如何将第二列从正数更改为负数，反之亦然
aa.txt
31OCT2013:00:00:00,220.10,"O","090500 13549951"
31OCT2013:00:00:00,-0.32,"I","090500 13549964"
答案应该是
aa.txt
31OCT2013:00:00:00,-220.10,"O","090500 13549951"
31OCT2013:00:00:00,+0.32,"I","090500 13549964"
如果我使用awk
'BEGIN{FS=OFS=","} {$2=-$2}1' aa.txt
结果是
1OCT2013:00:00:00,-220.1,"O","090500 13549951"
31OCT2013:00:00:00,+0.32,"I","090500 13549964"
在第一个结果中，符号正在更改，但缺少0，而不是-220.10，它显示为-220.1。
sed 's/^([^,]*,)-/1/
t
s/^[^,]*,/&-/' YourFile
基于-存在
考虑到您想要更改基于逗号的第二列，这可以是一种方法：
$ awk 'BEGIN{FS=OFS=","} {$2=-$2}1' aa.txt
31OCT2013:00:00:00,-220.1,"O","090500 13549951"
31OCT2013:00:00:00,0.32,"I","090500 13549964"
参见
31OCT2013:00:00:00,-220.1,"O","090500 13549951"
^
31OCT2013:00:00:00,0.32,"I","090500 13549964"
^
更新
根据算术行为，在以前的版本中，我们丢失了尾随的零。为了正确处理它，我们可以做：
awk 'BEGIN{FS=OFS=","} {if ($2>0) $2="-"$2; else sub("-", "", $2)}1' file
如果该值为正，它会在字符串前面添加一个短划线。否则，它只会删除短划线。
给定一个样本文件
$ cat a
31OCT2013:00:00:00,220.10,"O","090500 13549951"
31OCT2013:00:00:00,-0.32,"I","090500 13549964"
31OCT2013:00:00:00,0.300,"I","090500 13549964"
31OCT2013:00:00:00,-0.300,"I","090500 13549964"
31OCT2013:00:00:00,0,"I","090500 13549964"
它返回
$ awk 'BEGIN{FS=OFS=","} {if ($2>0) $2="-"$2; else sub("-", "", $2)}1' a
31OCT2013:00:00:00,-220.10,"O","090500 13549951"
31OCT2013:00:00:00,0.32,"I","090500 13549964"
31OCT2013:00:00:00,-0.300,"I","090500 13549964"
31OCT2013:00:00:00,0.300,"I","090500 13549964"
31OCT2013:00:00:00,0,"I","090500 13549964
使用sprintf格式化输出。
awk 'BEGIN{FS=OFS=","} {$2=sprintf("%+.2f",-$2)}1' aa.txt