Extracting Specific Rows From xlsx or txt File

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Joseph Salerno
Joseph Salerno el 6 de Oct. de 2023
Comentada: Anton Kogios el 10 de Oct. de 2023
I have an excle file with raw data output by another program, but it is formatted rather oddly. I would like to take every 3rd row of data and make it into a row on a matrix. I've tried using readmatrix, but I don't know how to make it take out multiple specific rows and place them into one matrix. I feel like this should be doable by using an array made with [14:3:764] which would tell readmatrix to start pulling data from row 14, and then pull every 3rd row until it reaches row 764. Ideally, these would also be stacked into a matrix so that row 1 data is row 1 in the matrix, row 2 data is row 2 data in the matrix, and so on. Should I be using something with Opts and DataRange? I'm completely new to data importing in Matlab, so any help is much appreciated.
Original Question: The data can also be output to a txt file, but I have no idea how to parse through it and turn the rows of data and text into a useful input for Matlab. Is there a way to extract the data for each timestep and place it in a structure with its respective timestep?

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Anton Kogios
Anton Kogios el 6 de Oct. de 2023
Editada: Anton Kogios el 6 de Oct. de 2023
Ran in:
I think you edited your question from what you had initially quite a bit (you should have just made a new question). Since I spent time on it, I think I found a solution to your initial question. This code reads the initial file that you sent with text, finds where the time and force data is, and assigns the data to a struct. Hope it helps. It's also pretty fast (<1.5 sec on my computer).
% Read data
unzip CompForceTest1.zip
fid = fopen('CompForceTest1','r');
txtDouble = fread(fid);
txtChar = char(txtDouble');
fclose(fid);
% Find indices
newlinesIdx = find(txtDouble == 10);
timesIdx = strfind(txtChar,'TIME:,');
forceIdx = strfind(txtChar,'Particle Compressive Force:,');
data = struct('Time',0,'Force',0);
% Cycle through and extract times and force data
for i = 1:length(timesIdx)
diffTime = (newlinesIdx-timesIdx(i));
diffForce = (newlinesIdx-forceIdx(i));
data(i).Time = str2double(txtChar((timesIdx(i)+5)+(1:min(diffTime(diffTime>0))-5)));
data(i).Force = str2num(txtChar((forceIdx(i)+27)+(1:min(diffForce(diffForce>0))-27)));
end
data
data = 1×251 struct array with fields:
Time Force
  2 comentarios
Joseph Salerno
Joseph Salerno el 9 de Oct. de 2023
I'm quite glad you caught the original. After doing a bit more searching, I thought it may just end up being far easier to accomplish with an excel file, so i changed it. I can add back the old text file and question for clarity, because this gives me exactly what I need and is far more efficient than going into and then back out of excel.
I do have some questions about how you did it though, mainly with the indices.
What is find(txtDouble == 10) doing?
I follow the timesIdx and forceIdx, but where are the 5 and 27 coming from later on when moving from timestep to timestep in the loop?
Thank you, this is extremely helpful.
Anton Kogios
Anton Kogios el 10 de Oct. de 2023
Ran in:
Hey Joseph, I'm glad this works for you!
(txtDouble == 10) finds the newline characters (which is equal to 10 in ASCII/as a double) as an array of logical values. I wanted to get the actual indices where there are new lines, so find(txtDouble == 10) returns the index of all non-zero elements in the array (in our case, all the logical 1s). A simple example:
find([true false false true true])
ans = 1×3
1 4 5
timesIdx and forceIdx are doing a similar thing which I think you understand, but the index they return is the start of the string they're looking for. So that's where the 5 and 27 come from - they are the length of 'TIME:,' and 'Particle Compressive Force:,' which are added to get the data after they occur.
Feel free to ask more questions, happy to help your understanding!

