使用R和轴断开符()绘制相当复杂的图表

Question

我有一个由4563个蛋白质氨基酸组成的数据集。用三种不同的处理方法和两种不同的氧化剂，对该蛋白质中的氨基酸进行氧化。我想在治疗的基础上在图表中画出这些氧化的位置。不同的线条尺寸代表不同的氧化剂浓度，线条类型(虚线和实线)代表不同类型的氧化剂。我想在每1000个氨基酸上打破轴心。我已经用excel和gimp创建了一个类似的模板(这相当耗时，而且可能不合适！)。模板中的0.33是行高。http://dl.dropbox.com/u/58221687/Chakraborty_Figure1.png。下面是数据集：http://dl.dropbox.com/u/58221687/AA-Position-template.xls

提前谢谢。Sourav

Answer 1

我将在基础图形中做到这一点，尽管我相信其他人可以在晶格或ggplot2中做同样或更好的事情。我认为你需要做的最主要的事情就是对你的数据进行重塑，并重新思考数据需要什么样的格式才能进行绘图。如果1)数据是长格式的，2)颜色、线型、宽度等变量作为额外的列可用，我就会用你的数据做这件事。如果您有这样的数据，那么您可以将其减少到只包括需要绘制线段的氨基酸。我已经模拟了一个与您的数据集类似的数据集。您应该能够修改此代码以适合您的情况:首先是数据集：

    set.seed(1)
    # make data.frame just with info for the lines you'll actually draw
    # your data was mostly zeros, no need for those lines
    position <- sort(sample(1:4563,45,replace = FALSE))
    # but the x position needs to be shaved down!
    # modulars are the real x positions on the plot:
    xpos <- position%%600
    # line direction appeared in your example but not in your text
    posorneg <- sample(c(-1,1),45,replace = TRUE,prob=c(.05,.95))
    # oxidant concentration for line width- just rescale the oxidant concentration
    # values you have to fall between say .5 and 3, or whatever is nice and visible
    oxconc   <- (.5+runif(45))^2
    # oxidant type determines line type- you mention 2
    # just assign these types to lines types (integers in R)
    oxitype  <- sample(c(1,2),45,replace = TRUE) 
    # let's say there's another dimension you want to map color to
    # in your example png, but not in your description.
    color <- sample(c("green","black","blue"),45,replace=TRUE)

    # and finally, which level does each segment need to belong to?
    # you have 8 line levels in your example png. This works, might take
    # some staring though:
    level <- 0
    for (i in 0:7){
    level[position %in% ((i*600):(i*600+599))] <- 8-i
    }

    # now stick into data.drame:
    AminoData <-data.frame(position = position, xpos = xpos, posorneg = posorneg, 
         oxconc = oxconc, oxitype = oxitype, level = level, color = color)

好的，想象一下你可以把你的数据简化成这样简单的东西。绘图的主要工具(在base中)将是segments()。它是矢量化的，所以不需要循环或花哨的东西：

    # now we draw the base plot:
    par(mar=c(3,3,3,3))
    plot(NULL, type = "n", axes = FALSE, xlab = "", ylab = "", 
         ylim =  c(0,9), xlim = c(-10,609))
    # horizontal segments:
    segments(0,1:8,599,1:8,gray(.5))
    # some ticks: (also not pretty)
    segments(rep(c((0:5)*100,599),8), rep(1:8,each=7)-.05, rep(c((0:5)*100,599),8), 
       rep(1:8,each=7)+.05, col=gray(.5))
    # label endpoints:
    text(rep(10,8)+.2,1:8-.2,(7:0)*600,pos=2,cex=.8)
    text(rep(589,8)+.2,1:8-.2,(7:0)*600+599,pos=4,cex=.8)
    # now the amino line segments, remember segments() is vectorized
    segments(AminoData$xpos, AminoData$level, AminoData$xpos, 
       AminoData$level + .5 * AminoData$posorneg, lty = AminoData$oxitype, 
       lwd = AminoData$oxconc, col = as.character(AminoData$color))
    title("mostly you just need to reshape and prepare\nyour data to do this easily in base")

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对于一些人来说，这可能太手工了，但这是我处理特殊情节的方式。