多谢大家了,另外一个问题awk 提取匹配

liaoboda

比如边边内容(标红),我要提取第一行和后边匹配KXXX
Ame_R016766        "PPP2R3A; protein phosphatase 2, regulatory subunit B'', alpha; K11583 protein phosphatase 2 (formerly 2A), regulatory subunit B''"
Ame_R013656        "CDH13; cadherin 13, H-cadherin (heart); K06808 cadherin 13, H-cadherin"
Ame_R018461        NOTCH4; notch 4; K02599 Notch
Ame_R018610        rho guanine nucleotide exchange factor 11-like; K12331 Rho guanine nucleotide exchange factor 11
Ame_R015728        GL19043 gene product from transcript GL19043-RA; K07295 nuclear receptor subfamily 2 group A

lionfun

回复 15# zooyo


   

xiaopan3322

回复  liaoboda
lionfun 发表于 2011-07-26 00:26

    感觉效率上会有点差

lionfun

回复 1# liaoboda

1. sed -rne 's@(Ame_[^ \t]+)[ \t]+.*(K[0-9]+).*@\1 \2@g;tc;s@(Ame_[^ \t]+)[ \t]+.*@\1 none@g;:c p' urfile
   
2. 
   
3. ps:t即条件转移，tc意思是当其前面的s@(Ame_[^ \t]+)[ \t]+.*(K[0-9]+).*@\1 \2@g这个命令替换成功，直接跳转到c标签，而不用执行s@(Ame_[^ \t]+)[ \t]+.*@\1 none@g这个替换，如果s@(Ame_[^ \t]+)[ \t]+.*(K[0-9]+).*@\1 \2@g替换失败，则tc不跳转，即会执行s@(Ame_[^ \t]+)[ \t]+.*@\1 none@g这个替换，p命令是打印当前pattern的内容。
   

复制代码

liaoboda

回复 9# zooyo

就是我的Ame 后边可能存在没有Kxxx的情况.

rdcwayx

1. grep -Eo "Ame_R[0-9]+| K[0-9]+ " infile |awk '{printf /^Ame/?RS $0:FS $0}'

复制代码

yinyuemi

回复 4# liaoboda

1. $ echo 'Ame_R016766        "PPP2R3A; protein phosphatase 2, regulatory subunitB'', alpha; K11583 protein phosphatase 2 (formerly 2A), regulatory subunit B''"
   
2. Ame_R013656        "CDH13; cadherin 13, H-cadherin (heart); K06808 cadherin 13, H-cadherin"
   
3. Ame_R018461        NOTCH4; notch 4; K02599 Notch
   
4. Ame_R018610        rho guanine nucleotide exchange factor 11-like; K12331 Rho guanine nucleotide exchange factor 11
   
5. Ame_R015728        GL19043 gene product from transcript GL19043-RA;nuclear receptor subfamily 2 group A' |\
   
6. awk '{printf $1 FS}{print /; K[0-9]+ /?gensub(/.*; (K[0-9]+).*/,"\\1","g",$0):"none"}'
   
7. Ame_R016766 K11583
   
8. Ame_R013656 K06808
   
9. Ame_R018610 K12331
   
10. Ame_R015728 none

复制代码