Let's say I have a cell array raweeg each cell of which is a matrix with time points in the first column and some markers in the second. I want to create a vector to store the time points, when the marker is not equal to -1. I found a non-elegant (and not working) way to create a vector of zeros of size 1x1 and then to append the following values in a loop.
P.S.: There are exactly 96 non-"-1" values and corresponding time points.
startpoints = zeros(1,1);
for i = length(raweeg{1,1}(:,1))
if raweeg{1,1}(i,2) ~= -1
startpoints(end+1,1) = raweeg{1,1}(i,1);
end
end
Thank you
Vectorize it like this, for a given cell of raweeg:
startpoints = raweeg{1,1}(raweeg{1,1}(:,2) ~= -1, 1);
This is called logical indexing.
Just be sure your markers are not generated with floating point computations, or the comparisons will likely fail often.
P.S. The problem in your code is the for loop statement, which should be:
for i = 1:length(raweeg{1,1}(:,1))
Or better, for i = 1:size(raweeg{1,1},1).
With out the "1:" part, it just has a single iteration, the last row.
Related
The problem is as follow:
I have a k-nearest index matrix. The column of the matrix denotes the number of an image (1:n). The row of index matrix denotes the index of k-nearest pixels correspoinding to the query pixel.
So I am using for-loops to find out whether a-th index is in column b: for example, 3 is in the first column (1-th pixel), then find out whether 1 is in 3-th column, if no, assign value, if yes, assign another value.
The code is here:
for pix = 1:row*col
for i = 1:k
position = find(Index(:, Index(i,pix)) == pix);
if isempty(position) difference = - weight(k, Index(i,pix)) ;
else difference = weight(i,pix) - weight(position, Index(i,pix)) ; end
end
end
Since row*col is very large. I hope to reduce the code into one for-loop. Is there any other method to speed up? Thanks in advance!
I am trying to remove rows from a matrix based on a condition. I have a 371000x5 double matrix, and a 371000x1 vector of dummies (1 and 0). I want to remove each row from the original matrix, where the value of the vector of dummies is 1.
I have tried the following, but it is taking very long:
for i = 1:size(matrix_num,1)
if missing_matrix(i,1) >=0
matrix_num(i,:) = [];
end
end
My Matlab has been busy for over 30 minutes now, so I am not even sure if the code is right. Is there a more efficient way to do this?
Additionally, I have to do the same action on a cell matrix (categorical data). Should I expect any huge difference from the numerical matrix?
The programmatic way of doing this is:
new_matrix = old_matrix(missing_vector==1,:)
for keeping lines with missing_vector 1
new_matrix = old_matrix(missing_vector==0,:)
for removing lines with missing_vector 1
For educational values, if you want the loop to work, don't do that row by row. Your solution causes the matrix to be copied and re-allocated on every row removed.
So, you would be better off if you calculate the resulting matrix size in advance:
new_matrix = zeros(sum(missing_vector), 5)
and then your iteration would work:
index_new=1
for index_old = 1:size(old_matrix,1)
if missing_vector(index_old) ==0
new_matrix(index_new,:) = old_matrix(index_old,:);
end
end
Try compact MATLAB code
matrix_num(missing_matrix>=0,:)=[]
Note : You must make a vector for missing_matrix variable. If this variable is matrix, you need other form of code .
As I know, you can use it in cell array too.
I am writing code which compares the data of a vector. It should count how many positions (acc) have equal values, and save a specific value in a vector of the same length of the quantity of positions (n_T(acc)).
My data vector is [30000 x 1]. For example, first 80 positions have the same value, next 60 positions have the same value, etc., next 5 positions have the same value.
The code works well if I use just 29996 values. I do not understand why when I try to use the complete vector MATLAB stays Busy.
Checking my data vector, I noticed that the last 5 positions are equivalent [29996:30000]. Could it be the reason, and what should I change?
Following is the code
%========================================================
%ac: data vector`
%acc1: accumulator which count how much positions have the same value
%n_T: vector which presents the values I need, in the same positions the data is equal
%m: show a value where i should begin
%========================================================
i=m; %previously used`
fv=length(ac)
while i<fv %29996
acc1=0;
for i=m+1:fv
if ac(i)==ac(i-1)
acc1=acc1+1; % count how much positions are equals
else
m=i;
break
end
end
mi=m-acc1; %define where the data n_T should begin
for i=mi:m
n_T(i)=tm/acc1; %create a vector with length [acc x1] begining in mi and finishing in m
end
m=i;
end
plot(n_T)
Does it work if you do this in a vectorized way? Not completely sure what you want the program to output.
% locate repeated elements
eq_els = ac(diff(ac) == 0);
% count number of repeated elements (unique)
unique_els = unique(eq_els);
num_equal_els = numel(unique_els);
% create variable-length lists for each unique element
each_eq_list = cell(num_equal_els,1);
for (k = 1:num_equal_els)
% each vector in the cell array is equal to the elements of ac that are equal to the current unique item
each_eq_list{[k]} = ac(ac == unique_els(k));
end
The length of each_eq_list{[k]} is the length of the total number of contiguous repeated values of repeated value k.
In every other language if I have a matrix, if I call a mono-dimensional index, the result will be an array.I don't know why in Matlab if you take a single index of a matrix, you'll get a single element, that's stupid.
Anyway in C:
mat[4][4];
mat[0] is an array.
In Matlab:
mat=[1 2; 3 4];
How do I take the first row of the matrix? mat(1) is 1, not [1 2].
EDIT: There is another problem, I have a problem with this function:
function str= split(string, del)
index=1;
found=0;
str=['' ; ''];
for i=1:length(string)
if string(i)==del
found=1;
index=1;
elseif found==1
str(2,index)=string(i);
index=index+1;
else
str(1,index)=string(i);
index=index+1;
end
end
end
This returns sometimes a matrix and sometimes an array.
For example if I use split('FF','.') I get 'FF' as result, but what if I want to return a matrix? I can't even choose the dimensione of the matrix, in this context a weak typed language is a big disvantage.
You have to say which columns you want. : stands for all indices in a dimension, so to take first row
mat(1,:)
It is not stupid, but useful. If you address a matrix with only one index, it implicitly gets converted to a vector. This gives you the option to use linear indices (see sub2ind).
This will extract the second row
vector = mat(2,:)
And This will extract the second column
vector = mat(:,2)
You can use
vector = mat(end,:)
To extract the last row
Hope this helps you
From Matrix Indexing in MATLAB:
When you index into the matrix A using only one subscript, MATLAB
treats A as if its elements were strung out in a long column vector,
by going down the columns consecutively
I just hope it doesn't look stupid to you anymore (along with the right answers from angainor and Marwan)
I have a set of data that is <106x25 double> but this is inside a struct and I want to extract the data into a matrix. I figured a simple FOR loop would accomplish this but I have hit a road block quite quickly in my MATLAB knowledge.
This is the only piece of code I have, but I just don't know enough about MATLAB to get this simple bit of code working:
>> x = zeros(106,25); for i = 1:106, x(i,:) = [s(i).surveydata]; end
??? Subscripted assignment dimension mismatch.
's' is a very very large file (in excess of 800MB), it is a <1 x 106 struct>. Suffice it to say, I just need to access a small portion of this which is s.surveydata where most rows are a <1 x 25 double> (a row vector IIRC) and some of them are empty and solely return a [].
s.surveydata obviously returns the results for all of the surveydata contained where s(106).surveydata would return the result for the last row. I therefore need to grab s(1:106).surveydata and put it into a matrix x. Is creating the matrix first by using x = zeros(106,25) incorrect in this situation?
Cheers and thanks for your time!
Ryan
The easiest, cleanest, and fastest way to write all the survey data into an array is to directly catenate it, using CAT:
x = cat(1,s.surveydata);
EDIT: note that if any surveydata is empty, x will have fewer rows than s has elements. If you need x to have the same amount of rows as s has elements, you can do the following:
%# find which entries in s have data
%# note that for the x above, hasData(k) contains the
%# element number in s that the k-th row of x came from
hasData = find(arrayfun(#(x)~isempty(x.surveydata),s));
%# initialize x to NaN, so as to not confuse the
%# real data with missing data entries. The call
%# to hasData when indexing makes this robust to an
%# empty first entry in s
x = NaN(length(s),length(s(hasData(1)).surveydata);
%# fill in only the rows of x that contain data
x(hasData,:) = cat(1,s(hasData).surveydata);
No, creating an array of zeroes is not incorrect. In fact it's a good idea. You don't have to declare variables in Matlab before using them, but for loops, pre-allocating has speed benefits.
x = zeros(size(s), size(s(1)));
for i = 1:106
if ~isempty(s(i).surveydata)
x(i, :) = s(i).surveydata;
end
end
Should accomplish what you want.
EDIT: Since OP indicated that some rows are empty, I accounted for that like he said.
what about this?
what s is?
if s(i).surveydata is scalar:
x = zeros(106,25);
for i = 1:106
x(i,1) = [s(i).surveydata];
end
I am guessing that is what you want tough it is not clear at all :
if s(i).surveydata is row vector:
x = zeros(106,25);
for i = 1:106
x(i,:) = [s(i).surveydata];
end
if s(i).surveydata is column vector:
x = zeros(106,25);
for i = 1:106
x(i,:) = [s(i).surveydata]';
end