問題描述

第一列（scaffold這一列）相同，則根據(jù)第AS列（AS:i:xx）數(shù)字 xx的大小，保留數(shù)字最大的行。如果數(shù)字大小相同則都保留。

舉例，輸入文件

scaffold_010679_1AL.2 16 chr1A 429400034 119 3272M * GACACAAGAGACTCTTTG * AS:i:3268 XS:i:2147 XF:i:0 XE:i:29 NM:i:1scaffold_010679_1AL.2 16 chr1A 429400034 119 3272M * GACACAAGAGACTCTTTG * AS:i:3268 XS:i:2147 XF:i:0 XE:i:29 NM:i:1 scaffold_010679_1AL.2 16 chr1A 429400034 119 3272M * GACACAAGAGACTCTTTG * AS:i:1268 XS:i:2147 XF:i:0 XE:i:29 NM:i:1scaffold_010679_1AL.3 16 chr1A 429397743 19 599S1730M1I279M * 0 0 TGCCGAGGTTTTTGA * AS:i:1998 XS:i:1877 XF:i:3 XE:i:20 NM:i:2 XN:i:1scaffold_010679_1AL.3 16 chr1A 429397743 19 599S1730M1I279M * 0 0 TGCCGAGGTTTTTGA * AS:i:1098 XS:i:1877 XF:i:3 XE:i:20 NM:i:2 XN:i:1

結(jié)果文件

scaffold_010679_1AL.2 16 chr1A 429400034 119 3272M * GACACAAGAGACTCTTTG * AS:i:3268 XS:i:2147 XF:i:0 XE:i:29 NM:i:1scaffold_010679_1AL.2 16 chr1A 429400034 119 3272M * GACACAAGAGACTCTTTG * AS:i:3268 XS:i:2147 XF:i:0 XE:i:29 NM:i:1 scaffold_010679_1AL.3 16 chr1A 429397743 19 599S1730M1I279M * 0 0 TGCCGAGGTTTTTGA * AS:i:1998 XS:i:1877 XF:i:3 XE:i:20 NM:i:2 XN:i:1

問題解答

# coding: utf-8from itertools import groupbywith open(’a.txt’) as f: data = [line for line in f] #因為數(shù)據(jù)的列數(shù)不相同, 只能以AS:i:為開頭來識別 #取第一列為key, AS:i:列為value lst = [(l.split()[0], _) for l in data for _ in l.split() if _.startswith(’AS:i:’)]#找出同key下的max(value) max_lst = [max(list(g)) for k, g in groupby(lst, lambda x: x[0])]#從原數(shù)據(jù)里找到同時包含key和value的行 print [line for line in data for _ in max_lst if _[0] in line and _[1] in line]回答2：

awk ’{n=gensub('.*AS:i:([0-9]+).*','1','g')}n>=k[$1]{c[$1]=n==k[$1]?c[$1]'n'$0:$0;k[$1]=n}END{for(i in c)print c[i]}’ file回答3：

grep '`sort -r -t '*' -k 3 b.txt | head -1 |awk -F '*' ’{split($3,a,' ');print a[1]}’`' b.txt

思路文件按星號*分列分3列，按照第三列降序排序，取出第一行，取出AS:i:最大數(shù)，grep搜索之，得到結(jié)果。

是我沒仔細看提問，失誤了~~結(jié)果不對