Ubuntu中文论坛

发表于： **2012-07-06 11:36**

有一文件内容如下：
pname,timestamp,mpe,dir,push,ee,total
Mexico,2012-06-25,16:00,0.52,0.52,1.04,0.25,1.29
Mexico,2012-06-25,16:01,0.00,0.00,0,0.3,0.3
Mexico,2012-06-25,16:02,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25,16:03,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25,16:04,0.00,0.00,0,0.266667,0.266667
Mexico,2012-06-25,16:05,0.00,0.00,0,0.25,0.25
Mexico,2012-06-25,16:06,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25,16:07,0.00,0.00,0,0.183333,0.183333
Mexico,2012-06-25,16:08,0.00,0.00,0,0.166667,0.166667
Mexico,2012-06-25,16:09,0.00,0.00,0,0.15,0.15
Mexico,2012-06-25,16:10,0.00,0.00,0,0.15,0.15,0.00,0.00,0,0.15,0.15
Mexico,2012-06-25,16:10,0.00,0.00,0,0.15,0.15
请问如何把第二行开始的所有行（一行为一条记录）的第二个逗号都替换为空格，即如2012-06-25,16:09替换为2012-06-25 16:09的形式。

急，请大家不吝赐教，O(∩_∩)O谢谢~~

发表于： **2012-07-06 11:52**

很明显是用正则~~~

发表于： **2012-07-06 12:08**

求详细解答

发表于： **2012-07-06 12:24**

代码：全选

[maplebeats@maplebeats ~]$ cat tes |sed 's/25,16/25 16/'
ime,timestamp,mpe,dir,push,ee,total
Mexico,2012-06-25 16:00,0.52,0.52,1.04,0.25,1.29
Mexico,2012-06-25 16:01,0.00,0.00,0,0.3,0.3
Mexico,2012-06-25 16:02,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25 16:03,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25 16:04,0.00,0.00,0,0.266667,0.266667
Mexico,2012-06-25 16:05,0.00,0.00,0,0.25,0.25
Mexico,2012-06-25 16:06,0.00,0.00,0,0.2,0.2
Mexico,2012-06-25 16:07,0.00,0.00,0,0.183333,0.183333
Mexico,2012-06-25 16:08,0.00,0.00,0,0.166667,0.166667
Mexico,2012-06-25 16:09,0.00,0.00,0,0.15,0.15
Mexico,2012-06-25 16:10,0.00,0.00,0,0.15,0.15,0.00,0.00,0,0.15,0.15
[maplebeats@maplebeats ~]$

sed的正则还不会呢，我再看看

发表于： **2012-07-06 12:35**

写了个很笨的正则，不过能用

代码：全选

sed 's/\([1-9]\{2\}\),\([1-9]\{2\}\)/\1 \2/'

发表于： **2012-07-06 13:04**

断言

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