Set vector equal to row of matrix matlab - matlab

I'm new to matlab and would like to create a new vector equal to a row of an nxn matrix that I already have. I know how to do it with loops but is there a command? Something like
vector1=matrix1(row5)
Thanks!

Yep this is easy, the syntax is:
row5Vector = matrix1(5,:);
Also to grab a column you could:
col5Vector = matrix1(:,5);

Related

Matlab Parse Error with a simple parenthesis?

I'm trying to finish a program and for some reason, the matrix I loaded into Matlab is messing with the ability to select the rows inside it. I'm trying to select all the rows in the matrix and see which values match the criteria for a Live setting. However I can select specific values/sections of the matrix in the command window without issue. Why is this happening? Any ideas?
It appears to only happen when in a for loop, I can do it just fine when it's on its own.
The syntax is: for x = start:stop. I think you are trying to do a for to the whole "A" matrix. You can split "A", according to its format (e.g. if is a table split in two variables).
bye
Richardd is right on; you're trying to iterate on a matrix, no good.
If I read you right, you're trying to run through your A matrix one column at a time, and see all the rows in that column? Assuming that is correct...
Your A matrix is 14x3, so you should go through your for loop 3 times, which is the size of your column dimension. Luckily, there is a function that MATLAB gives you to do just that. Try:
for iColumn = 1:size(A,2)
...
end
The size function returns the size of your array in a vector of [rows, columns, depth...] - it will go as many dimensions as your array. Calling size(A,2) returns only the size of your array in the column dimension. Now the for loop is iterating on columns.

Indexing data in matlab

I have imported a lot of data from an excel spreadsheet so that I have a 1x27 matrix.
I have imported data from excel using this
filename = 'for_matlab.xlsx';
sheet = 27;
xlRange = 'A1:G6';
all_data = {};
for i=1:sheet,
all_data{i} = xlsread(filename, i, xlRange);
end
However each element of this all_data matrix (which is 1x27) contains my data but I'm having trouble accessing individual elements.
i.e.
all_data{1}
Will give me the entire matrix but I need to perform multiplications on individual elements of this data
also
all_data(1)
just gives '5x6 double', i.e. the matrix dimensions.
Does anybody know how I can divide all elements of each row by the third element in each row and do this for all of my 'sub-matrices' (for want of a better word)
Assuming that all_data is a cell array and that each cell contains a matrix (with at least three columns):
result = cellfun(#(x) bsxfun(#rdivide, x, x(:,3)), all_data, 'uniformoutput', 0);
You are mixing terminology in matlab. what you have is 1x27 CELLS each of them containing a matrix.
If you access all_data{1} it will give you the whole matrix stored in the first cell.
If you want to access the elemets of that matrix then you need to do: all_data{1}(2,4). This example access the 2,4 element of the matrix in the first cell.
Definitely Luis Mendo has solved you problem, but be aware of the differences of Cells and matrixes in Matlab!
Okay I have found the answer now.
Basically you have to use both types of brackets because the data types are different
i.e. all_data{1}(1:4) or something like that anyway.
Cheers

Referring to coordinates of a 3d matrix in Matlab

In Matlab I'm trying to find points in a 3d matrix whose coordinates are smaller than some function.
If these coordinates are equal to some functions than I can write:
A(some_function1,some_function2,some_function3)=2;
But what if I want to do something like:
A(<some_function1,<some_function2,<some_function3)=2;
This isn't working - so what is the other way of finding such points without using "for" loop? Unfortunately with "for" loop my code takes a lot of time to compute. Thank you for your help!
How about something along the lines of
A( ceil(min(some_function1,size(A,1))),...
ceil(min(some_function2,size(A,2))),...
ceil(min(some_function3,size(A,3))) );
This will cap the indicies to the end of each array dimension
You can just use regular indexing to achieve this:
A(1:floor(some_function1),1:floor(some_function2),1:floor(some_function3)) = 2;
assuming you check / ensure that floor(some_function*) is smaller than the dimensions of A
Try:
A(1:size(A,1)<some_function1, 1:size(A,2)<some_function2, 1:size(A,3)<some_function3) = 2
I hope I got your question correctly.

Adding a dimension to a matrix in Matlab

I need to add a new matrix to a previously existant matrix, but on his dimension coordinate.
I know this is hard to understand, so let's see it on a example:
I've a matrix like this:
480x640x3
And I want to add the following one:
480x640x6
The result has be this: (6+3 = 9)
480x640x9
As you can see it adds but on the 3rd dimension.
For concatenating along higher dimensions, use the function CAT:
newMatrix = cat(3,matrix1,matrix2);
I would say that gnovice's answer is probably the best way to go, but you could do it this way too:
matrix1(:,:,4:9) = matrix2;

What's the best way to iterate through columns of a matrix?

I want to apply a function to all columns in a matrix with MATLAB. For example, I'd like to be able to call smooth on every column of a matrix, instead of having smooth treat the matrix as a vector (which is the default behaviour if you call smooth(matrix)).
I'm sure there must be a more idiomatic way to do this, but I can't find it, so I've defined a map_column function:
function result = map_column(m, func)
result = m;
for col = 1:size(m,2)
result(:,col) = func(m(:,col));
end
end
which I can call with:
smoothed = map_column(input, #(c) (smooth(c, 9)));
Is there anything wrong with this code? How could I improve it?
The MATLAB "for" statement actually loops over the columns of whatever's supplied - normally, this just results in a sequence of scalars since the vector passed into for (as in your example above) is a row vector. This means that you can rewrite the above code like this:
function result = map_column(m, func)
result = [];
for m_col = m
result = horzcat(result, func(m_col));
end
If func does not return a column vector, then you can add something like
f = func(m_col);
result = horzcat(result, f(:));
to force it into a column.
Your solution is fine.
Note that horizcat exacts a substantial performance penalty for large matrices. It makes the code be O(N^2) instead of O(N). For a 100x10,000 matrix, your implementation takes 2.6s on my machine, the horizcat one takes 64.5s. For a 100x5000 matrix, the horizcat implementation takes 15.7s.
If you wanted, you could generalize your function a little and make it be able to iterate over the final dimension or even over arbitrary dimensions (not just columns).
Maybe you could always transform the matrix with the ' operator and then transform the result back.
smoothed = smooth(input', 9)';
That at least works with the fft function.
A way to cause an implicit loop across the columns of a matrix is to use cellfun. That is, you must first convert the matrix to a cell array, each cell will hold one column. Then call cellfun. For example:
A = randn(10,5);
See that here I've computed the standard deviation for each column.
cellfun(#std,mat2cell(A,size(A,1),ones(1,size(A,2))))
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Of course, many functions in MATLAB are already set up to work on rows or columns of an array as the user indicates. This is true of std of course, but this is a convenient way to test that cellfun worked successfully.
std(A,[],1)
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Don't forget to preallocate the result matrix if you are dealing with large matrices. Otherwise your CPU will spend lots of cycles repeatedly re-allocating the matrix every time it adds a new row/column.
If this is a common use-case for your function, it would perhaps be a good idea to make the function iterate through the columns automatically if the input is not a vector.
This doesn't exactly solve your problem but it would simplify the functions' usage. In that case, the output should be a matrix, too.
You can also transform the matrix to one long column by using m(:,:) = m(:). However, it depends on your function if this would make sense.