Donald, regarding your new "Answer", you can do it like this: N=[10 100 1000]; for n = 1 : length(N) first=1; second=1; for k = 3 : N(n) %The index has to have two terms removed because it starts with 1 and 1 already next = first+second; %The current term in the series is a summation of the previous two terms first = second; %Each term must by iterated upwards by an index of one second = next; %The term that previously was second is now referred to as next end % Print to command window next end

The paths need to be enclosed in single or double quotes. Then when you put them into a cell array don't use quotes there. And you need to use fullfile to construct the full path. And don't use i (the imaginary constant) as a loop iterator. Untested code: % Paths where the files are located P1 = 'C:\Users\me\Documents\My Info'; P2 = 'C:\Users\me\Documents\My data'; %Array of paths Array = {P1, P2}; for k = 1:length(Array) thisFolder = Array{k}; % Find all Excel files in this folder. filePattern = fullfile(thisFolder, '*.xls*'); fileList = dir(filePattern) for f = 1 : numel(fileList) fullFileName = fullfile(thisFolder, fileList(f).name); fprintf('Processing %s.\n', fullFileName); % Pull data data = readmatrix(fullFileName); % Write into am output file baseFileName = sprintf('New %s', fileList(f).name) outputFileName = fullfile(thisFolder, baseFileName); writeMatrix(data, outputFileName); end end Also, please read the FAQ: What is a cell array for a good explanation of how to use cell arrays and when to use braces, parentheses, and brackets. To learn other fundamental concepts, invest 2 hours of your time here: MATLAB Academy - Free 2 hour training

Subscript ‘in’ here — values_change = values_change + in(i); and then subscript ‘values’ (the preallocation is optional, however a good habit to adopt, sinc it produces much more efficnet matrix operations in this snd similar applications). See if that produces the desired result. in = [8 8 8 8]; values = NaN(1,numel(in)+1); values_change = 29; values(1) = values_change; for i=1:length(in) values_change = values_change + in(i); values(i+1) = values_change; end disp(values) .

The loop, if one is appropriate, depends on what you want to do. This is an example of how to address them: numlist = {3,6,18}; for k1 = 1:length(numlist) fprintf('Number at position %d = %6.2f\n', k1, numlist{k1}) end

The problem is that you are not using the index N. A = (Array(1,1)*Z)/Array(1,2); This line doesn't change, so you will get the same output 15 times. A nested loop would look like this: [row,col] = size(B); for N = 1:col for M = 1:row A = (B(M,N)*Z)/B(1,2); disp (A) end end The first loop will start at column 1, then the second loop goes through all rows. After that it repeats with column 2 and so on if you have more columns. Depending on how you want to manipulate the entries you have to use both indexes accordingly.

The code below gives you what you want. The only difference is that you need to write my_data{1} instead of my_data1. [~,~,data]=xlsread('ex[1].xlsx') indices=cell2mat(data(:,4)); my_data=cell(max(indices),1) for n=1:max(indices) my_data{n}=data(indices==n,:); end my_data my_data{1}

It's to show you how matlab syntax supports vectorised operations. Internally to the Matlab interpreter there is a loop, but you don't need one in the script. Use fscanf or dlmread to read the data into a vector, then use logical indexing to test your interval. e.g. PointsLessThanOne = vector<1; TheSmallValues = vector(PointsLessThanOne); https://uk.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

You could do something like this a = [1 2; 3 4]; for k = a(:).' % reshape a into a row vector if k>2 fprintf('%d is greater than 2\n',k) end end but I don't really think it's as readable. There are probably other options. Bear in mind that using length() will break your example if a is not a vector. Use numel() instead.