expressão regular e correspondência de padrões

Question

EQ963472        29264   .       G       A       212.0   .       DP=170;VDB=0.0253;AF1=1;AC1=2;DP4=0,0,79,83;MQ=60;FQ=-282;ANN=A|stop_gained|HIGH|AFLA_072280|AFLA_072280|transcript|EED56534|protein_coding|
EQ963472        31777   .       C       T       222.0   .       DP=179;VDB=0.0245;AF1=1;AC1=2;DP4=0,0,66,95;MQ=60;FQ=-282;ANN=T|stop_gained|HIGH|AFLA_072310|AFLA_072310|transcript|EED56537|protein_coding|
EQ963472        58523   .       G       A       222.0   .       DP=161;VDB=0.0269;AF1=1;AC1=2;DP4=0,0,71,83;MQ=60;FQ=-282;ANN=A|start_lost|HIGH|AFLA_072370|AFLA_072370|transcript|EED56543|protein_coding|1
EQ963472        171022  .       A       C       222.0   .       DP=164;VDB=0.0253;AF1=1;AC1=2;DP4=0,0,90,65;MQ=60;FQ=-282;ANN=C|stop_lost&splice_region_variant|HIGH|AFLA_072870|AFLA_072870|transcript|EED5
EQ963472        174382  .       C       T       136.0   .       DP=159;VDB=0.0253;AF1=1;AC1=2;DP4=0,0,65,76;MQ=60;FQ=-282;ANN=T|stop_gained|HIGH|AFLA_072890|AFLA_072890|transcript|EED56595|protein_coding|
EQ963472        185314  .       T       C       77.0    .       DP=168;VDB=0.0259;AF1=1;AC1=2;DP4=0,0,6,2;MQ=60;FQ=-51;ANN=C|stop_lost|HIGH|AFLA_072940|AFLA_072940|transcript|EED56600|protein_coding|1/1|c
EQ963472        188490  .       C       T       217.0   .       DP=175;VDB=0.0267;AF1=1;AC1=2;DP4=0,1,86,78;MQ=60;FQ=-282;PV4=0.48,8.8e-08,1,1;ANN=T|stop_gained|HIGH|AFLA_072960|AFLA_072960|transcript|EED

Buscar IDs com AFLA_ * e ids de coluna da primeira correspondência usando correspondência padrão e expressão regular

Eu tentei buscar apenas os IDs do AFLA usando o seguinte: grep -o "AFLA_ [0-9]" A1.SNP.contig.snpeff_high.out | menos Isto resulta em: AFLA_0 AFLA_0 AFLA_0 AFLA_0 Podemos concatenar as duas primeiras colunas com um _ para obter um ID único.

A saída do arquivo deve ter duas primeiras colunas e o AFLA_ *

a saída deve ser:

EQ963472_29264 AFLA_072280 AFLA_072280

    

Answer 1

The output from the file should have first two columns and the AFLA_* ids

Solução

awk :

awk -F'[[:space:]]+|\|AFLA_' -v pfx="ALFA_" 
     '{ print $1,$2,pfx $9,pfx substr($10,0,index($10,"|")-1) }' yourfile

A saída:

EQ963472 29264 ALFA_072280 ALFA_072280
EQ963472 31777 ALFA_072310 ALFA_072310
EQ963472 58523 ALFA_072370 ALFA_072370
EQ963472 171022 ALFA_072870 ALFA_072870
EQ963472 174382 ALFA_072890 ALFA_072890
EQ963472 185314 ALFA_072940 ALFA_072940
EQ963472 188490 ALFA_072960 ALFA_072960

• -F'[[:space:]]+|\|AFLA_' - considerando espaço (s) e |AFLA_ seqüência como separadores de campo

Answer 2

Com o seguinte grep "|AFLA_*" foo | cut -f1 -d" "

vejo resultado como

EQ963472
EQ963472
EQ963472
EQ963472
EQ963472
EQ963472
EQ963472