svg.selectAll(".link")
      .data(links)
      .enter().append("path")
      .attr("class", "link")
      .attr("d", diagonal);

      var circle = svg.selectAll(".circle")
      .data(nodes)
      .enter()
      .append("g")
      .attr("class", "circle");

      var el = circle.append("circle")
      .attr("cx", function(d) return d.x)
      .attr("cy", function(d) return d.y)
      .attr("r", radius)
      .style("fill", function(d) return color(d.name))
      .append("title").text(function(d) return d.name);


      There is sample of this at http://jsfiddle.net/zmagdum/qsEbd/:



      enter image description here



      However, it looks like the behavior of curves close to nodes are opposite of desired. I would like them to start straight horizontally at the nodes and make a curve in the middle. Is there a trick to do this?







      javascript svg d3.js curve






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 17 '15 at 17:11









      VividD

      8,50565097




      8,50565097










      asked Feb 21 '13 at 16:53









      John SmithJohn Smith

      121126




      121126






















          2 Answers
          2






          active

          oldest

          votes


















          26














          This solution is based on excellent @bmdhacks solution, however, I believe mine is slightly better, since it doesn't require swapping x and y within data itself.



          The idea is that you can use diagonal.source() and diagonal.target() to swap x and y:



          var diagonal = d3.svg.diagonal()
          .source(function(d) return "x":d.source.y, "y":d.source.x; )
          .target(function(d) return "x":d.target.y, "y":d.target.x; )
          .projection(function(d) return [d.y, d.x]; );


          All x y swapping is now encapsulated within the code above.



          The result:



          enter image description here



          Here is also jsfiddle.






          share|improve this answer

























          • Thanks for giving this example showing that the expected return values for source and projection are different.

            – David J.
            Sep 10 '14 at 14:35











          • What if there are multiple links between two nodes? How can I show it without overlapping?

            – Jerry
            May 17 '17 at 11:06


















          8














          Note that in the blocks example, the x and y values are swapped in the links. This would normally draw the links in the wrong place, but he's also supplied a projection function that swaps them back.



          var diagonal = d3.svg.diagonal()
          .projection(function(d) return [d.y, d.x]; );


          Here's your jsfiddle with this technique applied:
          http://jsfiddle.net/bmdhacks/qsEbd/5/






          share|improve this answer























          • What to do if i want to make link between two children? Like first child linked with second child in first row after root?

            – Sohail Ahmad
            Jul 24 '13 at 10:52










          Your Answer






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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          26














          This solution is based on excellent @bmdhacks solution, however, I believe mine is slightly better, since it doesn't require swapping x and y within data itself.



          The idea is that you can use diagonal.source() and diagonal.target() to swap x and y:



          var diagonal = d3.svg.diagonal()
          .source(function(d) return "x":d.source.y, "y":d.source.x; )
          .target(function(d) return "x":d.target.y, "y":d.target.x; )
          .projection(function(d) return [d.y, d.x]; );


          All x y swapping is now encapsulated within the code above.



          The result:



          enter image description here



          Here is also jsfiddle.






          share|improve this answer

























          • Thanks for giving this example showing that the expected return values for source and projection are different.

            – David J.
            Sep 10 '14 at 14:35











          • What if there are multiple links between two nodes? How can I show it without overlapping?

            – Jerry
            May 17 '17 at 11:06















          26














          This solution is based on excellent @bmdhacks solution, however, I believe mine is slightly better, since it doesn't require swapping x and y within data itself.



          The idea is that you can use diagonal.source() and diagonal.target() to swap x and y:



          var diagonal = d3.svg.diagonal()
          .source(function(d) return "x":d.source.y, "y":d.source.x; )
          .target(function(d) return "x":d.target.y, "y":d.target.x; )
          .projection(function(d) return [d.y, d.x]; );


          All x y swapping is now encapsulated within the code above.



          The result:



          enter image description here



          Here is also jsfiddle.






          share|improve this answer

























          • Thanks for giving this example showing that the expected return values for source and projection are different.

            – David J.
            Sep 10 '14 at 14:35











          • What if there are multiple links between two nodes? How can I show it without overlapping?

            – Jerry
            May 17 '17 at 11:06













          26












          26








          26







          This solution is based on excellent @bmdhacks solution, however, I believe mine is slightly better, since it doesn't require swapping x and y within data itself.



          The idea is that you can use diagonal.source() and diagonal.target() to swap x and y:



          var diagonal = d3.svg.diagonal()
          .source(function(d) return "x":d.source.y, "y":d.source.x; )
          .target(function(d) return "x":d.target.y, "y":d.target.x; )
          .projection(function(d) return [d.y, d.x]; );


          All x y swapping is now encapsulated within the code above.



