0

输入 ：

    Process1 |5f|03/29/13 15:32:59
    Process1 |6f|03/29/13 17:59:03
    Pro cess1 |s7|03/29/13 17:59:05
    Pro cess1 |1234|03/29/13 20:33:43
    Pro cess1 |L|03/29/13 20:33:54
    A Pro cess |45g|03/29/13 20:34:59
    A Pro cess |f4|03/29/13 21:10:22
    B Pro cess |Es|03/30/13 00:11:25
    B Pro cess |23|03/30/13 00:47:20
    B Pro cess |E23|03/30/13 01:03:07

预期输出：

Process1 |5f|03/29/13 15:32:59
 |6f|03/29/13 17:59:03
Pro cess1|s7|03/29/13 17:59:05
 |1234|03/29/13 20:33:43
 |L|03/29/13 20:33:54
A Pro cess |45g|03/29/13 20:34:59
 |f4|03/29/13 21:10:22
B Pro cess |Es|03/30/13 00:11:25
 |23|03/30/13 00:47:20
 |E23|03/30/13 01:03:07

逻辑：仅保留第一列的一个实例，并将其他实例替换为空白

我试过这个，但没有用（在循环中使用下面的代码）：

sed -e 's/Process1/ /g' -e 's/ /Process1/1' Input

（将 Process1 替换为“-”，然后再次将“-”的第一个实例替换为 Process1。

正如“cnicutar”所给出的，我将 awk 修改为：

awk -F "|" '{a[$1]++; if (a[$1] > 1) $1=""; 打印;}'文件

abv 的输出：

Process1 |5f|03/29/13 15:32:59
 6f 03/29/13 17:59:03
Pro cess1 |s7|03/29/13 17:59:05
 1234 03/29/13 20:33:43
 L 03/29/13 20:33:54
A Process |45g|03/29/13 20:34:59
 f4 03/29/13 21:10:22
B Pro cess |Es|03/30/13 00:11:25
 23 03/30/13 00:47:20
 E23 03/30/13 01:03:07

期望的输出（插入 |）

Process1 |5f|03/29/13 15:32:59
     |6f|03/29/13 17:59:03
Pro cess1 |s7|03/29/13 17:59:05
     |1234|03/29/13 20:33:43
     |L|03/29/13 20:33:54
A Process |45g|03/29/13 20:34:59
     |f4|03/29/13 21:10:22
B Pro cess |Es|03/30/13 00:11:25
     |23 | 03/30/13 00:47:20
     |E23| 03/30/13 01:03:07
4

1 回答 1

2

稍微横向的解决方案怎么样：

[cnicutar@ariel ~]$ awk '{a[$1]++; if (a[$1] > 1) $1=""; print;}' file
Process1 |5f|03/29/13 15:32:59
 |6f|03/29/13 17:59:03
 |s7|03/29/13 17:59:05
 |1234|03/29/13 20:33:43
 |L|03/29/13 20:33:54
AProcess |45g|03/29/13 20:34:59
 |f4|03/29/13 21:10:22
BProcess |Es|03/30/13 00:11:25
 |23|03/30/13 00:47:20
 |E23|03/30/13 01:03:07

根据编辑，您可能需要：

awk -F'|' '{OFS="|"; a[$1]++; if (a[$1] > 1) $1=" "; print;}' file
于 2013-04-12T06:30:25.523 回答