将重叠的数字范围合并为连续的范围

Question

我正在尝试将一系列基因组坐标合并为连续范围，并具有跨间隙合并的附加选项。

例如，如果我有基因组范围[[0, 1000], [5, 1100]]，我希望结果是[0, 1100]。如果偏移选项设置为100，而输入为[[0, 1000], [1090, 1000]]，我将再次希望结果为[0, 1100]。

我已经实现了这样做的一种方式，依次通过对齐步骤，并尝试合并前一个结束位置和下一个开始位置，但它失败了，因为实际结果有不同的长度。例如，我有结果[[138, 821],[177, 1158], [224, 905], [401, 1169]]在我的列表中按开始位置排序。答案应该是[138, 1169]，但我得到了[[138, 1158], [177, 905], [224, 1169]]。显然，我需要考虑的不仅仅是前一个结尾和下一个开始，但我还没有找到一个好的解决方案(最好不是一个巨大的if语句的巢)。有人有什么建议吗?

def overlap_alignments(align, gene, overlap):
    #make sure alignments are sorted first by chromosome then by start pos on chrom
    align = sorted(align, key = lambda x: (x[0], x[1]))
    merged = list()
    for i in xrange(1, len(align)):
        prv, nxt = align[i-1], align[i]
        if prv[0] == nxt[0] and prv[2] + overlap >= nxt[1]:
            start, end = prv[1], nxt[2]
            chrom = prv[0]
            merged.append([chrom, start, end, gene])
    return merged

Answer 1

Python自带电池

from itertools import chain
flatten = chain.from_iterable
LEFT, RIGHT = 1, -1
def join_ranges(data, offset=0):
    data = sorted(flatten(((start, LEFT), (stop + offset, RIGHT))
            for start, stop in data))
    c = 0
    for value, label in data:
        if c == 0:
            x = value
        c += label
        if c == 0:
            yield x, value - offset
if __name__ == '__main__':
    print list(join_ranges([[138, 821], [900, 910], [905, 915]]))
    print list(join_ranges([[138, 821], [900, 910], [905, 915]], 80))

结果:

[(138, 821), (900, 915)]
[(138, 915)]

如何工作:我们这样标记每个起点和终点，然后我们排序，然后我们简单地计数向上为每个起点，向下为每个终点。如果访问了相同数量的起始点和结束点，则有一个闭合(连接)范围。

Answer 2

那么，如何跟踪每个开始和结束以及每个位置所属的范围的数量呢?

def overlap_alignments(align, overlap):
    # create a list of starts and ends
    stends = [ (a[0], 1) for a in align ]
    stends += [ (a[1] + overlap, -1) for a in align ]
    stends.sort(key=lambda x: x[0])
    # now we should have a list of starts and ends ordered by position,
    # e.g. if the ranges are 5..10, 8..15, and 12..13, we have
    # (5,1), (8,1), (10,-1), (12,1), (13,-1), (15,-1)
    # next, we form a cumulative sum of this
    s = 0
    cs = []
    for se in stends:
        s += se[1]
        cs.append((se[0], s))
    # this is, with the numbers above, (5,1), (8,2), (10,1), (12,2), (13,1), (15,0)
    # so, 5..8 belongs to one range, 8..10 belongs to two overlapping range,
    # 10..12 belongs to one range, etc
    # now we'll find all contiguous ranges
    # when we traverse through the list of depths (number of overlapping ranges), a new
    # range starts when the earlier number of overlapping ranges has been 0
    # a range ends when the new number of overlapping ranges is zero 
    prevdepth = 0
    start = 0
    combined = []
    for pos, depth in cs:
        if prevdepth == 0:
            start = pos
        elif depth == 0
            combined.append((start, pos-overlap))
        prevdepth = depth
    return combined

画起来容易解释起来难。(是的，累加和可以用更短的空格写，但我觉得这样更清楚。)

为了图解地解释这一点，让我们取输入([5,10]，[8,15]，[12,13]，[16,20])和重叠=1。

.....XXXXXo.............. (5-10)
........XXXXXXXo......... (8-15)
............Xo........... (12-13)
................XXXXo.... (16-20)
.....1112221221111111.... number of ranges at each position
.....----------------.... number of ranges > 0
.....---------------..... overlap corrected (5-20)