在R中旋转直方图或在条形图中覆盖密度

5 回答

• 5

  scatterBarNorm <- function(x, dcol="blue", lhist=20, num.dnorm=5*lhist, ...){
      ## check input
      stopifnot(ncol(x)==2)
      ## set up layout and graphical parameters
      layMat <- matrix(c(2,0,1,3), ncol=2, byrow=TRUE)
      layout(layMat, widths=c(5/7, 2/7), heights=c(2/7, 5/7))
      ospc <- 0.5 # outer space
      pext <- 4 # par extension down and to the left
      bspc <- 1 # space between scatter plot and bar plots
      par. <- par(mar=c(pext, pext, bspc, bspc),
                  oma=rep(ospc, 4)) # plot parameters
      ## scatter plot
      plot(x, xlim=range(x[,1]), ylim=range(x[,2]), ...)
      ## 3) determine barplot and height parameter
      ## histogram (for barplot-ting the density)
      xhist <- hist(x[,1], plot=FALSE, breaks=seq(from=min(x[,1]), to=max(x[,1]),
                                       length.out=lhist))
      yhist <- hist(x[,2], plot=FALSE, breaks=seq(from=min(x[,2]), to=max(x[,2]),
                                       length.out=lhist)) # note: this uses probability=TRUE
      ## determine the plot range and all the things needed for the barplots and lines
      xx <- seq(min(x[,1]), max(x[,1]), length.out=num.dnorm) # evaluation points for the overlaid density
      xy <- dnorm(xx, mean=mean(x[,1]), sd=sd(x[,1])) # density points
      yx <- seq(min(x[,2]), max(x[,2]), length.out=num.dnorm)
      yy <- dnorm(yx, mean=mean(x[,2]), sd=sd(x[,2]))
      ## barplot and line for x (top)
      par(mar=c(0, pext, 0, 0))
      barplot(xhist$density, axes=FALSE, ylim=c(0, max(xhist$density, xy)),
              space=0) # barplot
      lines(seq(from=0, to=lhist-1, length.out=num.dnorm), xy, col=dcol) # line
      ## barplot and line for y (right)
      par(mar=c(pext, 0, 0, 0))
      barplot(yhist$density, axes=FALSE, xlim=c(0, max(yhist$density, yy)),
              space=0, horiz=TRUE) # barplot
      lines(yy, seq(from=0, to=lhist-1, length.out=num.dnorm), col=dcol) # line
      ## restore parameters
      par(par.)
  }
  
  require(mvtnorm)
  X <- rmvnorm(1000, c(0,0), matrix(c(1, 0.8, 0.8, 1), 2, 2))
  scatterBarNorm(X, xlab=expression(italic(X[1])), ylab=expression(italic(X[2])))
  

  

  回复于 2024-05-06T07:47:03+08:00
• 0

  知道 hist() 函数使用更简单的绘图函数（如 rect() ）无形地返回重现其所需信息的所有信息可能会有所帮助 .

  vals <- rnorm(10)
      A <- hist(vals)
      A
      $breaks
      [1] -1.5 -1.0 -0.5  0.0  0.5  1.0  1.5
  
      $counts
      [1] 1 3 3 1 1 1
  
      $intensities
      [1] 0.2 0.6 0.6 0.2 0.2 0.2
  
      $density
      [1] 0.2 0.6 0.6 0.2 0.2 0.2
  
      $mids
      [1] -1.25 -0.75 -0.25  0.25  0.75  1.25
  
      $xname
      [1] "vals"
  
      $equidist
      [1] TRUE
  
      attr(,"class")
      [1] "histogram"
  

  您可以手动创建相同的直方图，如下所示：

  plot(NULL, type = "n", ylim = c(0,max(A$counts)), xlim = c(range(A$breaks)))
      rect(A$breaks[1:(length(A$breaks) - 1)], 0, A$breaks[2:length(A$breaks)], A$counts)
  

  使用这些部件，您可以随意翻转轴：

  plot(NULL, type = "n", xlim = c(0, max(A$counts)), ylim = c(range(A$breaks)))
      rect(0, A$breaks[1:(length(A$breaks) - 1)], A$counts, A$breaks[2:length(A$breaks)])
  

  对于与 density() 类似的自己动手，请参阅：Axis-labeling in R histogram and density plots; multiple overlays of density plots

  回复于 2024-05-06T07:47:03+08:00
• 2

  我不确定它是否有意义，但我有时想要使用没有任何包装的水平直方图，并且能够在图形的任何位置书写或绘图 .

