Question 1: I have a 1x15 array, comprising of positive integers and negative integers. I wish to implement a MATLAB code which keeps all positive integers and skips the cells with negative contents.
I have tried the following:
X = [1 2 3 4 5 -10 1 -5 4 6 8 9 2 4 -2];
[r c] = size(X);
for i=1:r
for j=1:c
if X(i,j)<0
X(i,j)=X(i,j+1)
end
end
end
The output should be:
X_new = [1 2 3 4 5 1 4 6 8 9 2 4]
How do I do this?
Question 2:
X = [1 2 3 4 5 -10 1 -5 4 6 8 9 2 4 -2]
Y = [5 3 8 9 4 5 6 7 4 7 9 5 2 1 4]
From Question 1,
X_new = [1 2 3 4 5 1 4 6 8 9 2 4]
I need to delete the corresponding values in Y so that:
Y_new = [5 3 8 9 4 6 4 7 9 5 2 1]
How do I perform this?
In MATLAB, manipulating arrays and matrices can be done much easier than for-loop solutions,
in your task, can do find and delete negative value in the array, simply, as follows:
Idx_neg = X < 0; % finding X indices corresponding to negative elements
X ( Idx_neg ) = []; % removing elements using [] operator
Y ( Idx_neg ) = []; % removing corresponding elements in Y array
Related
I am trying to generate random numbers between 1 and 6 using Matlab's randperm and calling randperm = 6.
Each time this gives me a different array let's say for example:
x = randperm(6)
x = [3 2 4 1 5 6]
I was wondering if it was possible to create pairs of random numbers such that you end up with x like:
x = [3 4 1 2 5 6]
I need the vector to be arranged such that 1 and 2 are always next to each other, 3 and 4 next to each other and 5 and 6 next to each other. As I'm doing something in Psychtoolbox and this order is important.
Is it possible to have "blocks" of random order? I can't figure out how to do it.
Thanks
x=1:block:t ; %Numbers
req = bsxfun(#plus, x(randperm(t/block)),(0:block-1).'); %generating random blocks of #
%or req=x(randperm(t/block))+(0:block-1).' ; if you have MATLAB R2016b or later
req=req(:); %reshape
where,
t = total numbers
block = numbers in one block
%Sample run with t=12 and block=3
>> req.'
ans =
10 11 12 4 5 6 1 2 3 7 8 9
Edit:
If you also want the numbers within each block in random order, add the following 3 lines before the last line of above code:
[~, idx] = sort(rand(block,t/block)); %generating indices for shuffling
idx=bsxfun(#plus,idx,0:block:(t/block-1)*block); %shuffled linear indices
req=req(idx); %shuffled matrix
%Sample run with t=12 and block=3
req.'
ans =
9 8 7 2 3 1 12 10 11 5 6 4
I can see a simple 3 step process to get your desired output:
Produce 2*randperm(3)
Double up the values
Add randperm(2)-2 (randomly ordered pair of (-1,0)) to each pair.
In code:
x = randperm(3)
y = 2*x([1 1 2 2 3 3])
z = y + ([randperm(2),randperm(2),randperm(2)]-2)
with result
x = 3 1 2
y = 6 6 2 2 4 4
z = 6 5 2 1 3 4
I have a vector and I would like to extract all the 4's from it:
x = [1 1 1 4 4 5 5 4 6 1 2 4 4 4 9 8 4 4 4 4]
so that I will get 4 vectors or a cell containing the 4 blocks of 4's:
[4 4], [4], [4 4 4], [4 4 4 4]
Thanks!
You can create cells from the appropriate ranges using arrayfun:
x = [1 1 1 4 4 5 5 4 6 1 2 4 4 4 9 8 4 4 4 4];
x = [0, x, 0]; D = diff (x==4); % pad array and diff its mask
A = find (D == 1); B = find (D == -1); % find inflection points
out = arrayfun (# (a,b) {x(a+1 : b)}, A, B) % collect ranges in cells
This should be pretty fast, using accumarray:
X = 4;
%// data
x = [1 1 1 4 4 5 5 4 6 1 2 4 4 4 9 8 4 4 4 4]
%// mask all 4
mask = x(:) == X
%// get subs for accumarray
subs = cumsum( diff( [0; mask] ) > 0 )
%// filter data and sort into cell array
out = accumarray( subs(mask), x(mask), [], #(y) {y} )
idx=find(x==4);
for (i= 1:length(idx))
if (i==1 || idx(i-1)!=idx(i)-1)if(i!=1) printf(",") endif; printf("[") endif;
printf("4");
if (i<length(idx)&&idx(i+1)==idx(i)+1) printf(",") else printf("]") endif
endfor
Note this won't give the actual vectors, but it will give the output you wanted. The above is OCtave code. I am pretty sure changing endfor and endif to end would work in MAtlab, but without testing in matlab, I am not positive.[edited in light of comment]
with regionprops we can set property PixelValues so the function returns 1s instead of 4s
x = [1 1 1 4 4 5 5 4 6 1 2 4 4 4 9 8 4 4 4 4]
{regionprops(x==4,'PixelValues').PixelValues}
if we set property PixelIdxList the function returns a cell of indices of 4s:
{regionprops(x==4,'PixelIdxList').PixelIdxList}
Update(without image processing toolbox):
this way we can get number of elements of each connected components:
c = cumsum(x~=4)
h=hist(,c(1):c(end));
h(1)=h(1)+~c(1);
result = h(h~=1)-1
Suppose I have the following array:
x = [a b
c d
e f
g h
i j];
I want to "swipe a window of two rows" progressively (one row at a time) along the array to generate the following array:
y = [a b c d e f g h
c d e f g h i j];
What is the most efficient way to do this? I don't want to use cellfun or arrayfun or for loops.
im2col is going to be your best bet here if you have the Image Processing Toolbox.
x = [1 2
3 4
5 6
7 8];
im2col(x.', [1 2])
% 1 2 3 4 5 6
% 3 4 5 6 7 8
If you don't have the Image Processing Toolbox, you can also easily do this with built-ins.
reshape(permute(cat(3, x(1:end-1,:), x(2:end,:)), [3 2 1]), 2, [])
% 1 2 3 4 5 6
% 3 4 5 6 7 8
This combines the all rows with the next row by concatenating a row-shifted version along the third dimension. Then we use permute to shift the dimensions around and then we reshape it to be the desired size.
If you don't have the Image Processing Toolbox, you can do this using simple indexing:
x =
1 2
3 4
5 6
7 8
9 10
y = x.'; %% Transpose it, for simplicity
z = [y(1:end-2); y(3:end)] %% Take elements 1:end-2 and 3:end and concatenate them
z =
1 2 3 4 5 6 7 8
3 4 5 6 7 8 9 10
You can do the transposing and reshaping in a simple step (see Suever's edit), but the above might be easier to read, understand and debug for beginners.
Here's an approach to solve it for a generic case of selecting L rows per window -
[m,n] = size(x) % Store size
% Extend rows by indexing into them with a progressive array of indices
x_ext = x(bsxfun(#plus,(1:L)',0:m-L),:);
% Split the first dim at L into two dims, out of which "push" back the
% second dim thus created as the last dim. This would bring in the columns
% as the second dimension. Then, using linear indexing reshape into the
% desired shape of L rows for output.
out = reshape(permute(reshape(x_ext,L,[],n),[1,3,2]),L,[])
Sample run -
x = % Input array
9 1
3 1
7 5
7 8
4 9
6 2
L = % Window length
3
out =
9 1 3 1 7 5 7 8
3 1 7 5 7 8 4 9
7 5 7 8 4 9 6 2
This question already has answers here:
Generate a matrix containing all combinations of elements taken from n vectors
(4 answers)
Closed 8 years ago.
