How to multiply each column in the matrix by different scientific notation values that should be imported from a txt file?

I have data in a txt file that I imported into Matlab.

The file has val(:,:,1), val(:,:,2), and so on all the way to val(:,:.256) and each array contains 128 data points.

I imported these data so that it’s in a 128x256 (column x row) matrix using the following:

filename = 'SeqLoop.data.kSpaceOS.txt'; 
fid = fopen(filename,'r');
S = textscan(fid,'%s','delimiter','\n');
fclose(fid); 
S = S{1}; 
S = str2double(S); 
idx = isnan(S); 
S = S(~idx); 
S = reshape(S,128,[]);

The first line of all the arrays contains a scientific notation that must be multiplied by the following cells in that array.

For example, the first line of val(:,:,2) is 1.0e-07 * followed by 128 data points.

The coding above does not multiply the data points by the scientific notation value.

And the value of the scientific notation changes for each array; i.e., the value of scientific notation for val(:,:,1), val(:,:,2), and so on are different.

How can I implement this into my coding so that the value of the scientific notations is accounted for in my matrix? (Please see attached for the txt file).

[Jan]

This is an extremely unuseful method to store values in a file. Why do you want to use such a strange layout?

[Cris LaPierre]

Do you have the option of going back and saving your original variable val to a mat file instead of a txt file?

[parslee]

The workspace includes a variable "SeqLoop", which contains a lot of variables including the variable "data".

The variable "data" contains a lot of variables including the variable "kSpaceOs", which is the data I would like to save.

When I try to save the data of "kSpaceOs" as a mat file by right clicking and pressing save as, it saves all the variables in SeqLoop.

I also tried the following:

save('data',SeqLoop.data.kSpaceOs) 

But it outputs an error message:

Error using save
Must be a text scalar.

Error in Turbo_Spin_Echo_T2_Gabrielle (line 167)
save('data',SeqLoop.data.kSpaceOs) 

The data of "kSpaceOS" is an output from a simulation, so I can't make changes to how the data is outputted, which is the reason why I decided to save the result as a txt file.

[Cris LaPierre]

Try this

val = SeqLoop.data.kSpaceOs;
save('data.mat','val')

[parslee]

It doesn't work. It still outputs it like the txt file above.

Could the reason as to why it's not working be due to the matrix being 128x1x256 complex double?

[Cris LaPierre]

No. If you are seeing it in a text file, you are not saving it to a *.mat file.

Answer 1

[David Goodmanson]

HI Gabrielle,

MODIFIED

Looks like you have constructing the 128x256 matrix of values covered. The code below converts S{1} to a character matrix and parses the result to obtain the exponents.

In the text file, the first header does not have the correct line to supply the exponent, and neither do headers 127-130, so I modified the text file before importing it, using e-07 for the first case and e-03 for the others.

filename = 'SeqLoop.data.kSpaceOS_128ETL_new.txt'; 
fid = fopen(filename,'r');
S = textscan(fid,'%s','delimiter','\n');
fclose(fid); 
S = S{1}; 
E = char(S);                     % note added line
S = str2double(S); 
idx = isnan(S); 
S = S(~idx);  
S = reshape(S,128,[]);

ind = any(E =='*',2);            % rows containing the exponent
E = E(ind,:);
E(E=='*') = ' ';
z = round(log10(str2num(E)))';   % find the exponent (includes its sign)
format compact
z
format
tenz = 10.^(z);

result = repmat(tenz,128,1).*S;

[parslee]

I'm a little confused about the line a = SeqLoop.

Shouldn't I be using the values in line S = S{1}; to extract the exponents?

[David Goodmanson]

Hi Gabrielle,

Thanks for asking about SeqLoop. S=S{1} gives an array of cells and I am in favor of cells but I did not particularly want to deal with them in this case. The original answer started over with the text file and reimported it as a character array in order to parse it. But your comment got me thinking about S{1} so now the code converts S{1} to a character array and parses that in a different way from before. I have modified the answer accordingly. Of course the original text file still has to be corrected as mentioned before.

[parslee]

Hi David,

Thank you so much for your reply.

I tried running the code, but for some reason, some of the cells are not displaying the correct exponents.

I used the modified txt file where I put 1.0e-0 * for headers 1 and 127-130.

And it makes sense based on the simulation that I'm running that the values for headers 1 and 127-130 are big compared to the rest.

Are they being not accounted for because the exponent is 0?

[David Goodmanson]

HI Gabrielle,

the single zero does make a difference, since the code was designed for two digits after the 'e', and one digit throws off the steps that use columns 7 and 9. I modified the answer so as to be more robust for entries such as 1.0e-0. Please let me know if it works.

[parslee]

Yes, it works! Thank you so much!