  这就是我编写以下函数的原因，下面提供了示例 . 如果有人知道这个包适合的包，请写信给我：berry-b at gmx.de

  请确保不要在工作区中使用变量hpos，因为它将被函数覆盖 . （是的，对于包，我需要在函数中插入一些安全部件） .

  horiz.hist <- function(Data, breaks="Sturges", col="transparent", las=1, 
  ylim=range(HBreaks), labelat=pretty(ylim), labels=labelat, border=par("fg"), ... )
    {a <- hist(Data, plot=FALSE, breaks=breaks)
    HBreaks <- a$breaks
    HBreak1 <- a$breaks[1]
    hpos <<- function(Pos) (Pos-HBreak1)*(length(HBreaks)-1)/ diff(range(HBreaks))
    barplot(a$counts, space=0, horiz=T, ylim=hpos(ylim), col=col, border=border,...)      
    axis(2, at=hpos(labelat), labels=labels, las=las, ...) 
    print("use hpos() to address y-coordinates") }
  

  举些例子

  # Data and basic concept
  set.seed(8); ExampleData <- rnorm(50,8,5)+5
  hist(ExampleData)
  horiz.hist(ExampleData, xlab="absolute frequency") 
  # Caution: the labels at the y-axis are not the real coordinates!
  # abline(h=2) will draw above the second bar, not at the label value 2. Use hpos:
  abline(h=hpos(11), col=2)
  
  # Further arguments
  horiz.hist(ExampleData, xlim=c(-8,20)) 
  horiz.hist(ExampleData, main="the ... argument worked!", col.axis=3) 
  hist(ExampleData, xlim=c(-10,40)) # with xlim
  horiz.hist(ExampleData, ylim=c(-10,40), border="red") # with ylim
  horiz.hist(ExampleData, breaks=20, col="orange")
  axis(2, hpos(0:10), labels=F, col=2) # another use of hpos()
  

  一个缺点：该功能不适用于作为具有不同宽度的条形的矢量提供的断点 .

  回复于 2024-05-06T07:47:03+08:00
• 16

  谢谢蒂姆和保罗 . 你让我更加思考并使用hist（）实际提供的东西 .

  这是我的解决方案（在Alex Pl的帮助下）：

  scatterBar.Norm <- function(x,y) {
   zones <- matrix(c(2,0,1,3), ncol=2, byrow=TRUE)
   layout(zones, widths=c(5/7,2/7), heights=c(2/7,5/7))
   xrange <- range(x)
   yrange <- range(y)
   par(mar=c(3,3,1,1))
   plot(x, y, xlim=xrange, ylim=yrange, xlab="", ylab="", cex=0.5)
   xhist <- hist(x, plot=FALSE, breaks=seq(from=min(x), to=max(x), length.out=20))
   yhist <- hist(y, plot=FALSE, breaks=seq(from=min(y), to=max(y), length.out=20))
   top <- max(c(xhist$density, yhist$density))
   par(mar=c(0,3,1,1))
   barplot(xhist$density, axes=FALSE, ylim=c(0, top), space=0)
   x.xfit <- seq(min(x),max(x),length.out=40)
   x.yfit <- dnorm(x.xfit, mean=mean(x), sd=sd(x))
   x.xscalefactor <- x.xfit / seq(from=0, to=19, length.out=40)
   lines(x.xfit/x.xscalefactor, x.yfit, col="red")
   par(mar=c(3,0,1,1))
   barplot(yhist$density, axes=FALSE, xlim=c(0, top), space=0, horiz=TRUE)
   y.xfit <- seq(min(y),max(y),length.out=40)
   y.yfit <- dnorm(y.xfit, mean=mean(y), sd=sd(y))
   y.xscalefactor <- y.xfit / seq(from=0, to=19, length.out=40)
   lines(y.yfit, y.xfit/y.xscalefactor, col="red")
  }
  

  举些例子：

  require(MASS)
  #Sigma <- matrix(c(2.25, 0.8, 0.8, 1), 2, 2)
  Sigma <- matrix(c(1, 0.8, 0.8, 1), 2, 2)
  mvnorm <- mvrnorm(1000, c(0,0), Sigma) ; scatterBar.Norm(mvnorm[,1], mvnorm[,2])
  

  不对称的Sigma导致相应轴的稍微更大的直方图 .

  代码留下故意“不雅”，以增加可理解性（当我稍后再次访问时，我自己......） .

  尼尔斯

  回复于 2024-05-06T07:47:03+08:00
• 3

  使用ggplot时，翻转轴非常有效 . 例如，参见this example，它显示了如何为箱形图执行此操作，但它对于我假设的直方图同样有效 . 在ggplot中，可以很容易地在ggplot2术语中叠加不同的绘图类型或几何 . 因此，结合密度图和直方图应该很容易 .

  回复于 2024-05-06T07:47:03+08:00