使用 sed 查找并划分文件中的所有数字 [英] Find and divide all numbers in a file using sed

问题描述

我正在尝试在 json 文件中查找所有数字，并在 mac 上使用 sed 将它们替换为原始数字的一半值.比如这里我搜索2010，替换成1005:

I am trying to find all numbers in a json file and replace them with a half value of the original number using sed on mac. For example, here I search for 2010 and replace it with 1005:

file="data.json"
sed -i '' -E 's,([^0-9]|^)2010([^0-9]|$),\1 1005\2,g' "$file"

我想找到所有数字实例，并用它们自己的一半值替换它们.它需要处理小数，例如:2009 将变为 1004.5，10.5 将变为 5.25.

I would like to find all number instances, and replace them with half values of themselves. It would need to work on decimals, eg: 2009 would become 1004.5, 10.5 would become 5.25.

我知道这可能需要每个单独的数字字符，所以可能需要找到两边都带有非数字字符的数字.

I'm aware this could take each individual number character, so perhaps it would need to find numbers with non-numerical characters either side of it.

edit:我希望它能够灵活处理所有形式的文本文件，而不仅仅是 JSON 文件.(.txt、.html、.rtf 等...)

edit: I would like it to be flexible and work on all forms of text files, not just JSON files. (.txt, .html, .rtf etc...)

推荐答案

您可以将 Perl 与带有 e 修饰符的正则表达式一起使用:

You may use Perl with a regex with e modifier:

perl -pe 's{(?<!\d)(\d+(?:\.\d+)?)(?!\d)}{$1/2}ge' file

要修改文件内联，添加-i选项:

To modify the file inline, add -i option:

perl -i -pe 's{(?<!\d)(\d+(?:\.\d+)?)(?!\d)}{$1/2}ge' file
perl -pi.bak -e 's{(?<!\d)(\d+(?:\.\d+)?)(?!\d)}{$1/2}ge' file # To save a backup of the original file

查看在线演示:

s="abc_2010_and+2009+or-10.5"
perl -pe 's{(?<!\d)(\d+(?:\.\d+)?)(?!\d)}{$1/2}ge' <<< "$s"
# => abc_1005_and+1004.5+or-5.25

(? 正则表达式匹配

• (?<!\d) - 不允许紧靠左边的数字
• (\d+(?:\.\d+)?) - 第 1 组 ($1):1+ 位数字后跟可选的 序列. 和 1+ 位数字
• (?!\d) - 不允许紧靠右侧的数字.

• (?<!\d) - no digit immediately to the left is allowed
• (\d+(?:\.\d+)?) - Group 1 ($1): 1+ digits followed with an optional sequence of . and 1+ digits
• (?!\d) - no digit immediately to the right is allowed.

RHS - $1/2 - 是一个将 Group 1 值除以 2 的表达式.它是通过在正则表达式的末尾添加 e 修饰符来实现的.

The RHS - $1/2 - is an expression that divides the Group 1 value with 2. It is achieved through adding e modifier at the end of the regex.

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