增加可视化重叠段

Question

我有一组对X点的绘制段沿x轴创建R中的自定义阅读地图的性能：

半绘制这些任务细分市场正在决定它们的y位置，因此没有两个重叠的细分市场在同一个y水平上。对于每个片段，我从第一个位置迭代y个级别，直到到达一个不包含与当前片段重叠的片段的位置。然后记录当前分段的结束位置并移至下一个分段。

实际的代码是一个函数如下：

# Dummy data 
# A list of start and end positions for each segment along the X axis. Sorted by start. 
# Passing the function few.reads draws a map in half a second. Passing it many.reads takes about half an hour to complete. 
few.reads <- data.frame(start=c(rep(10,150), rep(16,100), rep(43,50)), end=c(rep(30,150), rep(34,100), rep(57,50))); 
many.reads <- data.frame(start=c(rep(10,15000), rep(16,10000), rep(43,5000)), end=c(rep(30,15000), rep(34,10000), rep(57,5000))); 

#--- 
# A function to draw a series of overlapping segments (or "reads" in my along 
# The x-axis. Where reads overlap, they are "stacked" down the y axis 
#--- 
drawReads <- function(reads){ 

    # sort the reads by their start positions 
    reads <- reads[order(reads$start),]; 

    # minimum and maximum for x axis 
    minstart <- min(reads$start); 
    maxend <- max(reads$end); 

    # initialise yread: a list to keep track of used y levels 
    yread <- c(minstart - 1); 
    ypos <- c(); #holds the y position of the ith segment 

    #--- 
    # This iteration step is the bottleneck. Worst case, when all reads are stacked on top 
    # of each other, it has to iterate over many y levels to find the correct position for 
    # the later reads 
    #--- 
    # iterate over segments 
    for (r in 1:nrow(reads)){ 
     read <- reads[r,]; 
     start <- read$start; 
     placed <- FALSE; 

     # iterate through yread to find the next availible 
     # y pos at this x pos (start) 
     y <- 1; 
     while(!placed){ 

      if(yread[y] < start){ 
       ypos[r] <- y; 
       yread[y] <- read$end; 
       placed <- TRUE; 
      } 

      # current y pos is used by another segment, increment 
      y <- y + 1; 
      # initialize another y pos if we're at the end of the list 
      if(y > length(yread)){ 
       yread[y] <- minstart-1; 
      } 
     } 
    } 

    #--- 
    # This is the plotting step 
    # Once we are here the rest of the process is very quick 
    #--- 
    # find the maximum y pos that is used to size up the plot 
    maxy <- length(yread); 
    miny = 1; 


    reads$ypos <- ypos + miny; 

    print("New Plot...") 
    # Now we have all the information, start the plot 
    plot.new(); 
    plot.window(xlim=c(minstart, maxend+((maxend-minstart)/10)), ylim=c(1,maxy)); 

    axis(3,xaxp=c(minstart,maxend,(maxend-minstart)/10)); 
    axis(2, yaxp=c(miny,maxy,3),tick=FALSE,labels=FALSE); 

    print("Draw the reads..."); 
    maxy <- max(reads$ypos); 
    segments(reads$start, maxy-reads$ypos, reads$end, maxy-reads$ypos, col="blue"); 
} 

我的实际数据集是非常大的，并且包含最多可以有60万的区域读取，据我可以告诉。读取结果自然会堆叠在一起，因此很容易实现最糟糕的情况，即所有读取都相互重叠。绘制大量读取所花费的时间对我来说是不可接受的，所以我正在寻找一种方法来提高过程的效率。我可以用更快的东西来替换我的循环吗？有一种算法可以更快地安排读取吗？我现在真的想不出更好的方式来做这件事。

感谢您的帮助。

Answer 1

以贪婪的方式填充每个y级别。等级填满后，降低一级，永不回头。

伪代码：

y <- 1 
while segment-list.not-empty 
    i <- 1 
    current <- segment-list[i] 
    current.plot(y) 
    segment-list.remove(i) 
    i <- segment-list.find_first_greater(current.end) 
    while (i > 0) 
    current <- segment-list[i] 
    current.plot(y) 
    segment-list.remove(i) 
    y <- y + 1 

这并不一定产生在任何意义上的 “最优” 的情节，但至少它是为O（n log n）的。

Answer 2

你能不能对起始值进行排序吗？然后你从前到后浏览列表。对于每个项目，绘制它，然后对列表的其余部分进行二进制搜索，以查找第一个项目大于刚刚绘制的项目的结束坐标。如果没有找到，请增加Y.在绘制时删除每个项目。

排序为O（N lg N），二元搜索为O（lg N），因此总数为O（N lg N）。