ggplot：如何在 geom_polygon 内生成渐变填充

Question

这应该很容易，但我找不到方法。

tri_fill <- structure(
  list(x= c(0.75, 0.75, 2.25, 3.25), 
       y = c(40, 43, 43, 40)), 
  .Names = c("x", "y"),
  row.names = c(NA, -4L), class = "data.frame",Integrated=NA, Related=NA)

# install.packages("ggplot2", dependencies = TRUE)
require(ggplot2)


  ggplot(data=tri_fill,aes(x=x, y=y))+ 
      geom_polygon() + 
      scale_fill_gradient(limits=c(1, 4), low = "lightgrey", high = "red")

我想要的是沿 x 轴的渐变，但使用上面的方法我只得到一个带有渐变的图例和带有实心填充的多边形。

Answer 1

当你有一个相对简单的多边形时，这是一个可能的解决方案。我们创建了许多线段，而不是多边形，并通过渐变为它们着色。结果将因此看起来像具有渐变的多边形。

#create data for 'n'segments
n_segs <- 1000

#x and xend are sequences spanning the entire range of 'x' present in the data
newpolydata <- data.frame(xstart=seq(min(tri_fill$x),max(tri_fill$x),length.out=n_segs))
newpolydata$xend <- newpolydata$xstart


#y's are a little more complicated: when x is below changepoint, y equals max(y)
#but when x is above the changepoint, the border of the polygon
#follow a line according to the formula y= intercept + x*slope.

#identify changepoint (very data/shape dependent)
change_point <-  max(tri_fill$x[which(tri_fill$y==max(tri_fill$y))])

#calculate slope and intercept
slope <- (max(tri_fill$y)-min(tri_fill$y))/ (change_point - max(tri_fill$x))
intercept <- max(tri_fill$y)

#all lines start at same y
newpolydata$ystart <- min(tri_fill$y)

#calculate y-end
newpolydata$yend <- with(newpolydata, ifelse (xstart <= change_point,
                      max(tri_fill$y),intercept+ (xstart-change_point)*slope))

p2 <- ggplot(newpolydata) +
  geom_segment(aes(x=xstart,xend=xend,y=ystart,yend=yend,color=xstart)) +
  scale_color_gradient(limits=c(0.75, 4), low = "lightgrey", high = "red")
p2 #note that I've changed the lower border of the gradient.

编辑：如果只需要具有渐变的多边形，则上述解决方案有效，但是，正如评论中指出的那样，当您计划映射一个东西进行填充而另一个东西进行着色时，这可能会产生问题，因为每个 'aes' 只能使用一次。因此，我将解决方案修改为不绘制线条，而是绘制可以具有填充 aes 的（非常细的）多边形。

#for each 'id'/polygon, four x-variables and four y-variable
#for each polygon, we start at lower left corner, and go to upper left, upper right and then to lower right.


n_polys <- 1000
#identify changepoint (very data/shape dependent)
change_point <-  max(tri_fill$x[which(tri_fill$y==max(tri_fill$y))])

#calculate slope and intercept
slope <- (max(tri_fill$y)-min(tri_fill$y))/ (change_point - max(tri_fill$x))
intercept <- max(tri_fill$y)
#calculate sequence of borders: x, and accompanying lower and upper y coordinates
x_seq <- seq(min(tri_fill$x),max(tri_fill$x),length.out=n_polys+1)
y_max_seq <- ifelse(x_seq<=change_point, max(tri_fill$y), intercept + (x_seq - change_point)*slope)
y_min_seq <- rep(min(tri_fill$y), n_polys+1)

#create polygons/rectangles
poly_list <- lapply(1:n_polys, function(p){
  res <- data.frame(x=rep(c(x_seq[p],x_seq[p+1]),each=2),
                    y = c(y_min_seq[p], y_max_seq[p:(p+1)], y_min_seq[p+1]))
  res$fill_id <- x_seq[p]
  res
}
)

poly_data <- do.call(rbind, poly_list)

#plot, allowing for both fill and color-aes
p3 <- ggplot(tri_fill, aes(x=x,y=y))+
  geom_polygon(data=poly_data, aes(x=x,y=y, group=fill_id,fill=fill_id)) +
  scale_fill_gradient(limits=c(0.75, 4), low = "lightgrey", high = "red") +
  geom_point(aes(color=factor(y)),size=5)
p3

Answer 2

找到的答案关于对三角形进行热映射可以达到相同的效果，并且可以用于任何复杂的多边形形状。

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