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如何使用 awk 命令重复列中的数字序列

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如何使用 awk 命令重复列中的数字序列

我的输入文件包含这样的数据。

//address: 9050004 data: 1 id: 1 len: 0 size: 2
//address: 9050008 data: 1 id: 2 len: 0 size: 2
//address: 902e004 data: 1 id: 3 len: 0 size: 2
...
//address: 902e008 data: 1 id: 799 len: 0 size: 2
//address: 9096004 data: 1 id: 800 len: 0 size: 2
//address: 9096008 data: 1 id: 801 len: 0 size: 2
//address: 9097004 data: 1 id: 802 len: 0 size: 2
...
//address: 902e008 data: 1 id: 1599 len: 0 size: 2
//address: 9096004 data: 1 id: 1600 len: 0 size: 2
//address: 9096008 data: 1 id: 1601 len: 0 size: 2
//address: 9097004 data: 1 id: 1602 len: 0 size: 2

从上面的数据来看，id值不应超过 800。序列必须重复，如 1,2,...799,1,2...799。

所以我的输出应该如下所示

//address: 9050004 data: 1 id: 1 len: 0 size: 2
//address: 9050008 data: 1 id: 2 len: 0 size: 2
//address: 902e004 data: 1 id: 3 len: 0 size: 2
...
//address: 902e008 data: 1 id: 799 len: 0 size: 2
//address: 9096004 data: 1 id: 1 len: 0 size: 2
//address: 9096008 data: 1 id: 2 len: 0 size: 2
//address: 9097004 data: 1 id: 3 len: 0 size: 2
...
//address: 902e008 data: 1 id: 799 len: 0 size: 2
//address: 9096004 data: 1 id: 1 len: 0 size: 2
//address: 9096008 data: 1 id: 2 len: 0 size: 2
//address: 9097004 data: 1 id: 3 len: 0 size: 2

有人可以帮我在 awk 命令中获取上述命令吗？

答案1

要在达到 800 时回滚到 1，您必须将该列重新计算为((id - 1) % 799) + 1，其中%是模运算符：

$ awk '/^\/\// { $6 = (($6 - 1) % 799) + 1; print }' file
//address: 9050004 data: 1 id: 1 len: 0 size: 2
//address: 9050008 data: 1 id: 2 len: 0 size: 2
//address: 902e004 data: 1 id: 3 len: 0 size: 2
//address: 902e008 data: 1 id: 799 len: 0 size: 2
//address: 9096004 data: 1 id: 1 len: 0 size: 2
//address: 9096008 data: 1 id: 2 len: 0 size: 2
//address: 9097004 data: 1 id: 3 len: 0 size: 2
//address: 902e008 data: 1 id: 1 len: 0 size: 2
//address: 9096004 data: 1 id: 2 len: 0 size: 2
//address: 9096008 data: 1 id: 3 len: 0 size: 2
//address: 9097004 data: 1 id: 4 len: 0 size: 2

请注意，在您的预期输出中，idfor 地址902e008为 799。如果您将范围限制为 [1, 799]，则这是不可能的。

答案2

这对我有用

awk 'BEGIN{FS=OFS=" "}{i++;if (i<5) {$6=i} else {i=0};print}' aa.txt

只需将 i<5 更改为 i<800

测试：

$ cat aa.txt
//address: 9050004 data: 1 id: 1 len: 0 size: 2
//address: 9050008 data: 1 id: 2 len: 0 size: 2
//address: 902e004 data: 1 id: 3 len: 0 size: 2
//address: 9050004 data: 1 id: 4 len: 0 size: 2
//address: 9050008 data: 1 id: 5 len: 0 size: 2
//address: 902e004 data: 1 id: 6 len: 0 size: 2

$ awk 'BEGIN{FS=OFS=" "}{i++;if (i<5) {$6=i} else {i=0};print}' aa.txt
//address: 9050004 data: 1 id: 1 len: 0 size: 2
//address: 9050008 data: 1 id: 2 len: 0 size: 2
//address: 902e004 data: 1 id: 3 len: 0 size: 2
//address: 9050004 data: 1 id: 4 len: 0 size: 2
//address: 9050004 data: 1 id: 5 len: 0 size: 2
//address: 902e004 data: 1 id: 1 len: 0 size: 2

答案3

@Kusalananda 的 Perl 变体：

perl -pE 's/id: *\K(\d+)/($1-1)%799+1/e'

（数字与 PO 示例不完全相符）

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