Convert string with cosine/sine or other operations to number

I am reading multiple rows of strings from a text file and storing it in a cell. An example of a cell row is shown below:

A = 'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]'

My goal is to evaluate each XYZBC coordinates from string to a double format as shown below:

A = [-6.2228, -1.8209, 0.3972, -90, 0, 8000]

In the case of the X-coordinate, I use the regexprep command to replace the characters into a format Matlab recognises and use the eval command to evaluate the equation as shown exemplary below.

X_coord_new = regexprep(X_coord_old, {'X', '#629', '[', ']', 'COS', 'SIN'},...
                                      {'',    '0', '(', ')','cosd','sind'});
X_coord_final = eval(X_coord_new);

I cannot use the str2double command since there is infact math involved to solve the equation, e.g. cosine and sine as well as multiplication/addition/etc.

The problem is that the eval command, and infact the str2num command which also uses the eval command are very slow. Does anybody have any hints or tips of how I can evaluate my strings in a computationally efficient manner? Thank-you!

[Walter Roberson]

Is there a reason you are not using https://library.wolfram.com/infocenter/MathSource/577/ ?

[Stephen23]

You could modify this:

https://www.mathworks.com/matlabcentral/fileexchange/68458-evalequation

[Gregory Eberle]

@Walter Roberson

I am not familier with the link you sent. The presented .m file is not in Matlab synthax. I am actually trying to convert machine GCode coordinates into absolute coordinates.

@ Stephen Cobeldick

The EvalEquation.m works as desired by is unfortunately even slower than the eval or str2num commands from Matlab.

tic; eval( 'cosd(45)+1+2/(2*3)'); toc;
Elapsed time is 0.000526 seconds.

tic; str2num( 'cosd(45)+1+2/(2*3)'); toc;
Elapsed time is 0.012254 seconds.

tic; EvalEquation( 'cosd(45)+1+2/(2*3)'); toc;
Elapsed time is 0.029066 seconds.

I guess I will just have to stick with the eval/str2num commands from Matlab and make an effort not to use it them when not necessary.

[Walter Roberson]

I looked at gcode a couple of years ago to answer some questions. It did not have the structure you show: the coordinates had to be explicit.

https://www.mathworks.com/matlabcentral/answers/460610-extracting-numbers-from-a-g-code-file

Answer 1

[Mathieu NOE]

hello

so this is my "low level" retrieve function ; remains to do a loop on multiple lines

indexX = strfind(A,'X[');
indexY = strfind(A,'Y[');
indexZ = strfind(A,'Z[');
indexB = strfind(A,'B[');
indexC = strfind(A,'C[');

% 1st value to retrieve /% 
stringX = A(indexX:indexY-1);
index1 = strfind(stringX,'[');
index2 = strfind(stringX,'+ #');
valueX = str2num(stringX(index1(3)+1:index2(1)-1));

% 2nd value to retrieve /% 
stringY = A(indexY:indexZ-1);
index1 = strfind(stringY,'[');
index2 = strfind(stringY,'+ #');
valueY = str2num(stringY(index1(1)+1:index2(1)-1));

% 3rd value to retrieve /% 
stringZ = A(indexZ:indexB-1);
index1 = strfind(stringZ,'[');
index2 = strfind(stringZ,'+ #');
valueZ = str2num(stringZ(index1(3)+1:index2(1)-1));

% 4th value to retrieve /% 
stringB = A(indexB:indexC-1);
index1 = strfind(stringB,'[');
index2 = strfind(stringB,']');
valueB = str2num(stringB(index1(1)+1:index2(1)-1));

% 5th value to retrieve /% 
stringC = A(indexC:end);
index1 = strfind(stringC,'[');
index2 = strfind(stringC,']');
valueC = str2num(stringC(index1(1)+1:index2(1)-1));

data = [valueX, valueY, valueZ, valueB, valueC];

[Gregory Eberle]

This is essentially what I did.

The problem arises when having to execute the command "str2num" five times for each row. I am confronted with up to 1000 rows and as mentioned in the original question, I am interested in a solution that avoids the "eval" and "str2num" commands since they are slow.

[Mathieu NOE]

hello

maybe this is someting interesting for you

a created a dummy text file of 1000 lines (ok they are all the same...)

and created that code

str = fileread('Document1.txt');   %read entire file into string
parts = strtrim(regexp( str, '(\r|\n)+', 'split'));  %split by each line
columns = strtrim( regexp(parts{1}, '\s+', 'split'));  %columns
ncol = length(columns);  %number of columns
nrows = length(parts);  %number of rows
%now loop through parts
ind_col = [3 9 23 26 27 ];% retrieve only following cells (to update ccording to your needs)

M = cell( nrows, length(ind_col));  %pre-allocate empty cell array for data
str_blanks = blanks(100); % string of 100 blanks
str_all = [];

  for k=1:nrows,
    data = strtrim(regexp( parts{k}, '\s+', 'split'));   %split by spaces
    dd = regexp(data(ind_col),'[0-9]+.[0-9]+', 'match'); % extract numeric parts inside the cell
    
