我想制作一个只有一条垂直和一条横跨(0,0)点的水平网格线的图。我有以下脚本,但不幸的是轴刻度和网格线的原点没有正确对齐。这种差异的根源是什么?如何解决这个问题?
<!doctype html>
<meta charset="utf-8">
<script src="d3.min.js"></script>
<body>
</body>
<script>
var margin = {top: 60, right: 60, bottom: 60, left: 70},
width = 550,
height = 550;
var svg = d3.select('body').append('svg')
.attr('width', width)
.attr('height', height);
var xScale = d3.scaleLinear().domain([-1,1]).range([0+margin.left, width-margin.right]);
var yScale = d3.scaleLinear().domain([-1,1]).range([height - margin.bottom, 0 + margin.top]);
// Add the x Axis
svg.append("g")
.attr("transform", "translate(0," + (height - margin.top) + ")")
.attr("class", "axis")
.call(d3.axisBottom(xScale)
.ticks(4)
.tickSizeOuter(0)
);
svg.append("g")
.attr("transform", "translate(0," + (margin.top) + ")")
.attr("class", "axis")
.call(d3.axisTop(xScale)
.ticks(0)
.tickSizeOuter(0)
);
// Add the y Axis
svg.append("g")
.attr("transform", "translate(" + (margin.left) + ", 0)")
.attr("class", "axis")
.call(d3.axisLeft(yScale)
.ticks(4)
.tickSizeOuter(0)
);
svg.append("g")
.attr("transform", "translate(" + (width - margin.right) + ", 0)")
.attr("class", "axis")
.call(d3.axisRight(yScale)
.ticks(0)
.tickSizeOuter(0)
);
//grid lines
svg.append('line')
.attr('x1', xScale(0))
.attr('y1', height - margin.bottom)
.attr('x2', xScale(0))
.attr('y2', margin.top)
.style('stroke', 'grey')
.style('stroke-width', 1);
//grid lines
svg.append('line')
.attr('x1', margin.left)
.attr('y1', yScale(0))
.attr('x2', width - margin.right)
.attr('y2', yScale(0))
.style('stroke', 'grey')
.style('stroke-width', 1);
</script>
这是结果
答案 0 :(得分:2)
我最近遇到了同样的问题。下面是一个解决方案,但我觉得可能有更优雅的方式来做到这一点。如果你看一下轴刻度的x1和x2分量,你会发现它们的值都是0.5。 web inspector output
如果您将x1和x2行的值偏移0.5,它将正确排列:
//grid lines
svg.append('line')
.attr('x1', xScale(0) + 0.5)
.attr('y1', height - margin.bottom)
.attr('x2', xScale(0) + 0.5)
.attr('y2', margin.top)
.style('stroke', 'grey')
.style('stroke-width', 1);
以下是完整代码:js fiddle with gridlines