删除第一个标签和最后一个分号之间的一切

Question

我有它的线是那样的文件：

EF457507|S000834932  Root;Bacteria;"Acidobacteria";Acidobacteria_Gp4;Gp4 
EF457374|S000834799  Root;Bacteria;"Acidobacteria";Acidobacteria_Gp14;Gp14 
AJ133184|S000323093  Root;Bacteria;Cyanobacteria/Chloroplast;Cyanobacteria;Family I;GpI 
DQ490004|S000686022  Root;Bacteria;"Armatimonadetes";Armatimonadetes_gp7 
AF268998|S000340459  Root;Bacteria;TM7;TM7_genera_incertae_sedis 

我想打印第一个标签和最后一个分号之间的任何事情，就像

EF457507|S000834932  Gp4 
EF457374|S000834799  Gp14 
AJ133184|S000323093  GpI 
DQ490004|S000686022  Armatimonadetes_gp7 
AF268998|S000340459  TM7_genera_incertae_sedis 

我试图使用正则表达式，但它不工作，有没有办法使用Linux，awk或Perl做到这一点？

Answer 1

你可以使用sed：

sed 's/\t.*;/\t/' file 

## This matches a tab character '\t'; followed by any character '.' any number of 
## times '*'; followed by a semicolon; and; replaces all of this with a tab 
## character '\t'. 

sed 's/[^\t]*;//' file 

## Things inside square brackets become a character class. For example, '[0-9]' 
## is a character class. Obviously, this would match any digit between zero and 
## nine. However, when the first character in the character class is a '^', the 
## character class becomes negated. So '[^\t]*;' means match anything not a tab 
## character any number of times followed by a semicolon. 

或者awk：

awk 'BEGIN { FS=OFS="\t" } { sub(/.*;/,"",$2) }1' file 

awk '{ sub(/[^\t]*;/,"") }1' file 

结果：

EF457507|S000834932  Gp4 
EF457374|S000834799  Gp14 
AJ133184|S000323093  GpI 
DQ490004|S000686022  Armatimonadetes_gp7 
AF268998|S000340459  TM7_genera_incertae_sedis 

---

按照下面的评论，以“删除一切AFTE R中的最后一个分号”，与sed：

sed 's/[^;]*$//' file 

## '[^;]*$' will match anything not a semicolon any number of times anchored to 
## the end of the line. 

或者awk：

awk 'BEGIN { FS=OFS="\t" } { sub(/[^;]*$/,"",$2) }1' file 

awk '{ sub(/[^;]*$/,"") }1' file