d3.js中的交替刻度线填充

Question

我正在为刷子编写以下d3.js函数。这个功能对我来说很好，但我想交替刷子的刻度线填充，这样两个连续的刻度不在同一高度，因此不重叠。这保持了刻度的清晰度。任何人都可以通过以下功能建议正确的方法。

$scope.drawBrush = function (span) {

    var width = 400,
        height = 50;

    var x = d3.time.scale()
        .domain([new Date($scope.range1), new Date($scope.range2)])
        .range([0, width]);
    d3.select('#brushd').remove();
    var brush = d3.svg.brush()
        .x(x)
        .extent([new Date($scope.range1), new Date($scope.range2)])
        .on('brush', brushed);
    var svg = d3.select('#brush-div').append('svg')
        .attr('id', 'brushd')
        .attr('width', '400')
        .attr('height', '80')
        .style('margin-left', '0px')
        .append('g')
        .attr('transform', 'translate(' + 50 + ',' + 0 + ')');

    svg.append('rect')
        .attr('class', 'grid-background')
        .attr('width', width)
        .attr('height', height);

    svg.append('g')
        .attr('class', 'x grid')
        .attr('transform', 'translate(0,' + height + ')')
        .call(d3.svg.axis()
            .scale(x)
            .orient('bottom')
            .ticks(d3.time.hours, 12)
            .tickSize(-height)
            .tickFormat(''))
        .selectAll('.tick')
        .classed('minor', function (d) {
            return d.getHours();
        });

    svg.append('g')
        .attr('class', 'x axis')
        .attr('transform', 'translate(0,' + height + ')')
        .call(d3.svg.axis()
            .scale(x)
            .orient('bottom')
            .tickFormat(d3.time.format('%b-%y'))
            .tickPadding(0))
        .selectAll('text')
        .attr('x', 6)
        .style('text-anchor', null);

    var gBrush = svg.append('g')
        .attr('class', 'brush')
        .call(brush);

    gBrush.selectAll('rect')
        .attr('height', height);

    function brushed() {
        var extent0 = brush.extent(),
            extent1;
        // if dragging, preserve the width of the extent
        if (d3.event.mode === 'move') {
            var d0 = d3.time.day.round(extent0[0]),
                d1 = d3.time.day.offset(d0, Math.round((extent0[1] - extent0[0]) / 864e5));
            extent1 = [d0, d1];
        }

        // otherwise, if resizing, round both dates
        else {
            if (span === 'daily') {
                extent1 = extent0.map(d3.time.day.round);

                // if empty when rounded, use floor & ceil instead
                if (extent1[0] >= extent1[1]) {
                    extent1[0] = d3.time.day.floor(extent0[0]);
                    extent1[1] = d3.time.day.ceil(extent0[1]);
                }
            } else if (span === 'weekly') {
                extent1 = extent0.map(d3.time.week.round);

                // if empty when rounded, use floor & ceil instead
                if (extent1[0] >= extent1[1]) {
                    extent1[0] = d3.time.week.floor(extent0[0]);
                    extent1[1] = d3.time.week.ceil(extent0[1]);
                }
            } else {
                extent1 = extent0.map(d3.time.month.round);

                // if empty when rounded, use floor & ceil instead
                if (extent1[0] >= extent1[1]) {
                    extent1[0] = d3.time.month.floor(extent0[0]);
                    extent1[1] = d3.time.month.ceil(extent0[1]);
                }
            }
        }
        d3.select(this).call(brush.extent(extent1));
        $scope.getLimit(brush.extent()[0], brush.extent()[1]);

    }
};

Answer 1

在x轴上切换长和短刻度的一种方法，以实现此类目的：

......就像这样：

// flag to alternate between long and short ticks
var alternate_ticks = false;

var short_tick_length = 4;
var long_tick_length = 16;

// Alternate the tick line between long and short ticks
d3.selectAll("g.x.axis g.tick line")
    .attr("y2", function () {
        if (alternate_ticks) {
            alternate_ticks = false;
            return long_tick_length;
        } else {
            alternate_ticks = true;
            return short_tick_length;
        }
    });

var alternate_text = false;

// Alternate the tick label text to match up with the tick length
d3.selectAll("g.x.axis g.tick text")
    .attr("y", function () {
            if (alternate_text) {
                alternate_text = false;
                return long_tick_length + 1;
            } else {
                alternate_text = true;
                return short_tick_length + 1;
            }
    });

它是上面same approach as Lars K pointed out的更简单版本，但使用的是更简单的函数，只需在false和true之间切换，以确定相关集上的刻度长度和文本定位对象。

Answer 2

尝试使用此标记。

    // Alternate the tick label text to match up with the tick length
    d3.selectAll("g.x.axis g.tick text")
        .each(function (d, i) {
            //console.log(i % 2);
            if ((i % 2)) {  //even
                this.remove();
            }
    });