我正在尝试将一系列基因组坐标合并到连续范围内,还可以选择跨越间隙进行合并。
例如,如果我有基因组范围[[0, 1000], [5, 1100]]
,我希望结果为[0, 1100]
。如果偏移量选项设置为100
,输入为[[0, 1000], [1090, 1000]]
,我会再次希望结果为[0, 1100]
。
我已经实现了这样做的方法,它顺序逐步调整对齐并尝试合并前一个结束位置和下一个起始位置,但它失败了,因为实际结果有不同的长度。例如,我的列表中的结果[[138, 821],[177, 1158], [224, 905], [401, 1169]]
按起始位置排序。答案应为[138, 1169]
,但我得[[138, 1158], [177, 905], [224, 1169]]
。显然,我需要考虑的不仅仅是前一个结束和下一个开始,但我还没有找到一个好的解决方案(最好是一个不是if语句的巨大嵌套)。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
答案 0 :(得分:4)
那么,如何跟踪每个开始和结束以及每个职位所属的范围数量?
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
这比绘制更容易。 (是的,累积的总和可以用更短的空间写出来,但我觉得这样更清楚。)
为了以图形方式解释,我们取输入([5,10],[8,15],[12,13],[16,20])并重叠= 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)
答案 1 :(得分:4)
Python来batteries included:
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)]
工作原理:我们标记每个起点和终点,然后排序,之后我们只为每个起点计算 up ,并为每个终点计算 down 。如果我们访问了相同数量的起点和终点,我们就会有一个封闭的(加入的)范围。