Question

我使用radial.plot包中的plotrix函数创建了一个图表，说明了样本适合多个不同类别的情况。下面给出的示例的结果如下所示：

可以看出，我已经使用透明度来显示重叠的线条，例如样品1和2.我仍然认为这是一个丑陋的解决方法。我想做的是在这个模拟中略微偏移线条：

到目前为止我没有找到答案。有什么帮助吗？

library(RColorBrewer)
library(plotrix)
library(scales)

m <- matrix(0, 5, 5)
m[1,1] <- 10
m[2,1] <- 12
m[5,2] <- 15
m[2,3] <- 20
m[4,4] <- 12
m[3,5] <- 17

colors <- brewer.pal(nrow(m), "Dark2")
pdf(file = "overview-plot.pdf", width = 14, height = 10)
par(cex.axis=2)
radial.plot(m, 
            labels=c("A", "B", "C", "D", "E"),
            radlab = FALSE,
            mar = c(11,8,11,8),
            cex = 2,
            label.prop = 1.3,
            rp.type="rs",
            lty = 1,
            lwd = 5, 
            start = 0.65, 
            show.grid.labels=0, 
            point.symbols=19, 
            point.col=alpha(colors, 0.8), 
            line.col=alpha(colors, 0.5),
            boxed.radial = FALSE,
            show.radial.grid = FALSE,
            radial.lim=c(0,max(m)),
            grid.bg = "#f0f0f0")
legend(
  24,8,
  c(1:5),
  col=colors,
  lty=1, 
  lwd=5, 
  bty = "n",
  title = "Samples",
  cex = 2)
dev.off()

Answer 1

以下代码应生成类似于您想要的图表。它将矩阵转换为长度向量，并指定每个绘制长度的角度。基于针对组（“A”，“B”，......）绘制的长度的数量来调整角度。不绘制零长度。添加或修改的代码部分已标记。

库（RColorBrewer）库（plotrix）库（鳞）

m <- matrix(0, 5, 5)
m[1,1] <- 10
m[2,1] <- 12
m[5,2] <- 15
m[2,3] <- 20
m[4,4] <- 12
m[3,5] <- 17

#### added code ####
# vector of number of non-zero (plotted) lines for each group (label)
num_non_zero <- colSums(m > 0)

# angles of groups (in radians)
ang_gp <- 2*pi*seq(0,length(num_non_zero)-1,1)/length(num_non_zero)

# adjustment to angles for multiple entries in groups (in radians)
# increasing values will give more space
# no check for angle being too large and being plotted in another group
ang_adj <- 2*pi/50

# plotting angles used with one entry per line plotted (in radians)
ang_plot <- rep(0,sum(num_non_zero))

# values of lengths to plot
# which() returns values with numbering starting
# going down column 1
lens_plot <- m[which(m > 0)]

# counter used in loop
counter <- 1

# colors used in plotting as index of vectors
# modulo operator returns 0 when divisible by number of groups
# so assign this to the number of groups in this case
col_plot_inds <- which(m>0)%%nrow(m)
col_plot_inds[col_plot_inds == 0] <- nrow(m)

# calcuations of adjustment to angles for each line
# based on the number of groups
# loop over the number of groups
for(i in 1:length(num_non_zero)){

  # set of angle shifts, but not demeaned
  shifts_unscaled <- ang_adj*seq(0,num_non_zero[i]-1,1)

  # shifts demeaned
  shifts_scaled <- shifts_unscaled - mean(shifts_unscaled)

  # loop over the number if elements in each group
  for(j in 1:num_non_zero[i]){

    # assigning angle as group angle plus shift
    ang_plot[counter] <-ang_plot[counter]+ ang_gp[i] + shifts_scaled[j]

    # incrementing counter
    counter <- counter + 1
  }
}


colors <- brewer.pal(nrow(m), "Dark2")
pdf(file = "overview-plot.pdf", width = 14, height = 10)
par(cex.axis=2)
radial.plot(lens_plot, # using vector not matrix of value
            radial.pos=ang_plot, #  calcuated in loop
            labels=c("A", "B", "C", "D", "E"),
            radlab = FALSE,
            mar = c(11,8,11,8),
            cex = 2,
            label.prop = 1.3,
            rp.type="rs",
            lty = 1,
            lwd = 5, 
            start = 0.65, 
            show.grid.labels=0, 
            point.symbols=19, 
            point.col=alpha(colors[col_plot_inds], 0.8), # calling group index for colors
            line.col=alpha(colors[col_plot_inds], 0.5), # calling group index for colors
            boxed.radial = FALSE,
            show.radial.grid = FALSE,
            radial.lim=c(0,max(m)),
            grid.bg = "#f0f0f0",
            )

#### end of added code ####
legend(
  24,8,
  c(1:5),
  col=colors,
  lty=1, 
  lwd=5, 
  bty = "n",
  title = "Samples",
  cex = 2)
dev.off()

如何偏移径向图中的线？

1 个答案: