bash - 解析 VCF 文件的 INFO 字段

Question

请帮我解析一个 VCF 文件。我贴的是一个真实的例子。

输入:

1   1014143 rs786201005 C   T   .   .   RS=786201005;RSPOS=1014143;dbSNPBuildID=144;SSR=0;SAO=1;VP=0x050068000605000002110100;GENEINFO=ISG15:9636;WGT=1;VC=SNV;PM;PMC;NSN;REF;ASP;LSD;OM;CLNALLE=1;CLNHGVS=NC_000001.11:g.1014143C>T;CLNSRC=OMIM_Allelic_Variant;CLNORIGIN=1;CLNSRCID=147571.0003;CLNSIG=5;CLNDSDB=MedGen:OMIM:Orphanet;CLNDSDBID=C4015293:616126:ORPHA319563;CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification;CLNREVSTAT=no_criteria;CLNACC=RCV000162196.3
1   1014228 rs1921  G   A,C .   .   RS=1921;RSPOS=1014228;dbSNPBuildID=36;SSR=0;SAO=0;VP=0x050328000a0517053f000100;GENEINFO=ISG15:9636;WGT=1;VC=SNV;PM;PMC;S3D;SLO;NSM;REF;ASP;VLD;G5A;G5;HD;GNO;KGPhase1;KGPhase3;CLNALLE=1;CLNHGVS=NC_000001.11:g.1014228G>A;CLNSRC=.;CLNORIGIN=1;CLNSRCID=.;CLNSIG=2;CLNDSDB=MedGen;CLNDSDBID=CN169374;CLNDBN=not_specified;CLNREVSTAT=single;CLNACC=RCV000455759.1;CAF=0.6611,0.3389,.;COMMON=1
1   1014316 rs672601345 C   CG  .   .   RS=672601345;RSPOS=1014319;dbSNPBuildID=142;SSR=0;SAO=1;VP=0x050068001205000002110200;GENEINFO=ISG15:9636;WGT=1;VC=DIV;PM;PMC;NSF;REF;ASP;LSD;OM;CLNALLE=1;CLNHGVS=NC_000001.11:g.1014319dupG;CLNSRC=OMIM_Allelic_Variant;CLNORIGIN=1;CLNSRCID=147571.0002;CLNSIG=5;CLNDSDB=MedGen:OMIM:Orphanet;CLNDSDBID=C4015293:616126:ORPHA319563;CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification;CLNREVSTAT=no_criteria;CLNACC=RCV000148989.5
1   1014359 rs672601312 G   T   .   .   RS=672601312;RSPOS=1014359;dbSNPBuildID=142;SSR=0;SAO=1;VP=0x050068000605000002110100;GENEINFO=ISG15:9636;WGT=1;VC=SNV;PM;PMC;NSN;REF;ASP;LSD;OM;CLNALLE=1;CLNHGVS=NC_000001.11:g.1014359G>T;CLNSRC=OMIM_Allelic_Variant;CLNORIGIN=1;CLNSRCID=147571.0001;CLNSIG=5;CLNDSDB=MedGen:OMIM:Orphanet;CLNDSDBID=C4015293:616126:ORPHA319563;CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification;CLNREVSTAT=no_criteria;CLNACC=RCV000148988.5
1   1020183 rs539283387 G   C   .   .   RS=539283387;RSPOS=1020183;dbSNPBuildID=142;SSR=0;SAO=0;VP=0x050000000a05040026000100;GENEINFO=AGRN:375790;WGT=1;VC=SNV;NSM;REF;ASP;VLD;KGPhase3;CLNALLE=1;CLNHGVS=NC_000001.11:g.1020183G>C;CLNSRC=.;CLNORIGIN=1;CLNSRCID=.;CLNSIG=3;CLNDSDB=MedGen;CLNDSDBID=CN169374;CLNDBN=not_specified;CLNREVSTAT=single;CLNACC=RCV000424799.1;CAF=0.9904,0.009585;COMMON=1
1   1020216 rs764659938 C   G   .   .   RS=764659938;RSPOS=1020216;dbSNPBuildID=144;SSR=0;SAO=0;VP=0x050000000a05040002000100;GENEINFO=AGRN:375790;WGT=1;VC=SNV;NSM;REF;ASP;VLD;CLNALLE=1;CLNHGVS=NC_000001.11:g.1020216C>G;CLNSRC=.;CLNORIGIN=1;CLNSRCID=.;CLNSIG=0;CLNDSDB=MedGen;CLNDSDBID=CN221809;CLNDBN=cancer;CLNREVSTAT=single;CLNACC=RCV000422793.1

我需要一个输出:

1014143 rs786201005 C   T   CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
1014228 rs1921  G   A,C CLNSIG=2    CLNDBN=not_specified
1014316 rs672601345 C   CG  CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
1014359 rs672601312 G   T   CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
1020183 rs539283387 G   C   CLNSIG=3    CLNDBN=not_specified
1020216 rs764659938 C   G   CLNSIG=0    CLNDBN=not_provided

这意味着打印 2,3,4,5 列，然后解析最后一列并仅打印 CLNSIG 和 CLNDBN。问题是，这些值并不总是位于相同的位置。

我的尝试是:

awk -v OFS="\t"'{print $2,$3,$4,$5,$8}' input

...然后我不知道如何获取 CLNSIG 和 CLNDBN。

感谢您的任何想法。

Answer 1

1. 纯bash，通过使用bash来解析剩余的变量在 $h 中， parameter tranformation 输出:

   while read a b c d e f g h ; do 
       declare ${h//;/ }
       printf "%s\t%-10s\t%s\t%s\t%s\t%s\n" $b $c $d $e ${CLNSIG@A} ${CLNDBN@A}
   done < input
   

   输出:

   1014143 rs786201005 C   T   CLNSIG='5'  CLNDBN='Immunodeficiency_38_with_basal_ganglia_calcification'
   1014228 rs1921      G   A,C CLNSIG='2'  CLNDBN='not_specified'
   1014316 rs672601345 C   CG  CLNSIG='5'  CLNDBN='Immunodeficiency_38_with_basal_ganglia_calcification'
   1014359 rs672601312 G   T   CLNSIG='5'  CLNDBN='Immunodeficiency_38_with_basal_ganglia_calcification'
   1020183 rs539283387 G   C   CLNSIG='3'  CLNDBN='not_specified'
   1020216 rs764659938 C   G   CLNSIG='0'  CLNDBN='cancer'
   

2. POSIX shell、grep 和 printf 方法:

   while read a b c d e f g h ; do
       printf "%s\t%-10s\t%s\t%s\t%s\t%s\n" $b $c $d $e \
             $( echo "$h" | grep -o 'CLN\(SIG\|DBN\)=[^;]*' ) ; 
   done < input
   

   输出:

   1014143 rs786201005 C   T   CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
   1014228 rs1921      G   A,C CLNSIG=2    CLNDBN=not_specified
   1014316 rs672601345 C   CG  CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
   1014359 rs672601312 G   T   CLNSIG=5    CLNDBN=Immunodeficiency_38_with_basal_ganglia_calcification
   1020183 rs539283387 G   C   CLNSIG=3    CLNDBN=not_specified
   1020216 rs764659938 C   G   CLNSIG=0    CLNDBN=cancer