Question

这包括为堆积的条形图的每个条形图分配不同的颜色，因为目前在所有四个条形图中只有单一颜色，并且堆叠条形图中的颜色正在变化，但我试图为所有四个条形图分配不同的颜色酒吧以及堆积值。

这是代码

    var margin = {
            top: 20,
            right: 160,
            bottom: 35,
            left: 30
          };

          var width = 960 - margin.left - margin.right,
            height = 600 - margin.top - margin.bottom;

          var svg = d3.select("body")
            .append("svg")
            .attr("width", width + margin.left + margin.right)
            .attr("height", height + margin.top + margin.bottom)
            .append("g")
            .attr("transform", "translate(" + margin.left + "," + margin.top + ")");


          /* Data in strings like it would be if imported from a csv */

          var data = [{
              year: "A",
              redDelicious: "10",
              mcintosh: "5",
              oranges: "19"
            }, {
              year: "B",
              redDelicious: "12",
              mcintosh: "0",
              oranges: "15"
            }, {
              year: "C",
              redDelicious: "05",
              mcintosh: "0",
              oranges: "28"
            }, {
              year: "D",
              redDelicious: "14",
              mcintosh: "0",
              oranges: "12"
            },

          ];
          $("#btn").on("click", function(){
            d3.selectAll("svg > g > g").remove();
            data[1].mcintosh = (Number(data[1].mcintosh) + 1).toString();
            console.log(1,data);
            update();
          });
          update();
    function update(){
          var orangeData = data.map(function(d) {
            return {
              year: d.year,
              oranges: +d.oranges
            }
          });
          console.log(orangeData)



          // Transpose the data into layers
          var dataset = d3.layout.stack()(["redDelicious", "mcintosh"].map(function(skillset) {
            return data.map(function(d) {
              return {
                x: d.year,
                y: +d[skillset]
              };
            });
          }));
          console.log(dataset)

          // Set x, y and colors
          var x = d3.scale.ordinal()
            .domain(dataset[0].map(function(d) {
              return d.x;
            }))
            .rangeRoundBands([10, width - 10], 0.02);

          var y = d3.scale.linear()
            .domain([0, d3.max(dataset, function(d) {
              return d3.max(d, function(d) {
                return d.y0 + d.y;
              });
            })])
            .range([height, 0]);

          var colors = ["#b33040", "#d9d574"];
           var backcolors = ["red", "blue","green","pink"];


          // Define and draw axes
          var yAxis = d3.svg.axis()
            .scale(y)
            .orient("left")
            .ticks(5)
            .tickSize(-width, 0, 0)
            .tickFormat(function(d) {
              return d
            });

          var xAxis = d3.svg.axis()
            .scale(x)
            .orient("bottom");
          //  .tickFormat(d3.time.format("%Y"));

          svg.append("g")
            .attr("class", "y axis")
            .call(yAxis);

          svg.append("g")
            .attr("class", "x axis")
            .attr("transform", "translate(0," + height + ")")
            .call(xAxis);

            // Creating the Average Bar for the Semester
            svg.selectAll(".bar1").data(orangeData).enter().append("g")
            .attr("class", "bar1").append("rect")
            .attr("x", function(d) {
              return x(d.year) ; // center it
            })
            .attr("width", x.rangeBand()) // make it slimmer
            .attr("y", function(d) {
              return y(d.oranges);
            })
            .attr("height", function(d) {
              return height - y(d.oranges);
            });



          // Create groups for each series, rects for each segment in Stacked Bar 
          var groups = svg.selectAll("g.cost")
            .data(dataset)
            .enter().append("g")
            .attr("class", "cost")
            .style("fill", function(d, i) {
              return colors[i];
            });

          var rect = groups.selectAll("rect")
            .data(function(d) {
              return d;
            })
            .enter()
            .append("rect")
            .attr("x", function(d) {
              return x(d.x)  + 20 ;
            })
            .attr("y", function(d) {
              return y(d.y0 + d.y);
            })
            .attr("height", function(d) {
              return y(d.y0) - y(d.y0 + d.y);
            })
            .attr("width", x.rangeBand() -40 );
            }

这是工作fiddle

Answer 1

首先，设置颜色数组：

var colors = ['#7fc97f','#beaed4','#fdc086','#ffff99','#386cb0','#f0027f','#bf5b17','#666666'];

然后，在您的矩形中（不在您的组中），使用父级索引设置每个条形的填充：

.attr("fill", function(d, i, j) {
      return colors[(j*4)+i];
 });

这里，幻数“4”是组的数量。相应地更改它（如果你创建更多的条）。

这是你的小提琴：https://fiddle.jshell.net/747beqqc/

使用D3中的不同颜色更改每个堆积条形图的颜色

1 个答案: