Using regular expressions to parse numeric sequences

[maurobio]

Dear ALL,

I have sequences of numbers like these:

,(3-)5-100
8-250(-430)
60-110
(200-)500-3500

I want get the numbers outside parenthesis, splitting them into their minimum and maximum values, like this:

,(3-)5-100 --> 5 100
8-250(-430) --> 8 250
60-110 --> 60 110
(200-)500-3500 --> 500 3500

Is there a compact way of achieving this using regular expressions? Examples will be welcome!

Thanks in advance!

With best wishes,

[jamie]

Others will try to lure you onto RegExp  but that is so twisted and mind boggling that it just isn't worth it to  me..

 Write a simple little parser that scans the string..
 it needs to look for opening and closing (,) and numbers, math signs.. etc..

also I notice you may have some comma's there..

 you can use a Split function to break that up.

[lucamar]

If it's as simple as it seems (ignoring anything between parens and returning the rest, substituting operators by spaces and discarding other non-digits) I think a judicious use of Pos, PosEx(), Delete() and StringReplace() would do the trick quite nicely

[jamie]

Code: Pascal  [Select][+][-]

1. Function GetPairvalues(Str:String; Out left,Sep,Right:String):Boolean;
2. var
3.   I:integer=1;
4.   Q:Integer=0;
5.   D:Integer=1;
6. begin
7.   Left := '';sep:=''; Right := '';
8.   While I <= Length(Str) do
9.    Begin
10.     If Str[I] ='(' then Inc(Q) Else
11.       if Str[I] =')' Then
12.         Dec(Q) else // Raise an error if neg...
13.          If Q = 0 Then
14.          if not (Str[I]in ['-','+','*','/']) Then
15.            Case D of
16.             1:Left+=Str[I];
17.             2:Right+=Str[I];
18.            End Else
19.             Begin
20.              Sep:=Str[I];
21.              Inc(D);
22.             End;
23.       Inc(I);
24.      End;
25. End;
26.  
27. procedure TForm1.Button2Click(Sender: TObject);
28. var
29.   L,S,R:String;
30. begin
31.   GetPairValues('400-678(-50)',l,s,r);
32.   Caption := L+S+R;
33. end;    
34.  
35. You can improve on that if you wish  :D                              
36.  

[MarkMLl]

I broadly agree with what everybody else has already said, except that unlike Jamie I'd suggest calling the solution a state machine rather than a parser (which term is loaded with many implications).

The problem with a regex here is that it's difficult- if not impossible- to say "if this bit doesn't exist, return it as an empty string". The result of this is that you'd be starting off with at least five separate regexes to cover cases that look approximately like this:

Code: [Select]

12345
(12)345
12(34)5
123(45)
(123345)

then for each of those you'd be extracting the matched portions by index number. Of course, the regexes themselves would have both literal parentheses and grouping parentheses, and you'll notice that the above don't allow for any of the multiple-parentheses cases.

So you have variable called something like

Code: [Select]

var
  leftOfParentheses: string= '';
  insideParentheses: string= '';
  rightOfParentheses: string= '';
  parenthesesDepth: integer= 0;

You parse each string from the start, accumulating the characters into leftOfParentheses until you hit a left-parenthesis at which point you increment parenthesesDepth. You now continue, accumulating characters into insideParentheses and incrementing/decrementing parenthesesDepth until it becomes zero, at which point you accumulate into rightOfParentheses instead.

Then you look to see which of leftOfParentheses and rightOfParentheses is non-blank, and you can process those in a very similar fashion.

Unlike solution that applies patterns (regexen or otherwise) to each card as it's read, that's bomb-proof: it doesn't rely on your knowing all possible patterns in advance.

So to summarise, there are times when pattern matching is very useful. This isn't one of them.

MarkMLl

[maurobio]

Gentlemen,

Thank you very much for the insightful discussion and suggestions.

I do fully agree with @jamie in his remark that regular expressions are 'twisted and mind boggling'. Indeed! But they have the big advantage of being pretty concise and once one is working, it can do a lot in less lines.

But I do understand that my problem is not amenable to treatment by a regex. I will try the code kindly provided by @jamie, and hope that it will be useful.

