removing parentheses around digits using regular expressions

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Patrick Mboma 2012-10-5

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Dear all, I am slowly making progress on my learning of regular expressions. At the moment, I am trying to solve the following problem: replace all occurrences of (n) with n, where n is a number, provided that no alphabetical letter occurs before the first parenthesis. As an example,

str='(2)+p_5*(3)-(0.3)'

would become

2+p_5*3-0.3

I wrote the following

regexprep(str,'(\W)($)([.012345789]+)($)','$1$3')

but it does not solves the problem if one of the expressions to change occurs at the beginning as in the example above. More concretely, the answer I get from running this is

(2)+p_5*3-0.3

which is not the expected result.

Thanks in advance for any help

Pat.

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Matt Fig 2012-10-5

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编辑：Matt Fig 2012-10-5

在 MATLAB Online 中打开

regexprep(str,'($)([\d*\.]+)($)','$2')

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Walter Roberson 2012-10-5

For example, 1+()*2 the empty string inside the () would match because all parts of \d*\.?\d* are optional, so the expression would be converted to 1+*2

Also, I note that the problem description does not disallow numeric characters before the (), so 1+Henkel2(7)*5 would be converted to 1+Henkel27*5

per isakson 2012-10-5

编辑：per isakson 2012-10-6

在 MATLAB Online 中打开

Yes.

Assuming the requirement is to match numbers only. The second expression below seems to be closer to a working one. However, what expression is taught in the book?

    >> regexprep( '(12),(.12),(12.),(1.2),(.),()' ...
                , '\((\d*\.?\d*)\)', '#$1'        )
    ans =
    #12,#.12,#12.,#1.2,#.,#
    >> regexprep( '(12),(.12),(12.),(1.2),(.),()'       ...
                , '\(((\d+\.?\d*)|(\d*\.\d+))\)', '#$1' )
    ans =
    #12,#.12,#12.,#1.2,(.),()

\w takes care of the case with Henkel2 - by intent or not. However, are there any cases, in which "(" should be replaced when preceded by a digit?

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更多回答（2 个）

Walter Roberson 2012-10-5

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Consider using a "look-behind"

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Patrick Mboma 2012-10-5

编辑：Patrick Mboma 2012-10-5

Hi Walter,

My knowledge of regular expressions is not that good yet... I don't know yet how to implement "look behinds".

Walter Roberson 2012-10-5

http://www.mathworks.com/help/matlab/matlab_prog/regular-expressions.html#brchk1t

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per isakson 2012-10-5

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编辑：per isakson 2012-10-5

在 MATLAB Online 中打开

    >> regexprep( str, '\(([\d.]+)\)', '$1' )
    ans =
    2+p_5*3-0.3
    str =
    (2)+p_5*(3)-(0.3)+exp(3)
    >> regexprep( str, '\(([\d.]+)\)', '$1' )
    ans =
    2+p_5*3-0.3+exp3

*\(* represents "("
\) represents ")"
[\d.]+ represents one or more digits and periods, e.g. "....0" and "2"
(expr) "Group regular expressions and capture tokens." The token may be refered to in the replacement string by $1 - "1" because it is the first

Every substring in the string that matches this expression is replaced, i.e. a number enclosed by parentheses is replaced by the number.

--- in response to a comment ---

    >> regexprep( str, '(?<!\w)\(([\d.]+)\)', '$1' )
    ans =
    2+p_5*3-0.3+exp(3)

better

>> regexprep( str, '(?<![a-zA-Z])$([\d.]+)$', '$1' )

because \w includes digits.

(?<![a-zA-Z]) "Look behind from current position and test if expr is not found." Where expr evaluates to a letter. Thus, if preceded by a letter there is no match.