bash - Moving the second column to first column using awk

Question

I would like do the following using awk:

Example of my input data with four columns and a number of rows:

10 20 30 40
50 30 60 80
90 12 40 20

Desired output:

score 2 · Accepted Answer

尝试类似：

awk '{print $1 " "  $2 "\n" $3 " " $4 "\n>"}'

输出是：

抱歉拖后腿>

score 1 · Accepted Answer

1

尝试awk '{ print $1" "$2"\n" $3" "$4"\n>" }'

于 2013-06-13T10:01:09.657 回答

score 1 · Accepted Answer

sed -r 's/(\S+\s+){2}/&\n/;$!a >' file

注意最后一行，没有多余的尾随>。

score 0 · Accepted Answer

我添加了一个纯bash解决方案（不调用任何外部实用程序）：

脚本：

while read a b c d; do echo -e "$a $b\n$c $d\n>"; done <infile

或者没有显式循环：

printf "%s %s\n%s %s\n>\n" $(<infile)

输入：

cat >infile <<XXX
10 20 30 40
50 30 60 80
90 12 40 20
XXX

输出：

4 回答 4