remove null terms from a string and adjust signs

Hi,

I have a string with more than thousands of terms in it. A small sample part of the string is

'...(G5)*((G6)*(G2+G4)+(G4)*(G2))+(G2)*((0)*(G4+G6)+(0)*(G3+G7)+(G4+G6)*(G1))+(G4)*((0)*(G6)+(G6)*(G7)-(0)*(G6))-(G9)*((G40+G60)*(0)-(G24+G26)*(G3)+(G14+G16)*(0))...'

I do not want to convert this string to symbolic expressions and I want to process it as a string because the location of each term is meaningful and important and the conversion to symbolic expression disregards it.

What I need to do is to remove all null terms from the string and adjust signs when negatives (or combination of negatives and positives) are multiplied together. the answer for the given part of the main string should be:

'...(G5)*((G6)*(G2+G4)+(G4)*(G2))+(G2)*((G4+G6)*(G1))+(G4)*((G6)*(G7))+(G9)*((G24+G26)*(G3))...'

Could you please show me how to do it using a very efficient code that I can plug in my m file?

Answer 1

[John BG]

Now, I humbly claim my answer as accepted:

 str1='(G5)*((G6)*(G2+G4)+(G4)*(G2))+(G2)*((0)*(G4+G6)+(0)*(G3+G7)+(G4+G6)*(G1))+(G4)*((0)*(G6)+(G6)*(G7)-(0)*(G6))-(G9)*((G40+G60)*(0)-(G24+G26)*(G3)+(G14+G16)*(0))';

loc=strfind(str1,'(0)');loc2=loc+1;loc3=loc2+1;   % find locations of '(0)' only

loc4=strfind(str1,'*(0)');
ch4={0}
for k=1:length(loc4)
    ch4=[ch4 {0}]
end

for k=1:1:length(loc4)
    if str1(loc4(k)-1)==')'                         % spotting all characters related to each multiplying term
         no=0;nc=1;
         p4=loc4(k)-1;
         ch4{k}=p4;
         while (nc>no) && (p4>0)       %
                  p4=p4-1;
              if str1(p4)=='('
                    no=no+1;       end;
              if str1(p4)==')'
                    nc=nc+1;       end;
              ch4{k}=[p4 ch4{k}];
         end 
    end
    if str1(loc4(k)+4)=='+' || str1(loc4(k)+4)=='-'      % spotting sign after:   *(0)-   *(0)+
        ch4{k}=[ch4{k} loc4(k)+4]
    end
    if (str1(p4-1)=='+' || str1(p4-1)=='-')      % spotting sign before term:  +(..)*(0)   -(..)*(0)
        ch4{k}=[p4-1 ch4{k}]
    end
end
N4=0
for k=1:length(ch4)
    N=ch4{k}
    N4=[N4 N]
end    

N4(N4==0)=[]
loc42=N4    % contains all multiplying terms in front of *(0)
loc
loc43=loc4-3

loc5=strfind(str1,'(0)*');
ch5={0}
for k=1:length(loc5)
    ch5=[ch5 {0}]
end
for k=1:1:length(loc5)
    if str1(loc5(k)+4)=='('
         no2=1;nc2=0;
         p5=loc5(k)+4;
         ch5{k}=p5;
         while (no2>nc2) && (p5<length(str1))       %
                  p5=p5+1;
              if str1(p5)=='('
                    no2=no2+1;       end;
              if str1(p5)==')'
                    nc2=nc2+1;       end;
              ch5{k}=[ch5{k} p5];
         end 
    end

        if str1(loc5(k)-1)=='+' || str1(loc5(k)-1)=='-'      % spotting sign ahead:   -(0)*   +(0)*
          ch5{k}=[loc5(k)-1 ch5{k}]
      end
      if (str1(p5+1)=='+' || str1(p5+1)=='-')      % spotting sign before term:  (0)*(..)+   (0)*(..)-
          ch5{k}=[ch5{k} p5+1]
      end

end
N5=0
for k=1:length(ch5)
    N=ch5{k}
    N5=[N5 N]
end    
N5(N5==0)=[]
loc52=N5              % contains all multiplying terms ibehind all *(0)
loc53=loc5+3

% erasing all null related characters
str1(union(union(union(union(union(union(union(loc,loc2),loc3),loc42),loc52),loc4),loc5),loc53))=[];

so far the sample seems clean:

str1 =
(G5)*((G6)*(G2+G4)+(G4)*(G2))+(G2)*((G4+G6)*(G1))+(G4)*((G6)*(G7))-(G9)*((G24+G26)*(G3))

would it be possible to run this along the entire sequence?

If you find this answer of any help solving your question,

please mark it as accepted answer,

thanks in advance

John

[e-mail]

[S H]

Your code only removes (0) from string. It does not remove what is multiplied by (0). As numerical example, your code changes 2+0*3 to 2+*3 which is not correct.

[John BG]

you are right, let me have a look.

[S H]

This question can be solved by following steps:

1. identify one of the (0)* or *(0) terms.

2. count the following or leading parentheses until match if found

3. remove the whole thing and apply sign adjustment

Answer 2

[Jos (10584)]

str1 = strrep(str,'(0)*','') 
str2 = strrep(str1,'*(0)','')