Question

是否有任何曲目包可以绘制这样的和弦布局:(此可视化也称为 和弦图 ）

Chrod Diagramm

Answer 1

几年前我写过以下内容，但从未真正使用过它：随意根据自己的需要调整它，甚至将它变成一个完整的包。

# Return a line in the Poincare disk, i.e.,
# a circle arc, perpendicular to the unit circle, through two given points.
poincare_segment <- function(u1, u2, v1, v2) {
    # Check that the points are sufficiently different
    if( abs(u1-v1) < 1e-6 && abs(u2-v2) < 1e-6 )
        return( list(x=c(u1,v1), y=c(u2,v2)) )
    # Check that we are in the circle
    stopifnot( u1^2 + u2^2 - 1 <= 1e-6 )
    stopifnot( v1^2 + v2^2 - 1 <= 1e-6 )
    # Check it is not a diameter
    if( abs( u1*v2 - u2*v1 ) < 1e-6 )
        return( list(x=c(u1,v1), y=c(u2,v2)) )
    # Equation of the line: x^2 + y^2 + ax + by + 1 = 0 (circles orthogonal to the unit circle)
    a <- ( u2 * (v1^2+v2^2) - v2 * (u1^2+u2^2) + u2 - v2 ) / ( u1*v2 - u2*v1 )
    b <- ( u1 * (v1^2+v2^2) - v1 * (u1^2+u2^2) + u1 - v1 ) / ( u2*v1 - u1*v2 ) # Swap 1's and 2's
    # Center and radius of the circle
    cx <- -a/2
    cy <- -b/2
    radius <- sqrt( (a^2+b^2)/4 - 1 )
    # Which portion of the circle should we draw?
    theta1 <- atan2( u2-cy, u1-cx )
    theta2 <- atan2( v2-cy, v1-cx )
    if( theta2 - theta1 > pi )
        theta2 <- theta2 - 2 * pi
    else if( theta2 - theta1 < - pi )
        theta2 <- theta2 + 2 * pi
    theta <- seq( theta1, theta2, length=100 )
    x <- cx + radius * cos( theta )
    y <- cy + radius * sin( theta )
    list( x=x, y=y )
}

# Sample data
n <- 10
m <- 7
segment_weight <- abs(rnorm(n))
segment_weight <- segment_weight / sum(segment_weight)
d <- matrix(abs(rnorm(n*n)),nr=n, nc=n)
diag(d) <- 0 # No loops allowed
# The weighted graph comes from two quantitative variables
d[1:m,1:m] <- 0
d[(m+1):n,(m+1):n] <- 0
ribbon_weight <- t(d) / apply(d,2,sum) # The sum of each row is 1; use as ribbon_weight[from,to]
ribbon_order <- t(apply(d,2,function(...)sample(1:n))) # Each row contains sample(1:n); use as ribbon_order[from,i]
segment_colour <- rainbow(n)
segment_colour <- brewer.pal(n,"Set3")
transparent_segment_colour <- rgb(t(col2rgb(segment_colour)/255),alpha=.5)
ribbon_colour <- matrix(rainbow(n*n), nr=n, nc=n) # Not used, actually...
ribbon_colour[1:m,(m+1):n] <- transparent_segment_colour[1:m]
ribbon_colour[(m+1):n,1:m] <- t(ribbon_colour[1:m,(m+1):n])

# Plot
gap <- .01
x <- c( segment_weight[1:m], gap, segment_weight[(m+1):n], gap )
x <- x / sum(x)
x <- cumsum(x)
segment_start <- c(0,x[1:m-1],x[(m+1):n])
segment_end   <- c(x[1:m],x[(m+2):(n+1)])
start1 <- start2 <- end1 <- end2 <- ifelse(is.na(ribbon_weight),NA,NA)
x <- 0
for (from in 1:n) {
  x <- segment_start[from]
  for (i in 1:n) {
    to <- ribbon_order[from,i]
    y <- x + ribbon_weight[from,to] * ( segment_end[from] - segment_start[from] )
    if( from < to ) {
      start1[from,to] <- x
      start2[from,to] <- y
    } else if( from > to ) {
      end1[to,from] <- x
      end2[to,from] <- y
    } else {
      # no loops allowed
    }
    x <- y
  }
}

par(mar=c(1,1,2,1))
plot(
  0,0, 
  xlim=c(-1,1),ylim=c(-1,1), type="n", axes=FALSE, 
  main="Two qualitative variables in polar coordinates", xlab="", ylab="")
for(from in 1:n) {
  for(to in 1:n) {
    if(from<to) {
      u <- start1[from,to]
      v <- start2[from,to]
      x <- end1  [from,to]
      y <- end2  [from,to]
      if(!is.na(u*v*x*y)) {
            r1 <- poincare_segment( cos(2*pi*v), sin(2*pi*v), cos(2*pi*x), sin(2*pi*x) )
            r2 <- poincare_segment( cos(2*pi*y), sin(2*pi*y), cos(2*pi*u), sin(2*pi*u) )
            th1 <- 2*pi*seq(u,v,length=20)
            th2 <- 2*pi*seq(x,y,length=20)
            polygon(
                c( cos(th1), r1$x, rev(cos(th2)), r2$x ),
                c( sin(th1), r1$y, rev(sin(th2)), r2$y ),
                col=transparent_segment_colour[from], border=NA
            )
      }
    }
  }
}
for(i in 1:n) {
  theta <- 2*pi*seq(segment_start[i], segment_end[i], length=100)
  r1 <- 1
  r2 <- 1.05
  polygon( 
    c( r1*cos(theta), rev(r2*cos(theta)) ),
    c( r1*sin(theta), rev(r2*sin(theta)) ),
    col=segment_colour[i], border="black"
  )
}

Two quantitative variables in polar coordinates

Answer 2

chorddiag包（仍在开发中）提供了交互式D3实现

chorddiag包允许使用htmlwidgets接口框架从R内部使用JavaScript可视化库D3（http://d3js.org）创建交互式和弦图。

示例

devtools::install_github("mattflor/chorddiag") library(chorddiag) ## example taken from the github site m <- matrix(c(11975, 5871, 8916, 2868, 1951, 10048, 2060, 6171, 8010, 16145, 8090, 8045, 1013, 990, 940, 6907), byrow = TRUE, nrow = 4, ncol = 4) haircolors <- c("black", "blonde", "brown", "red") dimnames(m) <- list(have = haircolors, prefer = haircolors) m # prefer # have black blonde brown red # black 11975 5871 8916 2868 # blonde 1951 10048 2060 6171 # brown 8010 16145 8090 8045 # red 1013 990 940 6907 groupColors <- c("#000000", "#FFDD89", "#957244", "#F26223") chorddiag(m, groupColors = groupColors, groupnamePadding = 40)

Answer 3

如果您不想特别绘制基因组数据，而是来自任何域的数据，我认为最近发布的包circlize: Circular Visualization in R提供了比RCircos更直接的方法。

circlize example

Answer 4

这看起来非常像Circos情节。 Circos是在Perl中实现的，但您可以使用R来塑造数据，以便将其提供给Circos。 BioStar有一个相关的问题：http://www.biostars.org/p/17728/

Answer 5

如果你熟悉ggplot，那么ggbio就是你的选择。

此处提供了文档： http://www.tengfei.name/ggbio/

绘制圆形图的功能是layout_circle（）。绘制基因组数据的另一个非常有用的功能是layout_karyogram（）。

R中的和弦图

5 个答案: