The other day I discovered the following bug in a couple of places in my MATLAB code
I wanted to enter the column vector in my MATLAB script
[a-b,
c-d
e-f]
where a,b,c,d,e,f are long expressions in some variables.
and I entered it in as
[ a -b ;
c -d ;
e -f]
Now MATLAB interprets the second matrix as a 3x2 matrix instead of a column vector.
Is there a way/command/function to force MATLAB to use only the comma and NOT any white space characters as a column separator for matrices ?
I don't think there is any way to force matlab to not treat white space this way, since it is interpretive language, and doing so may affect some built-in functions/third-party code.
However, you can use parentheses to group data - i.e. (a -b) will still be a single element of the matrix.
Well your second matrix does look like it's intended as a 3x2. However, if you do it like this it will be a column vector again:
[a - b;
c - d;
e - f]
which to me is a reasonable intuitive distinction between a minus b and a, negative b.
You can also use brackets as Ilya suggested.
Assuming you have a piece of code in which you only want to have column vectors and no matrix, there is a fairly quick solution:
replace {space}+ by +
replace {space}- by -
It is quite safe to do and unless you have complicated expressions in your vector it should do the trick.
Related
I have a row cell vector M, containing matrices in each cell. Every matrix m (matrix inside the big matrix M) is made of 2 channels (columns), of which I only want to use the first.
The approach I thought about was going through each m, check if it has 2 channels, and if that is the case delete the second channel.
Is there a way to just slice it in matlab? or loop it and obtain the matrix M as the matrix m would disappear.
First code is:
load('ECGdata.mat')
I have the below.
when I double-click in one of the variable , here is what I can see:
As you can see the length of each matrix in each cell is different. Now let's see one cell:
The loop I'm trying to get must check the shape of the matrix (I'm talking python here/ I mean if the matrix has 2 columns then delete the second) because some of the variables of the dataframe have matrix containing one column (or just a normal column).
In here I'm only showing the SR variable that has 2 columns for each matrix. Its not the case for the rest of the variables
You do not need to delete the extra "channel", what you can do is quite simple:
newVar = cellfun(#(x)x(:,1), varName, 'UniformOutput', false);
where varName is SR, VF etc. (just run this command once for each of the variables you load).
What the code above does is go over each element of the input cell (an Nx2 matrix in your example), and select the first column only. Then it stores all outputs in a new cell array. In case of matrices with a single column, there is no effect - we just get the input back.
(I apologize in advance if there is some typo / error in the code, as I am writing this answer from my phone and cannot test it. Please leave a comment if something is wrong, and I'll do my best to fix it tomorrow.)
The following error occurs quite frequently:
Subscript indices must either be real positive integers or logicals
I have found many questions about this but not one with a really generic answer. Hence I would like to have the general solution for dealing with this problem.
Subscript indices must either be real positive integers or logicals
In nearly all cases this error is caused by one of two reasons. Fortunately there is an easy check for this.
First of all make sure you are at the line where the error occurs, this can usually be achieved by using dbstop if error before you run your function or script. Now we can check for the first problem:
1. Somewhere an invalid index is used to access a variable
Find every variable, and see how they are being indexed. A variable being indexed is typically in one of these forms:
variableName(index,index)
variableName{index,index}
variableName{indices}(indices)
Now simply look at the stuff between the brackets, and select every index. Then hit f9 to evaluate the result and check whether it is a real positive integer or logical. Visual inspection is usually sufficient (remember that acceptable values are in true,false or 1,2,3,... BUT NOT 0) , but for a large matrix you can use things like isequal(index, round(index)), or more exactly isequal(x, max(1,round(abs(x)))) to check for real positive integers. To check the class you can use class(index) which should return 'logical' if the values are all 'true' or 'false'.
Make sure to check evaluate every index, even those that look unusual as per the example below. If all indices check out, you are probably facing the second problem:
2. A function name has been overshadowed by a user defined variable
MATLAB functions often have very intuitive names. This is convenient, but sometimes results in accidentally overloading (builtin) functions, i.e. creating a variable with the same name as a function for example you could go max = 9 and for the rest of you script/function Matlab will consider max to be a variable instead of the function max so you will get this error if you try something like max([1 8 0 3 7]) because instead of return the maximum value of that vector, Matlab now assumes you are trying to index the variable max and 0 is an invalid index.
In order to check which variables you have you can look at the workspace. However if you are looking for a systematic approach here is one:
For every letter or word that is followed by brackets () and has not been confirmed to have proper indices in step 1. Check whether it is actually a variable. This can easily be done by using which.
Examples
Simple occurrence of invalid index
a = 1;
b = 2;
c = 3;
a(b/c)
Here we will evaluate b/c and find that it is not a nicely rounded number.
Complicated occurrence of invalid index
a = 1;
b = 2;
c = 3;
d = 1:10;
a(b+mean(d(cell2mat({b}):c)))
I recommend working inside out. So first evaluate the most inner variable being indexed: d. It turns out that cell2mat({b}):c, nicely evaluates to integers. Then evaluate b+mean(d(cell2mat({b}):c)) and find that we don't have an integer or logical as index to a.
Here we will evaluate b/c and find that it is not a nicely rounded number.
Overloaded a function
which mean
% some directory\filename.m
You should see something like this to actually confirm that something is a function.
a = 1:4;
b=0:0.1:1;
mean(a) = 2.5;
mean(b);
Here we see that mean has accidentally been assigned to. Now we get:
which mean
% mean is a variable.
In Matlab (and most other programming languages) the multiplication sign must always be written. While in your math class you probably learned that you can write write a(a+a) instead of a*(a+a), this is not the same in matlab. The first is an indexing or function call, while the second is a multiplication.
>> a=0
a =
0
>> a*(a+a)
ans =
0
>> a(a+a)
Subscript indices must either be real
positive integers or logicals.
Answers to this question so far focused on the sources of this error, which is great. But it is important to understand the powerful yet very intuitive feature of matrix indexing in Matlab. Hence how indexing works and what is a valid index would help avoid this error in the first place by using valid indices.
At its core, given an array A of length n, there are two ways of indexing it.
Linear indexing: with subset of integers from 1 : n (duplicates allowed). 0 is not allowed, as Matlab arrays are 1-based, unless you use the method below. For higher-dimensional arrays, multiple subscripts are internally converted into a linear index, although in an efficient and transparent manner.
Logical indexing:wherein you use a n-length array of 0s and 1s, to pick those elements where indexing is true. In this case, unique(index) must have only 0 and 1.
So a valid indexing array into another array with n number of elements ca be:
entirely logical of the same size, or
linear with subsets of integers from 1:n
Keeping this in mind, invalid indexing error occurs when you mix the two types of indexing: one or more zeros occur in your linearly indexing array, or you mix 0s and 1s with anything other than 0s and 1s :)
There is tons of material online to learn this including this one:
http://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html
Might sound too simple to you but I need some help in regrad to do all folowings in one shot instead of defining redundant variables i.e. tmp_x, tmp_y:
X= sparse(numel(find(G==0)),2);
[tmp_x, temp_y] = ind2sub(size(G), find(G == 0));
X(:)=[tmp_x, tmp_y];
(More info: G is a sparse matrix)
I tried:
X(:)=ind2sub(size(G), find(G == 0));
but that threw an error.
How can I achieve this without defining tmp_x, tmp_y?
A couple of comments with your code:
numel(find(G == 0)) is probably one of the worst ways to determine how many entries that are zero in your matrix. I would personally do numel(G) - nnz(G). numel(G) determines how many elements are in G and nnz(G) determines how many non-zero values are in G. Subtracting these both would give you the total number of elements that are zero.
What you are doing is first declaring X to be sparse... then when you're doing the final assignment in the last line to X, it reconverts the matrix to double. As such, the first statement is totally redundant.
If I understand what you are doing, you want to find the row and column locations of what is zero in G and place these into a N x 2 matrix. Currently with what MATLAB has available, this cannot be done without intermediate variables. The functions that you'd typically use (find, ind2sub, etc.) require intermediate variables if you want to capture the row and column locations. Using one output variable will give you the column locations only.
You don't have a choice but to use intermediate variables. However, if you want to make this more efficient, you don't even need to use ind2sub. Just use find directly:
[I,J] = find(~G);
X = [I,J];
I am wondering if there is some way, how to multiple assign values to different variables according logical vector.
For example:
I have variables a, b, c and logical vector l=[1 0 1] and vector with values v but just for a and c. Vector v is changing its dimension, but everytime, it has the same size as the number of true in l.
I would like to assign just new values for a and c but b must stay unchanged.
Any ideas? Maybe there is very trivial way but I didn't figure it out.
Thanks a lot.
I think your problem is, that you stored structured data in an unstructured way. You assume a b c to have a natural order, which is pretty obvious but not represented in your code.
Replacing a b c with a vector x makes it a really easy task.
x(l)=v(l);
Assuming you want to keep your variable names, the simplest possibility I know would be to write a function:
function varargout=update(l,v,varargin)
varargout=varargin;
l=logical(l);
varargout{l}=v(l);
end
Usage would be:
[a,b,c]=update(l,v,a,b,c)
while reading someone's code I came across this:
data = [data.a, data.b, data.c, ...
data.x, data.y'];
why does the y have a single quote after it? does it have something to do with its data type? I got this error after removing it:
??? Error using ==> horzcat
CAT arguments dimensions are not consistent.
can someone please explain? thanks!
The single quotation mark is Matlab's transposition operator. If x is a row vector then x' is a column vector, and so forth.
If data.x and data.y are both row vectors, it's no surprise that your attempt to horizontally concatenate data.x and data.y' is unsuccessful, it's not a well-defined operation, since the former is (I guess) a row vector and the latter (if I guess correctly) a column vector.