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I would like to pick two different random elements from given array with their positions. Similar like with datasample, but with datasample there is a possibility of picking the same element twice.
I could use while loop or similar, but I suppose there is an easier way to do it.
Say you have a matrix A:n by m, you can choose two elements at random as following,
A=[2 7 8;5 4 6;8 3 11];%given array
[n m]=size(A);
x=2;%two different random elements
i=randperm(n,x)%row index for x elements
j=randperm(m,x)%column index for x elements
A(i(1),j(1)) %First random element
A(i(2),j(2)) %Second random element
If you try this you could get something like,
i =
2 3
j =
2 1
ans =
4
ans =
8
the code can be further simplified but just wanted to make it clear. Please let me know if you have any additional questions or require further clarification.
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In a sample code I have X(1:n, 1)
I don't understand how it works
n=3 in a sample
Is it a matrix? but isn't () used for indexing so how it's a double index?
I imagine that your X is a two-dimmensional matrix. Generalizing, it can be said that to access the elements of a matrix is done with: X(n,m).
The common case is to get one single number, in that case n and m are integer numbers. But you can also pass vectors to n and m positions and that way extract a submatrix from the original one.
As an example:
X = [1 2 3; 4 5 6; 7 8 9]
X(1:3,1) = [1; 4; 7]
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I am very new to Matlab and coding in general, so I apologize if this is a basic question.
I have a matrix of three columns (data1) where the first column refers to time (s).
I would like to make a new matrix (bout1) consisting of entire rows of the matrix data1 based on the values in the first column (so, for example, in a range from 30 s to 120 s).
I know how to extract the rows based on the row number:
bout1 = data1(361126:391643,:)
but not based on the values in a specific column.
You can use the find function (see here) to find the rows you need, like this:
time = data1(:, 1);
i = find(30 <= time & time <= 120);
bout1 = data1(i, :);
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I need to start with the highest value in the matrix (i,j). Then, go backwards to one of positions (i-1,j), (i, j-1), and (i-1, j-1) depending on the direction of movement used to construct the matrix. This method is used throughout until a matrix cell with zero value is reached.The above image shows what I want, I need to trace those marked in blue. I know there is a max function in matlab.
If I understand what you try to do...
Suppose M is your matrix:
result=[];
[~,ind]=max(M(:));
[r,c]=ind2sub(size(M),ind);
result=[result; r c];
while M(r,c)~=0
[~,f]=max([M(r-1,c),M(r,c-1),M(r-1,c-1)]);
switch f
case 1
r=r-1;
case 2
c=c-1;
case 3
r=r-1;
c=c-1;
end
result=[result; r c];
end
Then, result is a 2 x n matrix, where each row is the indexes of one step- from the upper to lower.
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I have a vector, size: normal(494020);
normal={ 'back.' 'buffer_overflow.' 'ftp_write.' 'guess_passwd.' 'imap.'};
The strings in this vector are randomly distributed. I want to know what is the index of 'back.' , and how many indices it covers. Same way for other strings. Please help.
Use strcmp to compare a cell array of strings with a string:
>> normal={ 'back.' 'buffer_overflow.' 'ftp_write.' 'guess_passwd.' 'imap.'};
>> strcmp('back.',normal)
ans =
1 0 0 0 0
So what you're after is
string='back.';
index_of_string=find(strcmp(string,normal),1);
this will return the first index k for which normal{k} is the same as string. So it will discard multiplicities.
If by "how many indices it cover" you mean that you need multiplicities, then just remove the ,1 from the call to find, then you'll get an index vector containing every index k for which normal{k} is equal to string.
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I want to generate an array of random numbers that are divisible by 3 in matlab. What is the way to do this?
try this:
a = randi(100, 5) * 3;
a = a(:);
use randi(100, 5) to create a pseudo random 5x5 matrix of integers between 1 and 100. multiply by 3 to make them all divisible by 3. use the colon operator : to convert it to a 25 long array.
As suggested by #tashuhka you can avoid the : by:
a = randi(100, [25,1]) * 3;
You can read all about it here