    % string manipulations (assuming we want output is array of strings)
        str = str_blanks; % init
    %     str_temp = strjoin( cat(2,dd{:}), ',') ;    % Transform a cell array of cell arrays into string / delimiter = comma 
        str_temp = strjoin( cat(2,dd{:}), ' ') ;    % Transform a cell array of cell arrays into string / delimiter = blank 
        ll = length(str_temp);
        str(1:ll) = str_temp;
        % concatenation
        str_all = [str_all; str];
  end

 toc

the result appears as an array of string (variable : )

the code executes in 0.163743 seconds which is maybe not too bad

still someone will find out how to further refine it

[Gregory Eberle]

Unfortunately in this approach you are basically ignoring the cosine and sine operations leading to erroneous results.

Simply put, I wish to evaluate the following string fast:

X = 'cosd(45) + 1';

Answer should be a numeric, i.e X = 1.7071.

[Mathieu NOE]

ok

forgot you wanted to evaluated the function !!

[Mathieu NOE]

lookig back at the txt file, can we assume that once we have extracted the numerical content, we can simply redo the math (equations) afterwards, knowing on which data we have to take the cos and sin ?

[Gregory Eberle]

As long as the order of operations is held, I don't see why not.

[Mathieu NOE]

I'll try this way but you have to guide me : how should I compute the outputs (to be defined ) according to you data line

A = 'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]'

X = ?

Y = ...

give me a template please

[Gregory Eberle]

Basically, one starts with a cell of strings such as the following:

A = {'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]';
     'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]';
     'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]';
     'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]';
     'X[COS[0.0000] * [-6.2518 + #629] - SIN[0.0000] * [-0.0375] + [0.0290]] Y[-1.8209 + #627] Z[SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]] B[-90.0000] C[0.0000]'};

And the goal is to end up with a matrix full of doubles corresponding to XYZBC such as the following:

A = [-6.2228, -1.8209, 0.3972, -90, 0;
     -6.2228, -1.8209, 0.3972, -90, 0;
     -6.2228, -1.8209, 0.3972, -90, 0;
     -6.2228, -1.8209, 0.3972, -90, 0;
     -6.2228, -1.8209, 0.3972, -90, 0];

[Mathieu NOE]

ok

so if i'm right , for example X is computed like this

cos(0.0000) * (-6.2518) - SIN(0.0000) * (-0.0375) + (0.0290)

that's the -6.2228 output

I assume the " + #629 " does not go into the calculation and must be discarded

same for Y and Z

B and C , no equation , just read the value next to the character

[Gregory Eberle]

Correct!

As a note, the angular units are in degrees so Matlab needs to have cosd/sind instead of cos/sin.

Answer 2

[Mathieu NOE]

so , hopefully now the final code ! had to fight a bit with regexp as I am not super fluent with this coding.

also I remembered that your initial request was to have as output :

A = [-6.2228, -1.8209, 0.3972, -90, 0, 8000]

the '8000' is not in your data file, so I assumed this could be the number of lines scanned. So I added the loop index as the 6th column in the output matrix A = [X Y Z B C k];

the 1000 lines txt scan and processing took : Elapsed time is 0.246920 seconds.

str = fileread('Document1000.txt');   %read entire file into string
parts = strtrim(regexp( str, '(\r|\n)+', 'split'));  %split by each line
columns = strtrim( regexp(parts{1}, '\s+', 'split'));  %columns
ncol = length(columns);  %number of columns
nrows = length(parts);  %number of rows


% preallocation
A = zeros(nrows,6); 


  for k=1:nrows,
    data = strtrim(regexp( parts{k}, '\s+', 'split'));   %split by spaces
    dd = regexp(data,'-?[0-9]+.[0-9]+', 'match'); % extract numeric parts inside the cell
    % https://stackoverflow.com/questions/15814592/how-do-i-include-negative-decimal-numbers-in-this-regular-expression
        str_temp = strjoin( cat(2,dd{:}), ',') ;    % Transform a cell array of cell arrays into string / delimiter = blank or comma
        data2 = str2num(str_temp); % convert string to num 
    % we got : data2 = 0   -6.2518  629  0   -0.0375    0.0290   -1.8209  627   0   -6.2518  629   0   -0.0375    0.4347  -90      0       
     X = cosd(data2(1))*data2(2) - sind(data2(4))*(data2(5)) + data2(6); % X = cos(0.0000) * (-6.2518) - SIN(0.0000) * (-0.0375) + (0.0290) :
     Y = data2(7); % Y[-1.8209 + #627]
     Z = sind(data2(9))*data2(10) + cosd(data2(12))*(data2(13)) + data2(14); % Z = [SIN[0.0000] * [-6.2518 + #629] + COS[0.0000] * [-0.0375] + [0.4347]]
     B = data2(15);
     C = data2(16);
     A(k,:) = [X Y Z B C k];

  end