折叠序号到bash中的范围 [英] Collapse sequential numbers to ranges in bash

问题描述

我正在尝试将连续数字折叠到bash范围内.例如，如果我的输入文件是

I am trying to collapse sequential numbers to ranges in bash. For example, if my input file is

1
2
3
4
15
16
17
18
22
23
45
46
47

我希望输出为:

1 4
15 18
22 23
45 47

如何用awk或sed在单行命令中执行此操作?

How can I do this with awk or sed in a single line command?

感谢您的帮助！

推荐答案

$ awk 'NR==1{first=$1;last=$1;next} $1 == last+1 {last=$1;next} {print first,last;first=$1;last=first} END{print first,last}' file
1 4
15 18
22 23
45 47

说明

• NR==1{first=$1;last=$1;next}

在第一行上，初始化变量first和last，然后跳到下一行.

On the first line, initialize the variables first and last and skip to next line.

$1 == last+1 {last=$1;next}

如果此行从上一行开始继续执行，请更新last并跳至下一行.

If this line continues in the sequence from the last, update last and jump to the next line.

print first,last;first=$1;last=first

如果我们到达这里，我们将在序列中有所休息.打印出最后一个序列的范围，然后重新初始化新序列的变量.

If we get here, we have a break in the sequence. Print out the range for the last sequence and reinitialize the variables for a new sequence.

END{print first,last}

到达文件末尾后，打印最终序列.

After we get to the end of the file, print the final sequence.

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