[英]Get y coordinate of point along SVG path with given an x coordinate
I am using raphael.js to draw a simple SVG line graph like this:我正在使用 raphael.js 绘制一个简单的 SVG 线图,如下所示:
When the user hovers the graph, id like to display a popover pointing to the line at the X position of the cursor, and at the Y position where the line is for that X position like so:当用户悬停图形时, id 喜欢显示一个弹出框,指向光标X位置的线,以及Y位置的线所在的X位置,如下所示:
I need to take the path and find the Y coordinate for a given X coordinate.我需要走路径并找到给定X坐标的Y坐标。
Based on @Duopixel's D3 solution, I wrote the following function for my own use, in pure javascript using DOM API:基于@Duopixel 的 D3 解决方案,我使用 DOM API 在纯 javascript 中编写了以下函数供我自己使用:
function findY(path, x) {
var pathLength = path.getTotalLength()
var start = 0
var end = pathLength
var target = (start + end) / 2
// Ensure that x is within the range of the path
x = Math.max(x, path.getPointAtLength(0).x)
x = Math.min(x, path.getPointAtLength(pathLength).x)
// Walk along the path using binary search
// to locate the point with the supplied x value
while (target >= start && target <= pathLength) {
var pos = path.getPointAtLength(target)
// use a threshold instead of strict equality
// to handle javascript floating point precision
if (Math.abs(pos.x - x) < 0.001) {
return pos.y
} else if (pos.x > x) {
end = target
} else {
start = target
}
target = (start + end) / 2
}
}
If you know all the points of your path, it might be more performant to search the d
attribute of your path for the specific x coordinate you are looking for, and retrieve the y coordinate using regexp:如果您知道路径的所有点,那么在路径的
d
属性中搜索您要查找的特定 x 坐标并使用正则表达式检索 y 坐标可能会更高效:
const regex = new RegExp(`${x} ((\d*.\d*))`)
const match = regex.exec(d)
If you want to find the y of and arbitrary x coordinate not in your paths d attribute, you could loop through all the coordinates of your path and find the x coordinate that is closest to the one you're looking for.如果您想找到不在路径 d 属性中的 y 坐标和任意 x 坐标,您可以遍历路径的所有坐标并找到最接近您要查找的坐标的 x 坐标。 Not sure if that would be faster than stepping through the path and calling
getPointAtLength
though.不确定这是否比单步执行路径并调用
getPointAtLength
更快。
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