I noticed that both @jamie and @MarkML treated the problem as one of interpreting mathematical expressions, therefore considering more general solutions than what is really necessary. These numeric sequences are not expressions, they are data, and refer to measurements taken from several parts; so, for example, the sequence (200-)500-3500 represents the minimum (500) and the maximum (3500) values of a measurement, with (200-) representing an extreme lower value. Therefore, such sequences will ever be in that format (just with an occasional debris character like a starting comma). No character other than '-' will ever appear in them, and the value between parenthesis will only appear at the start (as an extreme lower value) or the end (as an extreme upper value) of a given sequence.

With best wishes,

[MarkMLl]

I noticed that both @jamie and @MarkML treated the problem as one of interpreting mathematical expressions, therefore considering more general solutions than what is really necessary. These numeric sequences are not expressions, they are data, and refer to measurements taken from several parts; so, for example, the sequence (200-)500-3500 represents the minimum (500) and the maximum (3500) values of a measurement, with (200-) representing an extreme lower value. Therefore, such sequences will ever be in that format (just with an occasional debris character like a starting comma). No character other than '-' will ever appear in them, and the value between parenthesis will only appear at the start (as an extreme lower value) or the end (as an extreme upper value) of a given sequence.

No, I am specifically NOT treating it as a mathematical expression. If I were, a parser would be in order... and I specifically said that for this you need a state machine NOT a parser.

You are- of course- entirely at liberty to ignore my suggestion. But I'd much prefer that you didn't mischaracterise it as something it isn't.

MarkMLl

[maurobio]

@MarkMLI,

It was not surely my intention to mischaracterise your suggestion! 

In a rush, I didn't read your entire post (because, I admit, I go straight for the code). Please accept my most sincere apologies .

With best wishes,

[Jurassic Pork]

hello,
maurobio, if your values  always looks like your examples, regex isn't very complicated  to use :

Code: Pascal  [Select][+][-]

1. procedure TForm1.BtRegexClick(Sender: TObject);
2. const
3.   exptest: array[0..3] of string=(',(3-)5-100',
4.                             '8-250(-430)',
5.                             '60-110',
6.                             '(200-)500-3500');
7. var
8.   R:TRegExpr;
9.   ex : string;
10. begin
11.   R := TRegExpr.Create;
12.   try
13.    for ex in exptest do
14.      begin
15.       R.Expression:= '([0-9]+\-[0-9]+)';
16.       if R.Exec(ex) then  Memo1.Append(StringReplace(R.Match[1],'-',' ',[rfReplaceAll]));
17.      end;
18.   finally
19.     R.Free;
20.   end;
21. end;

Result in attachment

Friendly, J.P

[maurobio]

@Jurassic Park,

Thank you very much for your code, it works nice (just as that of @jamie, although his does not use regular expressions).

With best wishes,

[Jurassic Pork]

hello,

@Jurassic Park,

I am Jurassic Pork not Park see my avatar 
With regex you can extract the min and max values at the same time.
Example :

Code: Pascal  [Select][+][-]

1. procedure TForm1.BtGoRegexClick(Sender: TObject);
2. const
3.   exptest: array[0..3] of string=(',(3-n)5-100','8-250(-430)','60-110','(200-)500-3500');
4. var
5.   R:TRegExpr;
6.   ex : string;
7.   row : integer;
8. begin
9.   R := TRegExpr.Create;
10.   try
11.    row := 1;
12.    for ex in exptest do
13.      begin
14.       R.Expression:= '([0-9]+)\-([0-9]+)';
15.       if R.Exec(ex) then
16.         begin
17.          StringGrid1.Cells[0,row] := ex;
18.          StringGrid1.Cells[1,row] := R.Match[1];
19.          StringGrid1.Cells[2,row] := R.Match[2];
20.          row += 1;
21.         end;
22.      end;
23.   finally
24.     R.Free;
25.   end;
26. end;  

Result :  look at Attachment (click to see animation)

Friendly, J.P

[maurobio]

Hi, @Jurassic PORK,

Thank you very much for that other fine example. I am glad to see that, after all, my problem has a simple solution with regular expressions!

Sorry for my mistake in your nickname - at least I placed you in the correct geological period (not in the Triassic, or the Silurian!) 

With best wishes,