removing parentheses around digits using regular expressions

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.

Answer 1

[Matt Fig]

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

[Patrick Mboma]

Hi Matt,

Your solution does not work on this example

 (2)+p_5*(3)-(0.3)+exp(3)

no change should occur if the opening parenthesis is preceded by an alphabetical letter.

Thanks,

Pat

[Matt Fig]

Try this one:

regexprep(str,'(?<![\w])(\()([\d*\.]+)(\))','$2')

[Walter Roberson]

nb: the (?<![\w]) is an example of look-behind.

[Matt Fig]

Yes, and it seems to work!

str='(2)+p_5*(3)-(0.3)+sin(0)-cos(6.7)+(3)^(2-3) + (55.55)'
regexprep(str,'(?<![\w])(\()([\d*\.]+)(\))','$2')
ans =

2+p_5*3-0.3+sin(0)-cos(6.7)+3^(2-3) + 55.55

[Patrick Mboma]

This is very helpful. Actually the (.) should occur at most once and so a fully correct expression would be

regexprep(str,'(?<![\w])(\()(\d*\.?\d*)(\))','$2')

Thanks a lot Matt.

Pat.

[per isakson]

(\d*\.?\d*) matches a single "."

[Walter Roberson]

(\d*\.?\d*) also matches the empty string.

It does not, however, match infinity, or any irrational number.

[per isakson]

"matches the empty string", what does that mean with regular expressions? In what type of cases could matching an empty string cause an effect?

[Walter Roberson]

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]

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?

Answer 2

[Walter Roberson]

Consider using a "look-behind"

[Patrick Mboma]

Hi Walter,

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

[Walter Roberson]

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

Answer 3

[per isakson]

    >> 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.

[Patrick Mboma]

Hi Per, This would not work. Here is a counter-example

 '(2)+p_5*(3)-(0.3)+exp(3)'

your solution would give

 '2+p_5*3-0.3+exp3'

[per isakson]

Yes, it does. What should it give? OK preceded by letter

[per isakson]

What about atan2()?

[Matt Fig]

The solution I posted in the comments to my answer handles atan2.

str='(2)+p_5*(3)-(0.3)+cos(5.7)+(3)^(2-3) + (.55) + atan2(9)';
regexprep(str,'(?<![\w])(\()([\d*\.]+)(\))','$2')
ans =

       2+p_5*3-0.3+cos(5.7)+3^(2-3) + .55 + atan2(9)

[per isakson]

Yes, and that is because "\w" stands for "[A-Za-z0-9]". However, it is difficult to know whether including "0-9" might have any unintended side effects.

I find it difficult to construct robust expressions. However, I have never tried to learn regular expressions in a systematic way.

[Patrick Mboma]

I am currently doing just that, with you guys' help.

Thanks to everybody.

Pat.

[per isakson]

See Mastering Regular Expressions and Regular Expressions Cookbook