Question

如何将第二列从正面更改为负面，反之亦然

aa.txt文件

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文件

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。

Answer 1

sed 's/^\([^,]*,\)-/\1/
t
s/^[^,]*,/&-/' YourFile

基于-存在

Answer 2

考虑到您想要基于逗号更改第二列，这可能是一种方式：

$ 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"

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，在之前的版本中，我们丢失了尾随零。为了正确处理它，我们可以做到：

awk 'BEGIN{FS=OFS=","} {if ($2>0) $2="-"$2; else sub("-", "", $2)}1' file

如果值为正，则在字符串前添加短划线。否则，它只是删除破折号。

给出样本文件

$ 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"

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

Answer 3

使用sprintf格式化输出。

awk 'BEGIN{FS=OFS=","} {$2=sprintf("%+.2f",-$2)}1' aa.txt

将第二列从正变为负，反之亦然

3 个答案:

更新