In MATLAB I have a vector with size 1-by-3. Now I need to insert an element to this vector but sometimes
this number must be first element of this vector, sometimes second and so on.
Does anyone know how I could do this?
Thanks
Your question is a bit vague, but if you mean you need to insert a new element into an existing vector, here's how it can be done:
>> insertAfter = 1; % insert element after first
>> newVec = cat(2, v(1:insertAfter), newElement, v( (insertAfter+1):end ) );
Insert element I to vector V at location N
V = [V(1:N-1) I V(N:end)]
Test
V = zeros(1,3);
I = 1;
N = 2;
V = [V(1:N-1) I V(N:end)]
V =
0 1 0 0
There are more than a few ways to do this, so you'll just have to take your pick. Here's one that I would prefer for in-place insertion of the scalar newEl at location ii of vector v:
v(ii:end+1) = [newEl v(ii:end)];
clear all
clc
v1 = [ 3 2 8 9 ] % The first vector
q=length(v1) % The length of the first vector
v2=1:q+1 % Creating a new vector with length old + 1
v2(1:q)=v1 % Changing the first part of the vector to the old (v1) vector
v1=v2 % To go back to the same name of the first vector
Related
I have several matrices <1x1000> containing integers such as:
matrix = [0,0,0,0,0,30,30,30,40,40,50,50,50,40,0,0,0,30,30,30]
I want to print (disp, and later plot) them like this: 30,40,50,40,30. Basically ignore the duplicates if they come after each other.
Another example:
matrix = [0,0,0,0,10,10,10,10,50,50,50,50,10,10,10,50,50] shall give: 10,50,10,50
Help is very much appreciated!
Use this:
[~,c]=find([NaN diff(matrix)]);
output=matrix(c);
output = output(output~=0)
and to plot the output, simply use: plot(output)
Result = 0;
% loop over all nonzero values in matrix
for Element = matrix
if Element == Result(end)
% skip if equal
continue
else
% add new value
Result(end+1) = Element;
end
end
% discard zero entries
Result = Result(Result ~= 0);
All solutions provided so far use either loops or the function find which are both inefficient.
Just use matrix indexation:
[matrix((matrix(1:end-1)-matrix(2:end))~=0), matrix(end)]
ans =
0 30 40 50 40 0 30
By the way in your example are you discarting the 0s even if they come in repeated sequences?
Lets call the output matrix um then
um(1) = matrix(1);
j = 1;
for i=2: length(matrix)
% Ignore repeating numbers
if (um(j) ~= matrix(i))
j = j + 1;
um(j) = matrix(i);
end
end
% Remove zeros
um = um(um~=0);
This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
Closed 6 years ago.
This is a basic program but since I'm new to MATLAB, I'm not able to figure out the solution.
I have a column vector "Time" in which I want to print value "1" in first 147 cells, followed by "2" in 148 to 2*147 cells and so on. For that, I have written the following script:
Trial>> c=1;
Trial>> k=0;
Trial>> for i = c:146+c
Time(i,1)=1+k;
c=i;
k=k+1;
end
I know I need to iterate the loop over "Time(i,1)=1+k;" before it executes the next statement. I tried using break but that's not supposed to work. Can anyone suggest me the solution to get the desired results?(It was quite simple in C with just the use of curly braces.)
I am sure you don't want to run c=i; in every iteration.
My code should work for you:
x = 10; % Replace 10 by the max number you need in your array.
k = 1;
for i = 1 : x * 147
Time(i, 1) = k;
if rem(i, 147) == 0
k = k + 1;
end
end
This is the prime example of a piece of code that should be vectorized can help you understand vectorization. Your code can be written like this:
n = 147;
reps = 10; %% Replace this by the maximum number you want your matrix to have
Time = reshape(bsxfun(#plus, zeros(n,1), 0:reps), 1, []);
Explanation:
Let A be a column vector (1 column, n rows), and B be a row vector (1 row, m columns.
