Reading multiple txt data

Hello

I am a kinda newbie in matlab and want to know how to read multiple txt-files in my function. I am using Matlab R2017a, windows 10.

Whole code below. It is reading only 1 txt-file(last one), but i want to read all 74 files

txtfiles = dir('C:\Users\User\Documents\MATLAB\signal\signal*.txt');
ttrigger = dir('C:\Users\User\Documents\MATLAB\signal\trigger*.txt');

N = length(txtfiles) ;
for k = 1:N
     thisfile = txtfiles(k).name ;
     Tthisfile = ttrigger(k).name;
     s = csvread(thisfile);
     t = csvread(Tthisfile);
     r=0;
     j=1;
     n=0;
     loc=0;
     q=0;
     lock=0;
     inPulseS=false;
     inPulseT=false;
     for i=1:length(s)
         if t(i)>3 && inPulseT==false
             n=n+1;
             loc(n)=i;
             inPulseT=true;
         elseif t(i)<3
             inPulseT=false;
         end
    
         if s(i)<=-5 && inPulseS==false
             q=q+1;
             lock(q)=i;
             inPulseS=true;
         elseif s(i)>-5
             inPulseS=false;
         end
     end
     g=1;
     for v=1:length(loc)-1
         next=lock(lock>loc(v) & lock<loc(v+1));
         if length(next)==1
             diff(g)=next-loc(v);
             g = g+1; 
         end
     end

     filter=diff(diff>100 & diff<400);
     kk=0;
     rr=1;
     for h=1:length(filter)-1
         if filter(h)<200
             kk(rr)=0;
             rr=rr+1;
         else
             kk(rr)=1;
             rr=rr+1;
         end
     end
end

Answer 1

[dpb]

txtfiles = dir('C:\Users\User\Documents\MATLAB\signal\signal*.txt');
ttrigger = dir('C:\Users\User\Documents\MATLAB\signal\trigger*.txt');
N = length(txtfiles) ;
for k = 1:N
     thisfile = txtfiles(k).name ;
     Tthisfile = ttrigger(k).name;
     s = csvread(thisfile);

You have a specific directory in the file name spec for the dir() call, but only use the name in the loop to open the file(s). Unless the current working directory is the same as that, above, it will not find the file (and even if there happens to be one of the same name, it won't be the same one as found).

rtdir='C:\Users\User\Documents\MATLAB\signal');       % set the target folder name
sdir=dir(fullfile(rtdir,'signal*.txt'));              % dir() signal files
tdir=dir(fullfile(rtdir,'trigger*.txt'));             % dir() trigger files
N = length(sdir);                % assumes same number of each...
for k = 1:N
  s=csvread(fullfile(sdir(k).folder,sdir(k).name));  % read fully-qualified name
  t=csvread(fullfile(tdir(k).folder,tdir(k).name));  % read fully-qualified name
  ...

The above still overwrites the s,t variables on each pass which is why it looks like it only reads one...if the above caveat on directory location is met, then it will operate on al N (of course, it still operates on N even if N is just one).

Do whatever it is you wish to do with each set of files (including save whatever results you want for each) before going on to the next. You don't say what the expected result really is...

[Noob]

Ok, thanks. It worked. Also I need to collect all kk values in one. Could you help me with that?

[dpb]

     filter=diff(diff>100 & diff<400);
     kk=0;
     rr=1;
     for h=1:length(filter)-1
         if filter(h)<200
             kk(rr)=0;
             rr=rr+1;
         else
             kk(rr)=1;
             rr=rr+1;
         end
     end

can be replaced with

kk=~(diff(diff>100 & diff<400)<200);

To save each vector for each set of files, just use a cell array--here's a good place for one as the length of each vector is likely different.

kk(k)={~(diff(diff>100 & diff<400)<200)};  % save each in cell array element k

NB: diff is the name of the builtin difference operator; it is not recommended to alias builtin Matlab functions. Use another name for the variable diff. dif isn't taken or any other desired perturbation that you like. There should be a warning in the editor of this; it is good practice to try to eliminate as many of those orange markers as you can or at least understand what the consequences are if choose to ignore them. This is one that really shouldn't be ignored as general rule.

[Noob]

I mean zeros and ones of kk. I cant replace last loop. I need 0 and 1 to collect in one place. And thank you for explaining diff

[dpb]

Did you try it???? It is a vector of 0's and 1's; you can easily replace the loop with a logical addressing expression:

%d=randi(2000,1,40);    % make up some sample data
d=[1630        1812         254        1827        1265         196;
    557        1094        1916        1930         316        1942;
   1915         971        1601         284         844        1832;
   1585        1919        1312          72        1699        1868;
   1358        1516        1487         785        1311         343;
   1413          64         554          93         195        1647;
   1390         635        1901          69];

Let's compare with above data--

>>   filter=d(d>100 & d<400);
     kk=0;
     rr=1;
     for h=1:length(filter)
         if filter(h)<200
             kk(rr)=0;
             rr=rr+1;
         else
             kk(rr)=1;
             rr=rr+1;
         end
     end
>> kk
kk =
     1     0     1     1     1     0
>> mykk={~(d(d>100 & d<400)<200)}
mykk =
  1×1 cell array
    {1×6 logical}
>> mykk{:}
ans =
  1×6 logical array
   1   0   1   1   1   0
>> all(kk==mykk{:})
ans =
  logical
   1
>> 

which shows are identical...

NB: Your original code only returns a vector of length 5, however, not 6 as your upper loop runs over only filter(1:end) so will miss the last element every time. I'm thinking this is probably not deliberate but if is, just use the indexing expression on the array inside the logical expression to duplicate behavior.

NBSecond: The locations in the vector have nothing to do with the original positions of those elements that satisfy the test, though, so it's not clear what the intent of it is? And, of course, there may well be a different length vector for each case as well...if you want the location in the original vector, you should return the positions found in the original instead of the truncated vector. The logical addressing vector is only good in comparison to the shorter subset...which might be ok excepting you're not saving it to compare into. As noted, it just isn't at all clear as to what this might mean in the end.