如何在 Bash 中将 `4-7` 替换为 `4,5,6,7`

问题描述

目标

我有一个长字符串 s ，它代表由逗号和破折号分隔的一系列数字(见下文).当几个数字相互跟随时，两个极端数字被写入并用破折号分隔.例如，系列4,5,6,7 写为4-7.我的目标是扩展这个字符串，让所有数字都用逗号分隔(4-7 应该变成 4,5,6,7).

I have a long string s which represents series of numbers separated by commas and dashes (see below). When several numbers follow each other, the two extreme numbers are written and are separated by a dash. For example, the series 4,5,6,7 is written as 4-7. My goal is to expand this string to have all numbers separated by comas (4-7 should become 4,5,6,7).

我做了什么

这是一个字符串的例子

s="4092-4093,4095-4097,4104,4107,4111,4125-4127"

我想首先用 {4..7} 替换 4-7 类型的模式(使用 sed 反向引用)

I wanted to first replace the patterns of the type 4-7 by {4..7} (using sed backreferening)

a="$(echo $s | sed 's/\([0-9]*\)-\([0-9]*\)/{\1..\2}/g')"

然后计算字符串以展开大括号

and then evaluate the string to expand the braces

b="$(eval echo $a)"

但是，当我运行最后一个命令时，扩展是以阶乘方式"完成的(导致 RAM 使用量激增).

However, when I run this last command, the expansion is done "in a factorial way" (leading to an explosion of the RAM usage).

问题

如何将字符串中 4-7 类型的模式替换为 4,5,6,7 ?

How can I replace the patterns of the kind 4-7 into 4,5,6,7 in my string?

版本

我使用 Mac OS X 10.11.3 并使用 Terminal 2.6.1 (361.1)

推荐答案

一个使用 GNU awk 的答案，它应该在大量输入时表现更好:

An answer using GNU awk which should perform better with large inputs:

#!/usr/bin/env gawk -f
{
    while ( match($0, /([0-9]+)-([0-9]+)/, arr) ) {
        s = arr[1]
        for (i=int(arr[1]) + 1; i<=int(arr[2]); i++) {
            s = s "," i
        }
        gsub(arr[1] "-" arr[2], s)
    }
    print
}

或者，在纯 bash 中(使用少量数据获得更好的性能):

---

Alternately, in pure bash (for better performance with small amounts of data):

s="4092-4093,4095-4097,4104,4107,4111,4125-4127"
re='([0-9]*)-([0-9]*)'
while [[ $s =~ $re ]]; do
  eval_str="printf -v replacement '%s,' {${BASH_REMATCH[1]}..${BASH_REMATCH[2]}}"
  eval "$eval_str"
  replacement=${replacement%,}
  s=${s//${BASH_REMATCH[0]}/$replacement}
done

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