javascript - 返回具有给定变量 x 的两个给定时间之间等距的数组

Question

目前我的代码如下:

 var dateArray = d3.time.scale()
            .domain([new Date(2013, 1, 1), new Date(2013, 1, 7)])
            .ticks(d3.time.days, 1);

在给定变量 x 的情况下，如何返回这两个日期之间间隔相等的日期数组？即 x = 10

Answer 1

从您的时间刻度中删除刻度:

var dateArray = d3.time.scale()
    .domain([new Date(2013, 1, 1), new Date(2013, 1, 7)]);

并将它与您的 x 变量一起使用:

var x = 10;
var spacedDates = dateArray.ticks(x);

根据文档，ticks():

Returns representative dates from the scale's input domain. The returned tick dates are uniformly spaced (modulo irregular time intervals, such as months and leap years), have human-readable values (such as midnights), and are guaranteed to be within the extent of the input domain.

但是，数组的长度不一定是传递给 ticks 的数字:

If count is a number, then approximately count ticks will be returned. (emphasis mine)

这是一个演示:

var dateArray = d3.time.scale()
    .domain([new Date(2013, 1, 1), new Date(2013, 1, 7)]);
		 
var x = 10;
var spacedDates = dateArray.ticks(x);
		 
console.log(spacedDates);

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>

编辑:要获取确切 日期数，您可以使用invert() 并填充您自己的数组:

for(var i = 0; i<x; i++){
    spacedDates.push(dateArray.invert(i));
}

查看演示:

var x = 10;

var dateArray = d3.time.scale()
    .domain([new Date(2013, 1, 1), new Date(2013, 1, 7)])
		.range([0, x-1]);
		
var spacedDates = [];
		
for(var i = 0; i<x; i++){
    spacedDates.push(dateArray.invert(i));
}

console.log(spacedDates);

<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>