          The result:



          enter image description here



          Here is also jsfiddle.






          share|improve this answer















          This solution is based on excellent @bmdhacks solution, however, I believe mine is slightly better, since it doesn't require swapping x and y within data itself.



          The idea is that you can use diagonal.source() and diagonal.target() to swap x and y:



          var diagonal = d3.svg.diagonal()
          .source(function(d) return "x":d.source.y, "y":d.source.x; )
          .target(function(d) return "x":d.target.y, "y":d.target.x; )
          .projection(function(d) return [d.y, d.x]; );


          All x y swapping is now encapsulated within the code above.



          The result:



          enter image description here



          Here is also jsfiddle.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Oct 5 '17 at 17:52

























          answered May 3 '14 at 15:14









          VividDVividD

          8,50565097




          8,50565097












          • Thanks for giving this example showing that the expected return values for source and projection are different.

            – David J.
            Sep 10 '14 at 14:35











          • What if there are multiple links between two nodes? How can I show it without overlapping?

            – Jerry
            May 17 '17 at 11:06

















          • Thanks for giving this example showing that the expected return values for source and projection are different.

            – David J.
            Sep 10 '14 at 14:35











          • What if there are multiple links between two nodes? How can I show it without overlapping?

            – Jerry
            May 17 '17 at 11:06
















          Thanks for giving this example showing that the expected return values for source and projection are different.

          – David J.
          Sep 10 '14 at 14:35





          Thanks for giving this example showing that the expected return values for source and projection are different.

          – David J.
          Sep 10 '14 at 14:35













          What if there are multiple links between two nodes? How can I show it without overlapping?

          – Jerry
          May 17 '17 at 11:06





          What if there are multiple links between two nodes? How can I show it without overlapping?

          – Jerry
          May 17 '17 at 11:06













          8














          Note that in the blocks example, the x and y values are swapped in the links. This would normally draw the links in the wrong place, but he's also supplied a projection function that swaps them back.



          var diagonal = d3.svg.diagonal()
          .projection(function(d) return [d.y, d.x]; );


          Here's your jsfiddle with this technique applied:
          http://jsfiddle.net/bmdhacks/qsEbd/5/






          share|improve this answer























          • What to do if i want to make link between two children? Like first child linked with second child in first row after root?

            – Sohail Ahmad
            Jul 24 '13 at 10:52















          8














          Note that in the blocks example, the x and y values are swapped in the links. This would normally draw the links in the wrong place, but he's also supplied a projection function that swaps them back.



          var diagonal = d3.svg.diagonal()
          .projection(function(d) return [d.y, d.x]; );


          Here's your jsfiddle with this technique applied:
          http://jsfiddle.net/bmdhacks/qsEbd/5/






          share|improve this answer























          • What to do if i want to make link between two children? Like first child linked with second child in first row after root?

            – Sohail Ahmad
            Jul 24 '13 at 10:52













          8












          8








          8







          Note that in the blocks example, the x and y values are swapped in the links. This would normally draw the links in the wrong place, but he's also supplied a projection function that swaps them back.



          var diagonal = d3.svg.diagonal()
          .projection(function(d) return [d.y, d.x]; );


          Here's your jsfiddle with this technique applied:
          http://jsfiddle.net/bmdhacks/qsEbd/5/






          share|improve this answer













          Note that in the blocks example, the x and y values are swapped in the links. This would normally draw the links in the wrong place, but he's also supplied a projection function that swaps them back.



          var diagonal = d3.svg.diagonal()
          .projection(function(d) return [d.y, d.x]; );


          Here's your jsfiddle with this technique applied:
          http://jsfiddle.net/bmdhacks/qsEbd/5/







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered May 15 '13 at 17:55









          bmdhacksbmdhacks

          13.4k62852




          13.4k62852












          • What to do if i want to make link between two children? Like first child linked with second child in first row after root?

            – Sohail Ahmad
            Jul 24 '13 at 10:52

















          • What to do if i want to make link between two children? Like first child linked with second child in first row after root?

            – Sohail Ahmad
            Jul 24 '13 at 10:52
















          What to do if i want to make link between two children? Like first child linked with second child in first row after root?

          – Sohail Ahmad
          Jul 24 '13 at 10:52





          What to do if i want to make link between two children? Like first child linked with second child in first row after root?

          – Sohail Ahmad
          Jul 24 '13 at 10:52

















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