parsing - awk中的字段分隔符

Question

我有以下“input.file”:

10      61694   rs546443136     T       G       .       PASS    RefPanelAF=0.0288539;AC=0;AN=1186;INFO=1.24991e-09      GT:DS:GP        0/0:0.1:0.9025,0.095,0.0025     0/0:0.1:0.9025,0.095,0.0025     0/0:0.1:0.9025,0.095,0.0025

我想要的输出文件是:

0.1, 0.1, 0.1

使用名为“parse.awk”的 awk 脚本:

BEGIN {FS = ":"}
{for (i = 4; i <= NF; i += 2) printf ("%s%c", $i, i +2 <= NF ? "," : "\n ");}

调用方式:

awk -f parse.awk <input.file >output.file

我目前的output.file如下:

0.1,0.1,0.1

即没有空格。

将 pasre.awk 更改为:

BEGIN {FS = ":"}
{for (i = 4; i <= NF; i += 2) printf ("%s%c", $i, i +2 <= NF ? ", " : "\n ");}

没有改变输出文件。对 parse.awk 进行哪些更改会产生所需的输出文件？

Answer 1

你可以使用这个awk:

awk -F: -v OFS=', ' '{
  for (i = 4; i <= NF; i += 2) printf "%s%s", $i, (i < NF-1 ? OFS : ORS)}' file

0.1, 0.1, 0.1