Question

在下面的小提示中，您可以注意到第一个栏的标签为no，即使其值为零。如何在堆积条形图中删除零值的标签？

CODE：

var data = [
    {
        "episode": "Ep. 01",
        "yes": "100",
        "no": "0"
    },
    {
        "episode": "Ep. 02",
        "yes": "70",
        "no": "30"
    },
    {
        "episode": "Ep. 03",
        "yes": "50",
        "no": "50"
    },
    {
        "episode": "Ep. 04",
        "yes": "90",
        "no": "10"
    },
    {
        "episode": "Ep. 05",
        "yes": "30",
        "no": "70"
    },
    {
        "episode": "Ep. 06",
        "yes": "60",
        "no": "40"
    },
    {
        "episode": "Ep. 07",
        "yes": "70",
        "no": "30"
    },
    {
        "episode": "Ep. 08",
        "yes": "50",
        "no": "50"
    },
    {
        "episode": "Ep. 09",
        "yes": "90",
        "no": "10"
    },
    {
        "episode": "Ep. 10",
        "yes": "30",
        "no": "70"
    }
];

var margin = {top: 20, right: 100, bottom: 30, left: 40},
    width = 600 - margin.left - margin.right,
    height = 500 - margin.top - margin.bottom;

var x = d3.scale.ordinal()
.domain(data.map( function(d) { return d.episode; }))
.rangeRoundBands([0, width], .1);

var y = d3.scale.linear()
.rangeRound([height, 0]);

var color = d3.scale.ordinal()
.range(["#B4D92A", "#FF3332"]);

var xAxis = d3.svg.axis()
.scale(x)
.orient("bottom")
.tickSize(10,10);

var yAxis = d3.svg.axis()
.scale(y)
.orient("left")
.tickFormat(d3.format(".0%"));

var svg = d3.select("body").append("svg")
.attr("width", width + margin.left + margin.right)
.attr("height", height + margin.top + margin.bottom + 100)
.append("g")
.attr("transform", "translate(" + margin.left + "," + margin.top + ")");

//d3.csv("data.csv", function(error, data) {
color.domain(d3.keys(data[0]).filter(function(key) { return key !== "episode"; }));

data.forEach(function(d) {
    var y0 = 0;
    d.ages = color.domain().map(function(name) { return {name: name, y0: y0, y1: y0 += +d[name]}; });
    d.ages.forEach(function(d) { d.y0 /= y0; d.y1 /= y0; });
});

//data.sort(function(a, b) { return b.ages[0].y1 - a.ages[0].y1; });

// var unique = data.map(function(d) { return d.episode.substring(0,5)+'...'; });

// x.domain(unique);

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

svg.selectAll(".ellipse")
.data(data)
.enter()
.append("ellipse")
.attr("cx", function(d) { return x(d.episode) + 14; })
.attr("cy", function(d) { return y(0); })
.attr("rx", 18)
.attr("ry", 5)
.style("fill", "#728220");

var episode = svg.selectAll(".episode")
.data(data)
.enter().append("g")
.attr("class", "episode")
.attr("transform", function(d) { return "translate(" + x(d.episode) + ",0)"; });

var tooltip = d3.select("body").append("div")   
.attr("class", "tooltip")               
.style("opacity", 0);

episode.selectAll("rect")
.data(function(d) { return d.ages; })
.enter().append("rect")
.attr("width", x.rangeBand() - 15)
.attr("y", function(d) { return y(d.y1); })
.attr("height", function(d) { return y(d.y0) - y(d.y1); })
.style("fill", function(d) { return color(d.name); })
.on("mouseover", function(d) { 
    var x=Number($(this).attr("height"))/45;
     if($(this).css("fill")=="rgb(255, 51, 50)" || $(this).css("fill")=="#ff3332" ){
    tooltip.html("YES: " + Number((10-x)*10) + "%<br/>NO: " + Number(x*10) + "%")
      }
    else
    {
        tooltip.html("YES: " + Number((x)*10) + "%<br/>NO: " + Number(10-x)*10 + "%")
    }

    tooltip.transition().duration(200).style("opacity", .9);      
    tooltip  
    .style("left", (d3.event.pageX) + "px")     
    .style("top", (d3.event.pageY - 28) + "px");    
})                  
.on("mouseout", function(d) {       
    tooltip.transition().duration(500).style("opacity", 0);   
});
/*.attr("rx", 5)
.attr("ry", 5);*/


var label = svg.selectAll(".episode")
.selectAll(".legend")
.data(function(d) { return d.ages; })
.enter().append("g")
.attr("class", "legend")
.attr("transform", function(d) { return "translate(" + x.rangeBand() / 2 + "," + y((d.y0 + d.y1) / 2) + ")"; });

label.append("text")
.attr("x", -15)
.attr("dy", "-0.35em")
.attr("transform", "rotate(-90)")
.text(function(d) { return d.name; });


function wrap(text, width) {
  text.each(function() {
    var text = d3.select(this),
        words = text.text().split(/\s+/).reverse(),
        word,
        line = [],
        lineNumber = 0,
        lineHeight = 1.1, // ems
        y = text.attr("y"),
        dy = parseFloat(text.attr("dy")),
        tspan = text.text(null).append("tspan").attr("x", 0).attr("y", y).attr("dy", dy + "em");
    while (word = words.pop()) {
      line.push(word);
      tspan.text(line.join(" "));
      if (tspan.node().getComputedTextLength() > width) {
        line.pop();
        tspan.text(line.join(" "));
        line = [word];
        tspan = text.append("tspan").attr("x", 0).attr("y", y).attr("dy", ++lineNumber * lineHeight + dy + "em").text(word);
      }
    }
  });
}


//});

Answer 1

有很多方法可以解决这个问题，在创建文本元素时请考虑这一点：

.style("display", function(d) { (test if value is zero) ? "none" : "inline"; }

在你的小提琴中实施：http://jsfiddle.net/V5fJ7/1/

或者，您可以在实际创建文本之前过滤您的选择，因此不会首先创建它们：

d3Selection.filter(function(d) { return (conditions of data you wish to keep); })
    .append("text")
    .etc..

您可以在此处看到：http://jsfiddle.net/V5fJ7/2/

D3条形图 - 删除零值标签

1 个答案: