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# Peter Jarosi

##### Last seen: 15 jours ago
13 total contributions since 2019

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how to extract all variables from a struct
I agree with Stephen but if you want a quick solution using an ugly trick: v = fieldnames(data_table); for x = 1 : length(v)...

environ un an ago | 0

How to get values of unknowns in the terms of one variable?
I recommend using fsolve() function. First of all, you cannot express your s's in explicit form but you can solve your non-li...

environ un an ago | 0

String Matching in an If Statement
I recommend reading more about it depending on your problem but something like this: Table3 = outerjoin(Table1, Table2, 'Keys',...

environ un an ago | 0

acces data in struct
You can use for loop indexing elements of your structure: myElements = fieldnames(Str); % or you can list your elements % myE...

environ un an ago | 1

| accepted

why this logical expression is wrong?
because of the accuracy problem of a floating-point system >> format longE >> sind(30) ans = 4.999999999999999e-0...

environ un an ago | 1

| accepted

matrix values from loop to raw vector?
I've fixed your code: clear; q=0.1; L=2.5;%m p=0.52; c1=0; c2=(1:10); C=zeros(4,40); for n=(1:10) C(:, 4*(n-1)+1:...

environ un an ago | 1

| accepted

how to create a function that takes a txt file with a lot of columns and saves a new file (while keeping the old one) with 3 specific columns from the original ..
You should have been more specific, so I could have given more accurate answer. For instance, a list of your filenames, names of...

environ un an ago | 1

| accepted

Export Cell-Data to .xlsx [Saves less data]
I recommend converting your cell-data to table and using function writetable(). filename='myExcelFile.xlsx'; myTable = cell2ta...

environ un an ago | 0

| accepted

Export Cell-Data to .xlsx [Saves less data]

environ un an ago | 1

Labeling the columns in an array
Do not convert your table into an array! Just use table object! Data=readtable('Filename.csv'); Later, if you want to refer it...

environ un an ago | 0

| accepted

Error Using a loop: Conversion to double from cell is not possible.
Perhaps what you need is struct. Try the following: s = struct(); for w = 1:10 PN = PName{w}; CS = CSize{w}; Project = [PN ...

environ un an ago | 1

| accepted