使用AWK将文件的第一列分成多列

Question

File looks like this, but with millions of lines ( TAB separated):

1_number_column_ranking_+   100 200 Target "Hello" 

I want to split the first column by the _ so it becomes:

1 number column ranking + 100 200 Target "Hello"

This is the code I have been trying:

awk -F"\t" '{n=split($1,a,"_");for (i=1;i<=n;i++) print $1"\t"a[i]}' 

But it's not quite what I need.
Any help is appreciated (the other threads on this topic were not helpful for me).

Answer 1

No need to split, just replace would do:

awk 'BEGIN{FS=OFS="\t"}{gsub("_","\t",$1)}1'

Eg:

$ cat file
1_number_column_ranking_+       100     200     Target "Hello"

$ awk 'BEGIN{FS=OFS="\t"}{gsub("_","\t",$1)}1' file
1       number  column  ranking +       100     200     Target "Hello"

gsub will replace all occurances, when no 3rd argument given, it will replace in $0.
Last 1 is a shortcut for {print} . (always true , implied {print} .)

Answer 2

Another awk, if the "_" appears only in the first column. Split the input field by regex "[_\\t]+" and just do a dummy operation like $1=$1 in the main section, so that $0 is reconstructed with OFS="\\t"

$ cat steveman.txt
1_number_column_ranking_+       100     200i    Target  "Hello"

$ awk -F"[_\t]" ' BEGIN { OFS="\t"} { $1=$1; print } ' steveman.txt
1       number  column  ranking +       100     200i    Target  "Hello"

$

Thanks @Ed, updated from -F"[_\\t]+" to -F"[_\\t]" that will avoid concatenating empty fields.