What bsxfun(#plus, A, B) will do here is to add all elements in A with all elements in B, like this:
A(1)+B(1) A(1)+B(2) A(1)+B(3) ... A(1)+B(m)
A(2)+B(1) A(2)+B(2) ............. A(2)+B(m)
............................................
A(n)+B(1) A(n)+B(2) .............. A(n)+B(m)
Now, for the two vectors we have: zeros(n,1), and 0:reps, this will give us;
0+0 0+1 0+2 0+reps
0+0 0+1 0+2 0+reps
% n rows of this
So, what we need to do now is place each column underneath each other, so that you will have the column with zeros first, then the row with ones, ... and finally the one with reps (147 in your case).
This can be achieved by reshaping the matrix:
reshape(bsxfun(#plus, zeros(n,1), 0:reps), [], 1);
^ ^ ^ ^
| | | Number of rows in the new matrix. When [] is used, the appropriate value will be chosen by Matlab
| | Number of rows in the new matrix
| matrix to reshape
reshape command
Another approach is using kron:
kron(ones(reps+1, 1) * 0:(n-1)
For the record, a review of your code:
You should always preallocate memory for matrices that are created inside loops. In this case you know it will become a matrix of dimensions ((reps+1)*n-by-1). This means you should do Time = zeros((reps+1)*n, 1);. This will speed up your code a lot.
You shouldn't use i and j as variable names in Matlab, as they denote the imaginary unit (sqrt(-1)). You can for instance do: for ii = 1:(n*147) instead.
You don't want c=i inside the loop, when the loop is supposed to go from c to c + 146. That doesn't make much sense.
You can use repmat,
x = 10; % Sequence length (or what ever it can be called)
M = repmat(1:x,147,1); % Replicate array 1:x for 147 columns
M = M(:); % Reshape the matrix so that is becomes a column vector.
I can assume that this is a task to practice for loops, but this will work.
An alternative solution may be to do
n = 147;
reps = 10;
a = ceil( (1:(n*reps)) / n);
You first construct an array with the length you want. Then you divide, and round of upwards. 1 to 147 will then become 1.
I have an array with n dimensions, and I have a sequence along one dimension at a certain location on all other dimensions. How do I find the location of this sequence? Preferably without loops.
I use matlab. I know what dimension it should be in, but the sequence isnt necessarily there. Find and == dont work. I could make an nd find function using crosscorrelation but Im guessing this is already implemented and I just dont know what function to call.
example:
ND = rand(10,10,10,10);
V = ND(randi(10),randi(10),randi(10),:);
[I1, I2, I3] = find(ND==V);
Edit: The sequence to be found spans the entire dimension it is on, I did not mention this in my original formulation of the problem. Knedlsepp`s solution solves exactly the problem I had, but Luis' solution solves a more general problem for when the sequence doesn't necessarily span the entire dimension.
As there are multiple ways to interpret your question, I will clarify: This approach assumes a 1D sequence of size: numel(V) == size(ND, dimToSearch). So, for V = [1,2] and ND = [1,2,1,2] it is not applicable. If you want this functionality go with Luis Mendo's answer, if not this will likely be faster.
This will be a perfect opportunity to use bsxfun:
We start with some example data:
ND = rand(10,10,10,10);
V = ND(3,2,:,3);
If you don't have the vector V given in the correct dimension (in this case [1,1,10,1]) you can reshape it in the following way:
dimToSearch = 3;
Vdims = ones(1, ndims(ND));
Vdims(dimToSearch) = numel(V);
V = reshape(V, Vdims);
Now we generate a cell that will hold the indices of the matches:
I = cell(1, ndims(ND));
At this point we compute the size of ND if it were collapsed along the dimension dimToSearch (we compute dimToSearch according to V, as at this point it will have the correct dimensions):
dimToSearch = find(size(V)>1);
collapsedDims = size(ND);
collapsedDims(dimToSearch) = 1;
Finally the part where we actually look for the pattern:
[I{:}] = ind2sub(collapsedDims, find(all(bsxfun(#eq, ND, V), dimToSearch)));
This is done in the following way: bsxfun(#eq, ND, V) will implicitly repmat the array V so it has the same dimensions as ND and do an equality comparison. After this we do a check with all to see if all the entries in the dimension dimToSearch are equal. The calls to find and ind2sub will then generate the correct indices to your data.
