Question

我有一个巨大的输入文件，看起来像这样，

contig  protein start end
con1    P1  140 602
con1    P2  140 602
con1    P3  232 548
con2    P4  335 801
con2    P5  642 732
con2    P6  335 779
con2    P7  729 812
con3    P8  17  348
con3    P9  16  348

我想删除同源P或冗余P，我假设它们分别具有相同的起始和终点以及具有较小起点或终点的那些。所以我的输出文件将是这样的， file.txt的

con1    P1  140 602
con1    P3  232 548
con2    P4  335 801
con2    P7  729 812

尝试过的脚本，出于某种原因，它不符合这两个条件，

from itertools import groupby
def non_homolog(hits):
    nonhomolog=[]
    overst = False
    for i in range(1,len(hits)):
        (p, c) = hits[i-1], hits[i]
        if p[2] <= c[2] and c[3] <= p[3]:
            if not overst: nonhomolog.append(c)
            nonhomolog.append(c)
            overst = True   
    return nonhomolog

fh = open('example.txt')
oh = open('nonhomologs.txt', 'w')
for qid, grp in groupby(fh, lambda l: l.split()[0]):
    hits = []
    for line in grp:
        hsp = line.split()
        hsp[2], hsp[3] = int(hsp[2]), int(hsp[3])
        hits.append(hsp)
    hits.sort(key=lambda x: x[2])
    if non_homolog(hits):
        for hit in hits:
            oh.write('\t'.join([str(f) for f in hit])+'\n')

Answer 1

尝试使用此尺寸：

# this code assumes Python 2.7
from itertools import groupby, izip
from operator import attrgetter

INPUT    = "file.txt"
HOMO_YES = "homologs.txt"
HOMO_NO  = "nonhomologs.txt"
MAX_DIFF = 5

class Row:
    __slots__ = ["line", "con", "protein", "start", "end"]

    def __init__(self, s):
        self.line    = s.rstrip()
        data         = s.split()
        self.con     = data[0]
        self.protein = data[1]
        self.start   = int(data[2])
        self.end     = int(data[3])

    def __str__(self):
        return self.line

def count_homologs(items, max_diff=MAX_DIFF):
    num_items  = len(items)
    counts     = [0] * num_items
    # first item
    for i, item_i in enumerate(items):
        max_start = item_i.start + max_diff
        max_end   = item_i.end   + max_diff
        # second item
        for j in xrange(i+1, num_items):
            item_j = items[j]
            if item_j.start > max_start:
                break
            elif item_j.end <= max_end:
                counts[i] += 1
                counts[j] += 1
    return counts

def main():
    with open(INPUT) as inf, open(HOMO_YES, "w") as outhomo, open(HOMO_NO, "w") as outnothomo:
        # skip header
        next(inf, '')
        rows = (Row(line) for line in inf)

        for con, item_iter in groupby(rows, key=attrgetter("con")):
            # per-con list of Rows sorted by start,end
            items = sorted(item_iter, key=attrgetter("start", "end"))
            # get #homologs for each item
            counts = count_homologs(items)
            # do output
            for c,item in izip(counts, items):
                if c:
                    outhomo.write(str(item) + "\n")
                else:
                    outnothomo.write(str(item) + "\n")

if __name__=="__main__":
    main()

对您的给定数据产生：

=== homologs.txt ===

con1    P1  140 602
con1    P2  140 602
con3    P9  16  348
con3    P8  17  348

=== nonhomologs.txt ===

con1    P3  232 548
con2    P6  335 779
con2    P4  335 801
con2    P5  642 732
con2    P7  729 812

使用python删除循环中的冗余

1 个答案: