I want to add a vector to just a single column of a matrix.
For example:
a = zeros(5,5);
b = ones(5,1);
I want to add such b only to the second column of a such that the resultant a is
a= [ 0 1 0 0 0;
0 1 0 0 0;
0 1 0 0 0;
0 1 0 0 0;
0 1 0 0 0;]
How can I do this? I have tried doing a+b but it adds one to all the columns.
a(:,2) = a(:,2)+b does this. Specifically, you index all rows, :, of the second column, 2, of a, and add the vector b to that. Read this post for details on various indexing methods.
rahnema1 mentioned that Python-like syntax of adding to or subtracting from an argument does not require that argument to be repeated. You can thus do:
a:(,2) += b
Related
My program allows to multiply a given B matrix with a z factor with some characteristics listed below, to give an H matrix. I would like to have a programming idea to do the inverse of what I programmed. That is to say with a given H matrix find the value of the B matrix.
For example with a matrix B = [-1 -1 ; 1 0]
I get with my code a matrix :
H = [ 0 0 0 0 0 0 ;
0 0 0 0 0 0 ;
0 0 0 0 0 0 ;
0 1 0 1 0 0 ;
0 0 1 0 1 0 ;
1 0 0 0 0 1 ]
I would like to have from a code H the value of the matrix B.
To specify the matrix H from the matrix B, it is necessary that:
Each coefficient -1 is replaced by a null matrix of dimension z*z;
Each coefficient 0 is replaced by an identity matrix of dimension z*z;
Each coefficient 1,2,...,z-1 is replaced by a circulating permutation matrix of dimension z*z shifted by 1,2,...,z-1 position to the right.
From a matrix B and the expansion factors z , we construct an extended binary H matrix with n-k rows and n columns.
My code :
clear;
close all;
B = [-1 -1 ; 1 0];
z = 3;
H = zeros(size(B)*z);
Y = eye(z);
for X1 = 1:size(B,1)
for X2 = 1:size(B,2)
X3 = B(X1,X2);
if (X3 > -1)
X4 = circshift(Y,[0 X3]);
else
X4 = zeros(z);
end
Y1 = (X1-1)*z+1:X1*z;
Y2 = (X2-1)*z+1:X2*z;
H(Y1,Y2) = X4;
end
end
[M,N] = size(H);
Any suggestions?
Assuming that all of the input matrices are well-formed, you can determine the mapping based on the first row of each block. For example, the block mapping to 1:
0 1 0
0 0 1
1 0 0
has a 1 in column 2 of row 1. Similarly, a one in column 1 maps to 0, and column 3 maps to 2. No ones in the row maps to -1. So we just need to find the column containing the 1 in the first row.
Annoyingly, find returns null when it doesn't find a nonzero value rather than 0 (which is what we would want in this case). We can adjust to this by adding a value to the matrix row that is only 1 when all of the others are 0.
If you have the Image Processing Toolbox, you can use blockproc to handle the looping for you:
B = blockproc(H, [z z], #(A)find([~any(A.data(1,:)) A.data(1,:)])-2);
Otherwise, just loop over the blocks and apply the function to each one.
I've got logical array(zeros and ones) 1500x700
I want to find "1" in every column and when there are more than one "1" in column i should choose the middle one.
Is that possible to do it? I know how to find "1", but don't know how to extract the middle "1" if there's couple of "1" in one column.
The find function returns the indices of your ones.
>> example=[1,0,0,1,0,1,1];
>> indices=find(example)
indices =
1 4 6 7
>> indices(floor(numel(indices)/2))
ans =
4
Do this for each column and you have a solution.
You can
Get the row and column indices of ones with find;
Apply accumarray with a custom function to get the middle row index for each column.
x = [1 0 0 0 0; 0 0 1 0 0; 1 0 1 0 0; 1 0 0 1 0]; % example
[ii, jj] = find(x); % step 1
result = accumarray(jj, ii, [size(x,2) 1], #(x) x(ceil(end/2)), NaN); % step 2
Note that:
For an even number of ones this gives the first of the two middle indices. If you prefer the average of the two middle indices replace #(x) x(ceil(end/2)) by #median.
For a column without ones this gives NaN as result. If you prefer a different value, replace the input fifth argument of accumarray by that.
Example:
x =
1 0 0 0 0
0 0 1 0 0
1 0 1 0 0
1 0 0 1 0
result =
3
NaN
2
4
NaN
I'm trying to remove the rows which has duplicates in sequence. I have only 2 possible values which are 0 and 1. I have nXm which n shows possible number of bits and m is not important for my question. My goal is to find an matrix which is nX(m-a). The rows a which has the property which includes duplicates in sequence. For example:
My matrix is :
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0]
I want to remove the rows has t duplicates in sequence for 0. In this question let's assume t is 3. So I want the matrix which:
B=[0 1 0 1 0 1;
0 0 1 0 0 1;
0 1 0 0 1 0]
2nd and 5th rows are removed.
I probably need to use diff.
So you want to remove rows of A that contain at least t zeros in sequence.
How about a single line?
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==-t, 2),:);
How this works:
Transform A to bipolar form (2*A-1)
Convolve each row with a sequence of t ones (conv2(...))
Keep only rows for which the convolution does not contain -t (~any(...)). The presence of -t indicates a sequence of t zeros in the corresponding row of A.
To remove rows that contain at least t ones, just change -t to t:
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==t, 2),:);
Here is a generalized approach which removes any rows which has given number of consecutive duplicates (not just zero. could be any number).
t = 3;
row_mask = ~any(all(~diff(reshape(im2col(A,[1 t],'sliding'),t,size(A,1),[]))),3);
out = A(row_mask,:)
Sample Run:
>> A
A =
0 1 0 1 0 1
0 0 1 5 5 5 %// consecutive 3 5's
0 0 1 0 0 1
0 1 0 0 1 0
1 1 1 0 0 1 %// consecutive 3 1's
>> out
out =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
How about an approach using strings? This is certainly not as fast as Luis Mendo's method where you work directly with the numerical array, but it's thinking a bit outside of the box. The basis of this approach is that I consider each row of A to be a unique string, and I can search each string for occurrences of a string of 0s by regular expressions.
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0];
t = 3;
B = sprintfc('%s', char('0' + A));
ind = cellfun('isempty', regexp(B, repmat('0', [1 t])));
B(~ind) = [];
B = double(char(B) - '0');
We get:
B =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
Explanation
Line 1: Convert each line of the matrix A into a string consisting of 0s and 1s. Each line becomes a cell in a cell array. This uses the undocumented function sprintfc to facilitate this cell array conversion.
Line 2: I use regular expressions to find any occurrences of a string of 0s that is t long. I first use repmat to create a search string that is full of 0s and is t long. After, I determine if each line in this cell array contains this sequence of characters (i.e. 000....). The function regexp helps us perform regular expressions and returns the locations of any matches for each cell in the cell array. Alternatively, you can use the function strfind for more recent versions of MATLAB to speed up the computation, but I chose regexp so that the solution is compatible with most MATLAB distributions out there.
Continuing on, the output of regexp/strfind is a cell array of elements where each cell reports the locations of where we found the particular string. If we have a match, there should be at least one location that is reported at the output, so I check to see if any matches are empty, meaning that these are the rows we don't want to remove. I want to turn this into a logical array for the purposes of removing rows from A, and so this is wrapped with a cellfun call to determine the cells that are empty. Therefore, this line returns a logical array where a 0 means that remove this row and a 1 means that we don't.
Line 3: I take the logical array from Line 2 and invert it because that's what we really want. We use this inverted array to index into the cell array and remove those strings.
Line 4: The output is still a cell array, so I convert it back into a character array, and finally back into a numerical array.
I am trying to write a svm code, but i am literally a beginner in matlab.
So in my code, in a for loop, i should store predictions. The data is like this:
testIdx = [1 1 1 0 0 0 0 0 1 0 1 1]'; % i wrote it like this but it says logical
and
pred = [1 1 1 0 1 0]'; % again logical
So i want to form a 12 length array and turn its 1st,2nd,3rd,9th,11th,12th elements into 1 1 1 0 1 0, and likewise rest of test elements into another set of 0/1s in other iteration.
If possible let it be a normal array, not logical. Thanks in advance
I did it myself old style but there must be a shorter direct way right?
Y = zeros ( size(testIdx,1), 1) ;
a=1;
for i = 1:size(testIdx,1)
if testIdx(i) ==1
Y(i) = pred(a);
a=a+1;
end
end
If you create testIdx and pred the way you specified, then they are double and not logical type. To use logical indexing, it is easiest if testIdx is converted to the logical type. Then you can simply use
testIdx = [1 1 1 0 0 0 0 0 1 0 1 1]';
pred = [1 1 1 0 1 0]';
Y = zeros(size(testIdx));
Y(logical(testIdx)) = pred;
With Y(logical(testIdx), you select all indexes which are set to 1 in the testIdx vector and then write predto these indexes.
I have a Vector 1xm, ShortMemory (SM), and a Matrix nxm, Agenda (AG).
SM only have positive integers and zeros.
Each column of AG only has one element equal to 1 and all the other elements of the same column equal to 0.
My objective is changing the position of the number 1 from a randomly choosen column from AG. The problem is that only columns that have a corresponding 0 in SM can be changed.
Example:
SM = [1 0 2];
AG = [1 1 0 ; 0 0 1 ; 0 0 0];
Randomly Generated number here
RandomColumn = 2;
The possible outcomes would be
AG = [1 0 0 ; 0 1 1 ; 0 0 0]; or AG = [1 0 0; 0 0 1 ; 0 1 0]; or AG = [1 1 0 ; 0 0 1 ; 0 0 0];
The Line that gets the 1 is also random but that's easy to do
I could do it by just getting random numbers between 1 and m but m can be very big in my problem and the number of zeros can be very small too, so it could potentially take alot of time. I could also do it with a cycle but it's Matlab and this is embeded on double cycle already.
Thanks
edit: Added commentary to the code for clarity.
edit: Corrected an error on possible outcome
My solution is based on following assumptions:
Objective is changing the position of the number 1 from a randomly choosen column from AG.
Only columns that have a corresponding 0 in SM can be changed.
Solution:
% input
SM = [1 0 2]
AG = [1 1 0 ; 0 0 1 ; 0 0 0]
% generating random column according to assumptions 1 and 2
RandomColumn1 = 1:size(AG,2);
RandomColumn1(SM~=0)=[];
RandomColumn1=RandomColumn1(randperm(length(RandomColumn1)));
RandomColumn=RandomColumn1(1);
% storing the current randomly chosen column before changing
tempColumn=AG(:,RandomColumn);
% shuffling the position of 1
AG(:,RandomColumn)=AG(randperm(size(AG,1)),RandomColumn);
% following checks if the column has remained same after shuffling. This while loop should execute (extremely) rarely.
while tempColumn==AG(:,RandomColumn)
tempColumn=AG(:,RandomColumn);
AG(:,RandomColumn)=AG(randperm(size(AG,1)),RandomColumn);
end
AG