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Voss
Voss el 6 de Oct. de 2023
Editada: Voss el 6 de Oct. de 2023
Ran in:
Here's one way. This method reads the file's contents into a cell array using readcell. The data on lines 14:3:end of the file end up in cells (12:2:end,3) of the cell array (that is, rows 12:2:end in column 3) because of blank lines in the file. All the numeric data from the file is stored in the cell array as character vectors, so use str2num to convert to a vector of numbers, one line at a time, storing the vectors as rows in a matrix.
unzip CompForceTest2.zip
ls
CompForceTest2.csv CompForceTest2.zip
filename = 'CompForceTest2.csv';
% read the file into a cell array C:
C = readcell(filename)
C = 510×3 cell array
{'Compressive Forces Test 2'} {[<missing> ]} {[<missing> ]} {'QUERY DESCRIPTIONS' } {[<missing> ]} {[<missing> ]} {'Q01' } {'Particle Compressive Force'} {[<missing> ]} {'GROUP/TYPE' } {',All' } {[<missing> ]} {'SELECTED OBJECTS' } {',All' } {[<missing> ]} {'RANGE' } {',None' } {[<missing> ]} {'UNITS' } {',N' } {[<missing> ]} {'EXTRACTED DATA' } {[<missing> ]} {[<missing> ]} {'TIME' } {',0' } {[<missing> ]} {'Q01' } {'Particle Compressive Force'} {',no data' } {'TIME' } {',0.0200013' } {[<missing> ]} {'Q01' } {'Particle Compressive Force'} {',0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,…'} {'TIME' } {',0.0400009' } {[<missing> ]} {'Q01' } {'Particle Compressive Force'} {',0,0.00028769485652446747,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0…'} {'TIME' } {',0.0600018' } {[<missing> ]} {'Q01' } {'Particle Compressive Force'} {',0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,…'}
% keep only the relevant stuff:
C = C(12:2:end,3);
% number of rows in C:
nrow_C = size(C,1);
% initialize a matrix M with that same number of rows and zero columns:
nrow_M = nrow_C;
ncol_M = 0;
M = NaN(nrow_M,ncol_M);
% loop over the rows of C
for row = 1:nrow_C
% convert the data in this cell into a numeric vector:
new_data = str2num(C{row});
% number of elements in the vector:
n_data = numel(new_data);
% going to store this vector as a row of M, but if this vector
% is bigger than the number of columns in M, need to add
% columns to M in order to store it:
if n_data > ncol_M
% add columns of NaNs to M:
M = [M NaN(nrow_M,n_data-ncol_M)];
% update the variable storing the number of columns of M:
ncol_M = n_data;
end
% put the new vector in place in the current row of M:
M(row,1:n_data) = new_data;
end
% see the result:
disp(M);
Columns 1 through 19 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0003 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0009 0 0 0 0 0 0 0 0 0 0 0 0.0023 0 0.0018 0 0 0 0 0 0 0 0.0205 0 0 0 0 0 0 0 0.0063 0 0 0 0 0 0 0 0 0.0001 0.0003 0 0 0 0.0001 0 0 0 0.0002 0 0.0129 0.0001 0 0.0070 0.0015 0 0 0.0089 0.0016 0 0 0 0.0016 0 0.0000 0 0 0.0026 0 0.0138 0.0002 0 0 0.0120 0.0002 0 0 0.0015 0 0.0073 0.0001 0 0 0.0026 0.0002 0 0.0006 0 0 0.0001 0.0192 0.0002 0 0.0002 0.0001 0.0066 0 0 0.0047 0 0 0.0036 0 0.0004 0 0.0111 0.0006 0.0106 0.0131 0 0.0177 0.0005 0.0017 0 0.0166 0.0130 0.0103 0.0002 0 0 0.0129 0.0111 0.0012 0.0082 0.4390 0 0.3372 0.0608 0 0 0.0455 0 0.5080 0 0 0.1900 0 0.1567 0.0923 0 0.0350 0 0.1602 0 0.0170 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0701 0 0 0.0840 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0001 0 0 0 0 0 0 0 0.1471 0 0.0000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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