I'm trying to build all possible arrays of length n of a vector of n elements with at least 2 integers in each position. I should be getting 2^n combinations, 16 in this case. My code is generating only half of them, and not saving the output to an array
allinputs = {[1 2] [2 3] [3 4] [5 6]}
A = []
the command I run is
inputArray = inputBuilder(A,[],allinputs,1)
for the function
function inputArray = inputBuilder(A,currBuild, allInputs, currIdx)
if currIdx <= length(allInputs)
for i = 1:length(allInputs{currIdx})
mybuild = [currBuild allInputs{currIdx}(i)];
inputBuilder(A,mybuild,allInputs,currIdx + 1);
end
if currIdx == length(allInputs)
A = [A mybuild];
%debug output
mybuild
end
if currIdx == 1
inputArray = A;
end
end
end
I want all 16 arrays to get output in a vector. Or some easy way to access them all. How can I do this?
EDIT:
Recursion may be a requirement because allinputs will have subarrays of different lengths.
allinputs = {[1] [2 3] [3 4] [5 6 7]}
with this array it will be 1*2*2*3 or 12 possible arrays built
Not sure exactly if this is what you want, but one way of doing what I think you want to do is as follows:
allinputs = {[1 2] [2 3] [3 4] [5 6]};
comb_results = combn([1 2],4);
A = zeros(size(comb_results));
for rowi = 1:size(comb_results, 1)
indices = comb_results(rowi,:);
for idxi = 1:numel(indices)
A(rowi, idxi) = allinputs{idxi}(indices(idxi));
end
end
This gives:
A =
1 2 3 5
1 2 3 6
1 2 4 5
1 2 4 6
1 3 3 5
1 3 3 6
1 3 4 5
1 3 4 6
2 2 3 5
2 2 3 6
2 2 4 5
2 2 4 6
2 3 3 5
2 3 3 6
2 3 4 5
2 3 4 6
combn is here.
I have a variable like this that is all one row:
1 2 3 4 5 6 7 8 9 2 4 5 6 5
I want to write a for loop that will find where a number is less than the previous one and put the rest of the numbers in a new row, like this
1 2 3 4 5 6 7 8 9
2 4 5 6
5
I have tried this:
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
m = zeros(size(test));
for i=1:numel(test)-1;
for rows=1:size(m,1)
if test(i) > test(i+1);
m(i+1, rows+1) = test(i+1:end)
end % for rows
end % for
But it's clearly not right and just hangs.
Let x be your data vector. What you want can be done quite simply as follows:
ind = [find(diff(x)<0) numel(x)]; %// find ends of increasing subsequences
ind(2:end) = diff(ind); %// compute lengths of those subsequences
y = mat2cell(x, 1, ind); %// split data vector according to those lenghts
This produces the desired result in cell array y. A cell array is used so that each "row" can have a different number of columns.
Example:
x = [1 2 3 4 5 6 7 8 9 2 4 5 6 5];
gives
y{1} =
1 2 3 4 5 6 7 8 9
y{2} =
2 4 5 6
y{3} =
5
If you are looking for a numeric array output, you would need to fill the "gaps" with something and filling with zeros seem like a good option as you seem to be doing in your code as well.
So, here's a bsxfun based approach to achieve the same -
test = [1 2 3 4 5 6 7 8 9 2 4 5 6 5] %// Input
idx = [find(diff(test)<0) numel(test)] %// positions of row shifts
lens = [idx(1) diff(idx)] %// lengths of each row in the proposed output
m = zeros(max(lens),numel(lens)) %// setup output matrix
m(bsxfun(#le,[1:max(lens)]',lens)) = test; %//'# put values from input array
m = m.' %//'# Output that is a transposed version after putting the values
Output -
m =
1 2 3 4 5 6 7 8 9
2 4 5 6 0 0 0 0 0
5 0 0 0 0 0 0 0 0