R 中的 Riverplot 包 - 网格线或轮廓覆盖的输出图

我在 R Riverplot (v0.5) 中制作了桑基图，输出在 RStudio 中看起来很小，但在导出或放大时，颜色具有黑色轮廓或网格线。

我认为这可能是因为形状的轮廓与我想要用于填充的透明度不匹配？

我可能需要找到一种方法来完全消除轮廓（而不是使它们半透明），因为我认为它们也是值为零的流仍然显示为细线的原因。

我的代码在这里：

#loading packages
library(readr)
library("riverplot", lib.loc="C:/Program Files/R/R-3.3.2/library")
library(RColorBrewer)

#loaing data
Cambs_flows <- read_csv("~/RProjects/Cambs_flows4.csv")

#defining the edges
edges = rep(Cambs_flows, col.names = c("N1","N2","Value"))
edges    <- data.frame(edges)
edges$ID <- 1:25

#defining the nodes
nodes <- data.frame(ID = c("Cambridge","S Cambs","Rest of E","Rest of UK","Abroad","to Cambridge","to S Cambs","to Rest of E","to Rest of UK","to Abroad"))
nodes$x = c(1,1,1,1,1,2,2,2,2,2)
nodes$y = c(1,2,3,4,5,1,2,3,4,5)

#picking colours
palette = paste0(brewer.pal(5, "Set1"), "90")

#plot styles
styles = lapply(nodes$y, function(n) {
  list(col = palette[n], lty = 0, textcol = "black")
})

#matching nodes to names
names(styles) = nodes$ID

#defining the river
r <- makeRiver( nodes, edges,
                node_labels = c("Cambridge","S Cambs","Rest of E","Rest of UK","Abroad","to Cambridge","to S Cambs","to Rest of E","to Rest of UK","to Abroad"),
                node_styles = styles)

#Plotting
plot( r, plot_area = 0.9)

我的数据在这里

dput(Cambs_flows)
structure(list(N1 = c("Cambridge", "Cambridge", "Cambridge", 
"Cambridge", "Cambridge", "S Cambs", "S Cambs", "S Cambs", "S Cambs", 
"S Cambs", "Rest of E", "Rest of E", "Rest of E", "Rest of E", 
"Rest of E", "Rest of UK", "Rest of UK", "Rest of UK", "Rest of UK", 
"Rest of UK", "Abroad", "Abroad", "Abroad", "Abroad", "Abroad"
), N2 = c("to Cambridge", "to S Cambs", "to Rest of E", "to Rest of UK", 
"to Abroad", "to Cambridge", "to S Cambs", "to Rest of E", "to Rest of UK", 
"to Abroad", "to Cambridge", "to S Cambs", "to Rest of E", "to Rest of UK", 
"to Abroad", "to Cambridge", "to S Cambs", "to Rest of E", "to Rest of UK", 
"to Abroad", "to Cambridge", "to S Cambs", "to Rest of E", "to Rest of UK", 
"to Abroad"), Value = c(0L, 1616L, 2779L, 13500L, 5670L, 2593L, 
0L, 2975L, 4742L, 1641L, 2555L, 3433L, 0L, 0L, 0L, 6981L, 3802L, 
0L, 0L, 0L, 5670L, 1641L, 0L, 0L, 0L)), class = c("tbl_df", "tbl", 
"data.frame"), row.names = c(NA, -25L), .Names = c("N1", "N2", 
"Value"), spec = structure(list(cols = structure(list(N1 = structure(list(), class = c("collector_character", 
"collector")), N2 = structure(list(), class = c("collector_character", 
"collector")), Value = structure(list(), class = c("collector_integer", 
"collector"))), .Names = c("N1", "N2", "Value")), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

罪魁祸首是一条线riverplot::curveseg。我们可以破解这个函数来修复它，或者还有一个非常简单的解决方法，不需要破解这个函数。事实上，在许多情况下，简单的解决方案可能是最好的，但首先我解释一下如何破解该功能，以便我们理解为什么该解决方法也有效。如果您只想要简单的解决方案，请滚动到此答案的末尾：

更新：下面建议的更改现已在 Riverplot 版本 0.6 中实现

要编辑该函数，您可以使用

trace(curveseg, edit=T)

然后找到函数末尾附近的一行：

polygon(c(xx[i], xx[i + 1], xx[i + 1], xx[i]), c(yy[i], 
      yy[i + 1], yy[i + 1] + w, yy[i] + w), col = grad[i], 
      border = grad[i])

我们可以在这里看到包作者选择不通过lty参数为polygon（更新：参见这个答案 https://stackoverflow.com/a/42294324/2761575解释包作者为什么这样做）。通过添加更改此行lty = 0（或者，如果你愿意，border = NA）并且它按照 OP 案例的预期工作。 （但请注意，如果您想渲染 pdf，这可能效果不佳 - 请参阅here https://stackoverflow.com/a/42294324/2761575)

polygon(c(xx[i], xx[i + 1], xx[i + 1], xx[i]), c(yy[i], 
      yy[i + 1], yy[i + 1] + w, yy[i] + w), col = grad[i], 
      border = grad[i], lty=0)

作为旁注，这也解释了评论中报道的有些奇怪的行为：“如果你运行它两次，第二次绘图看起来不错，尽管导出它并且线条又回来了”. When lty未在调用中指定polygon，它使用的默认值是lty = par("lty")。最初，默认par("lty")是一条实线，但运行一次 Riverplot 函数后，par("lty")在调用期间设置为 0riverplot:::draw.nodes因此，当riverplot第二次运行。但如果您随后尝试导出图像，打开新设备会重置par("lty")为其默认值。

更新功能的另一种方法与此编辑是使用assignInNamespace用您自己的版本覆盖包函数。像这样：

curveseg.new = function (x0, x1, y0, y1, width = 1, nsteps = 50, col = "#ffcc0066", 
          grad = NULL, lty = 1, form = c("sin", "line")) 
{
  w <- width
  if (!is.null(grad)) {
    grad <- colorRampPaletteAlpha(grad)(nsteps)
  }
  else {
    grad <- rep(col, nsteps)
  }
  form <- match.arg(form, c("sin", "line"))
  if (form == "sin") {
    xx <- seq(-pi/2, pi/2, length.out = nsteps)
    yy <- y0 + (y1 - y0) * (sin(xx) + 1)/2
    xx <- seq(x0, x1, length.out = nsteps)
  }
  if (form == "line") {
    xx <- seq(x0, x1, length.out = nsteps)
    yy <- seq(y0, y1, length.out = nsteps)
  }
  for (i in 1:(nsteps - 1)) {
    polygon(c(xx[i], xx[i + 1], xx[i + 1], xx[i]), 
            c(yy[i], yy[i + 1], yy[i + 1] + w, yy[i] + w), 
            col = grad[i], border = grad[i], lty=0)
    lines(c(xx[i], xx[i + 1]), c(yy[i], yy[i + 1]), lty = lty)
    lines(c(xx[i], xx[i + 1]), c(yy[i] + w, yy[i + 1] + w), lty = lty)
  }
}

assignInNamespace('curveseg', curveseg.new, 'riverplot', pos = -1, envir = as.environment(pos))

现在是简单的解决方案，不需要更改函数：

只需添加一行par(lty=0)在你策划之前！！！