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What would be the Regular Expression to retrieve number in front of % symbol in string?

Answer 1

+5 A:

Easiest way would be to let the math parser do the dividing:

s/([0-9.]+)%/($1\/100)/g

Which just replaces "25%" with "(25/100)" ... so if the user typed "25%*10" they'll now have "(25/100)*10" which when evaluated gives them the right answer.

Alternatively, in Perl, you could do:

s{([0-9.]+)%}{$1/100}eg

which would have Perl calculate the division by a hundred and pass that to the math parser

Sharkey 2009-04-19 07:09:16

What about 12.5%?

Andy 2009-04-19 07:38:35

Or (30+20)%? These get a bit harder. But that's not what the question is asking, I suppose.

Andy 2009-04-19 07:39:49

@Andy - This is indeed a problem. Your examples take us from regex land into the parser land. Ideally, the parser should take a maximal mathematical expression in front of the % symbol, evaluate it and divide it by 100.0, to get a real fractional value. I believe this is impossible using a regular expression and requires something like a recursive descent parser.

Yuval F 2009-04-19 10:13:27

Be careful with \d in Perl, as of 5.8 it means any UNICODE character that has the digit property. So unless you want to match U+1815 (MONGOLIAN DIGIT FIVE), you need to use [0-9].

Chas. Owens 2009-04-19 15:02:02

@Chas: That's a really good point that I hadn't considered ...@Andy: Yeah, 12.5% would be pretty common, good point.I'll edit the post from (\d+) to ([0-9.]+)@Yuval: Yeah, if you try to go too far with these things you'd be better off modifying the parser.

Sharkey 2009-04-19 21:22:43

@Iraimbilanja: Thanks for that, I hadn't noticed the runaway formatting.

Sharkey 2009-04-19 21:24:40

@Sharkey: Can I get an example of using this to automatically do the replacement? I'm using C#. I've tried "Regex regex = new Regex(@"s/([0-9.]+)%/($1\/100)/g"); regex.IsMatch(equation);" and it returns false. I probably should've added the c# relation in the question.

John Rennemeyer 2009-04-19 23:27:45

Using the following seems to work in simple testing: string finalExpression = Regex.Replace(expression, @"([0-9.]+)%", "($1/100)");If anyone sees anything wrong with this, please let me know.Thanks

John Rennemeyer 2009-04-19 23:38:22

@John: That looks right, although I'm no C# expert. This http://www.devhood.com/tutorials/tutorial_details.aspx?tutorial_id=523 might be useful:

Sharkey 2009-04-20 06:40:25

Answer 2

+1 A:

The regular expression to find a number before a '%' sign is /(\d+)%/.

However, mathematical expressions are not a regular language, so a regular expression is likely not the right tool. You should instead tell your parser to interpret '%' as "take whatever comes before this and divide it by 100". "Whatever comes before this" can then be any mathematical expression, and you won't run into operator precedence problems, resp. you could sort these out in the syntax tree parser.

Mixing text substitution with real language features often violates the principle of least astonishment. When a user sees that '%' divides whatever comes before by 100, he might try to use expressions like (23+42)%, but that will just produce a syntax error. Also, you need a more elaborate regex if you have something like 1.34e-14%, but these things would just sort themselves out when you use the tree parser.

Svante 2009-04-19 14:10:42

That's very true, and the syntax error, if you're passing it back from the parser to the user, is likely to be confusing ... with the regexp above, "(23+42)%" won't be substituted and "1.34e-14%" will be substituted to "1.34e-(14/100)" which is not terribly useful either.On the other hand, "23%+42%" would work fine.Yeah, if you need this level of complexity, you'd have the modify the parser.

Sharkey 2009-04-19 21:30:43

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What would be the Regular Expression to retrieve number in front of % symbol in string?

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