Question

我有一组x点用于沿x轴绘制线段以在R中创建自定义读取贴图：

example read map

绘制这些段的一半任务是确定它们的y位置，以便没有两个重叠的段在同一个y级别上。对于每个段，我从第一个位置迭代y个级别，直到我到达一个尚未包含与当前位置重叠的段的位置。然后我记录当前段的结束位置并移动到下一个段。

实际代码的功能如下：

# 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");   
}

我的实际数据集非常大，并且包含的区域最多可以包含600000个读数。读取将自然地堆叠在一起，因此很容易实现最坏情况，其中所有读取彼此重叠。绘制大量读取所需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流程的效率。我可以用更快的东西替换我的循环吗？是否有一种算法可以更快地安排读取？我现在真的想不出更好的方法。

感谢您的帮助。

Answer 1

以贪婪的方式填写每个y级别。在一个关卡完成后，向下一级，永远不会再回来。

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

这不一定会产生任何意义上的“最佳”情节，但至少它是O（n log n）。

Answer 2

你能否对起始值进行排序？然后，您从前到后浏览列表。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，使第一个项目大于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。

对于每个项目，排序为O（N lg N）并且二进制搜索为O（lg N），因此总数为O（N lg N）。

提高可视化重叠段的性能

2 个答案: