R中的R图像功能

Question

我在R中使用附加的图像功能.我更喜欢使用它作为速度的热图,因为我将它用于巨大的矩阵(~400000乘400).

我的功能中的问题是调色板的动态范围,在我的情况下它只有蓝色和黄色.我已尝试对colorramp行进行了几处更改,但没有一个给我所需的输出.

我试过的最后一个颜色渐变选项是在R中使用一个名为ColorRamps的漂亮包,它给出了合理的结果:

library("colorRamps")
ColorRamp = blue2green2red(400)
ColorLevels <- seq(min, max, length=length(ColorRamp))

但是,仍然没有matlab颜色渐变选项那么灵活.

我不太熟悉如何使它看起来更好并且具有更大的范围,例如附在照片中.

请告诉我是否可以更改我的图像功能以使我的图像看起来像照片中的图像.

我用于绘制图像的R函数,速度为rast = TRUE,如下所示:

# ----- Define a function for plotting a matrix ----- #
myImagePlot <- function(x, filename, ...){
  dev = "pdf"
  #filename = '/home/unix/dfernand/test.pdf'
  if(dev == "pdf") { pdf(filename, version = "1.4") } else{}
     min <- min(x)
     max <- max(x)
     yLabels <- rownames(x)
     xLabels <- colnames(x)
     title <-c()
  # check for additional function arguments
  if( length(list(...)) ){
    Lst <- list(...)
    if( !is.null(Lst$zlim) ){
       min <- Lst$zlim[1]
       max <- Lst$zlim[2]
    }
    if( !is.null(Lst$yLabels) ){
       yLabels <- c(Lst$yLabels)
    }
    if( !is.null(Lst$xLabels) ){
       xLabels <- c(Lst$xLabels)
    }
    if( !is.null(Lst$title) ){
       title <- Lst$title
    }
  }
# check for null values
if( is.null(xLabels) ){
   xLabels <- c(1:ncol(x))
}
if( is.null(yLabels) ){
   yLabels <- c(1:nrow(x))
}

layout(matrix(data=c(1,2), nrow=1, ncol=2), widths=c(4,1), heights=c(1,1))

 # Red and green range from 0 to 1 while Blue ranges from 1 to 0
 ColorRamp <- rgb( seq(0,1,length=256),  # Red
                   seq(0,1,length=256),  # Green
                   seq(1,0,length=256))  # Blue
 ColorLevels <- seq(min, max, length=length(ColorRamp))

 # Reverse Y axis
 reverse <- nrow(x) : 1
 yLabels <- yLabels[reverse]
 x <- x[reverse,]

 # Data Map
 par(mar = c(3,5,2.5,2))
 image(1:length(xLabels), 1:length(yLabels), t(x), col=ColorRamp, xlab="",
 ylab="", axes=FALSE, zlim=c(min,max), useRaster=TRUE)
 if( !is.null(title) ){
    title(main=title)
 }
# Here we define the axis, left of the plot, clustering trees....
#axis(BELOW<-1, at=1:length(xLabels), labels=xLabels, cex.axis=0.7)
# axis(LEFT <-2, at=1:length(yLabels), labels=yLabels, las= HORIZONTAL<-1,
# cex.axis=0.7)

 # Color Scale (right side of the image plot)
 par(mar = c(3,2.5,2.5,2))
 image(1, ColorLevels,
      matrix(data=ColorLevels, ncol=length(ColorLevels),nrow=1),
      col=ColorRamp,
      xlab="",ylab="",
      xaxt="n", useRaster=TRUE)

 layout(1)
  if( dev == "pdf") {
    dev.off() }
}
# ----- END plot function ----- #

Answer 1

您可以定义偏差colorRampPalette.我还调整了函数来定义颜色之间的步数color.palette:

#This is a wrapper function for colorRampPalette. It allows for the
#definition of the number of intermediate colors between the main colors.
#Using this option one can stretch out colors that should predominate
#the palette spectrum. Additional arguments of colorRampPalette can also
#be added regarding the type and bias of the subsequent interpolation.
color.palette <- function(steps, n.steps.between=NULL, ...){

 if(is.null(n.steps.between)) n.steps.between <- rep(0, (length(steps)-1))
 if(length(n.steps.between) != length(steps)-1) stop("Must have one less n.steps.between value than steps")

 fill.steps <- cumsum(rep(1, length(steps))+c(0,n.steps.between))
 RGB <- matrix(NA, nrow=3, ncol=fill.steps[length(fill.steps)])
 RGB[,fill.steps] <- col2rgb(steps)

 for(i in which(n.steps.between>0)){
  col.start=RGB[,fill.steps[i]]
  col.end=RGB[,fill.steps[i+1]]
  for(j in seq(3)){
   vals <- seq(col.start[j], col.end[j], length.out=n.steps.between[i]+2)[2:(2+n.steps.between[i]-1)]  
   RGB[j,(fill.steps[i]+1):(fill.steps[i+1]-1)] <- vals
  }
 }

     new.steps <- rgb(RGB[1,], RGB[2,], RGB[3,], maxColorValue = 255)
 pal <- colorRampPalette(new.steps, ...)
 return(pal)
}

这是两个例子(我已经挤压了青色和黄色之间的步数):

Z <- t(as.matrix(1:100))

pal.1 <- colorRampPalette(c("blue", "cyan", "yellow", "red"), bias=1)
pal.2 <- colorRampPalette(c("blue", "cyan", "yellow", "red"), bias=3)
pal.3 <- color.palette(c("blue", "cyan", "yellow", "red"), n.steps.between=c(10,1,10))

    x11()
par(mfcol=c(1,3))
image(Z, col=pal.1(100))
image(Z, col=pal.2(100))
image(Z, col=pal.3(100))

另外,如果你感兴趣的话,我写了一个绘制色标并使用相同参数的函数image.如果正确设置绘图布局,这也是绘制矩阵和相应色阶的快速方法.