如何在R中找出包围原点的直线约束区域及可视化？

核心需求 #

需要可视化由数据框中**截距（Intercept）和斜率（Slope）**定义的多条直线所围成的、包含原点(0,0)（可修改为指定点(a,b)）的区域。直线数量可变，部分直线因其他直线约束更强而冗余，需忽略这类冗余直线，找出包围目标点的约束直线交点，并用这些交点绘制凸包（convex hull），且方案需适配更大规模的数据框。

示例数据框 #

df <- data.frame("Line" = c("A", "B", "C", "D", "E", "F"),
                 "Intercept" = c(4, 3, -2.5, -1.5, -5, -.5),
                 "Slope" = c(-1, 1, 2.4, -.6, -.8, .6))

示例直线绘制代码（ggplot2） #

ggplot(data = df) +
  geom_vline(xintercept = 0) +
  geom_hline(yintercept = 0) +
  geom_abline(mapping = aes(intercept = Intercept, slope = Slope),
              colour = "red") +
  coord_cartesian(xlim = c(-6, 6), ylim = c(-6, 6))

核心问题 #

如何识别包围目标点的约束直线（自动过滤冗余直线），并计算这些直线的交点，同时保证方案可适配大规模数据框？

附加问题 #

geom_abline没有类似geom_line的group美学映射，如何在ggplot2中基于斜率截距或两点绘制可标识的直线？

更新说明 #

为将多边形中心改为指定点(a,b)而非原点，修改了@AllanCameron的innermost函数，代码如下：

innermost <- function(slopes, intercepts, a, b) {

  meetings <- function(slopes, intercepts) {
    meets_at <- function(i1, s1, i2, s2) {
      ifelse(s1 - s2 == 0, NA, (i2 - i1)/(s1 - s2))
    }
    xvals <- outer(seq_along(slopes), seq_along(slopes), function(i, j) {
      meets_at(intercepts[i], slopes[i], intercepts[j], slopes[j])
    })

    yvals <- outer(seq_along(slopes), seq_along(slopes), function(i, j) {
      intercepts + slopes *
        meets_at(intercepts[i], slopes[i], intercepts[j], slopes[j])
    })

    cbind(x = xvals[lower.tri(xvals)], y = yvals[lower.tri(yvals)])

  }

  xy <- meetings(slopes, intercepts)
  xy[,1] <- xy[,1] - a
  xy[,2] <- xy[,2] - b

  is_cut <- function(x, y, slopes, intercepts, a, b) {
    d <- sqrt(x^2 + y^2)
    slope <- y / x
    xvals <- (intercepts + slopes*a - b) / (slope - slopes)
    yvals <- xvals * slopes + (intercepts + slopes*a - b)
    ds <- sqrt(xvals^2 + yvals^2)
    any(d - ds > 1e-6 & sign(xvals) == sign(x) & sign(yvals) == sign(y))
  }

  xy <- xy[sapply(seq(nrow(xy)), function(i) {
    !is_cut(xy[i, 1], xy[i, 2], slopes, intercepts, a, b)
  }),]

  xy <- xy[order(atan2(xy[,2], xy[,1])),]
  
  xy[,1] <- xy[,1] + a
  xy[,2] <- xy[,2] + b
  
  as.data.frame(rbind(xy, xy[1,]))
}

内容的提问来源于stack exchange，提问作者nico.sch