Let d be the dimension along which to search. I'm assuming that the sought sequence V may be shorter than size(ND,d). So the sequence may appear once, more than once, or never along each dimension-d- "thread".
The following code uses num2cell to reshape ND into a cell array such that each dimension-d-thread is in a different cell. Then strfind is applied to each cell to determine matches with V, and the result is a cell array with the same dimensions as ND, but where the dimension d is a singleton. The contents of each cell tell the d-dimension-positions of the matches, if any.
Credit goes to #knedlsepp for his suggestion to use num2cell, which greatly simplified the code.
ND = cat(3, [1 2 1 2; 3 4 5 6],[2 1 0 5; 0 0 1 2] ); %// example. 2x4x2
V = 1:2; %// sought pattern. It doesn't matter if it's a row, or a column, or...
d = 2; %// dimension along which to search for pattern V
result = cellfun(#(x) strfind(x(:).', V(:).'), num2cell(ND,d), 'UniformOutput', 0);
This gives
ND(:,:,1) =
1 2 1 2
3 4 5 6
ND(:,:,2) =
2 1 0 5
0 0 1 2
V =
1 2
result{1,1,1} =
1 3 %// V appears twice (at cols 1 and 3) in 1st row, 1st slice
result{2,1,1} =
[] %// V doesn't appear in 2nd row, 1st slice
result{1,1,2} =
[] %// V appears appear in 1st row, 2nd slice
result{2,1,2} =
3 %// V appears once (at col 3) in 2nd row, 2nd slice
One not very optimal way of doing it:
dims = size(ND);
Vrep = repmat(V, [dims(1), dims(2), dims(3), 1]);
ND_V_dist = sqrt(sum(abs(ND.^2-Vrep.^2), 4));
iI = find(ND_V_dist==0);
[I1, I2, I3] = ind2sub([dims(1), dims(2), dims(3)], iI);
I have a matrix with constant consecutive values randomly distributed throughout the matrix. I want the indices of the consecutive values, and further, I want a matrix of the same size as the original matrix, where the number of consecutive values are stored in the indices of the consecutive values. For Example
original_matrix = [1 1 1;2 2 3; 1 2 3];
output_matrix = [3 3 3;2 2 0;0 0 0];
I have struggled mightily to find a solution to this problem. It has relevance for meteorological data quality control. For example, if I have a matrix of temperature data from a number of sensors, and I want to know what days had constant consecutive values, and how many days were constant, so I can then flag the data as possibly faulty.
temperature matrix is number of days x number of stations and I want an output matrix that is also number of days x number of stations, where the consecutive values are flagged as described above.
If you have a solution to that, please provide! Thank you.
For this kind of problems, I made my own utility function runlength:
function RL = runlength(M)
% calculates length of runs of consecutive equal items along columns of M
% work along columns, so that you can use linear indexing
% find locations where items change along column
jumps = diff(M) ~= 0;
% add implicit jumps at start and end
ncol = size(jumps, 2);
jumps = [true(1, ncol); jumps; true(1, ncol)];
% find linear indices of starts and stops of runs
ijump = find(jumps);
nrow = size(jumps, 1);
istart = ijump(rem(ijump, nrow) ~= 0); % remove fake starts in last row
istop = ijump(rem(ijump, nrow) ~= 1); % remove fake stops in first row
rl = istop - istart;
assert(sum(rl) == numel(M))
% make matrix of 'derivative' of runlength
% don't need last row, but needs same size as jumps for indices to be valid
dRL = zeros(size(jumps));
dRL(istart) = rl;
dRL(istop) = dRL(istop) - rl;
% remove last row and 'integrate' to get runlength
RL = cumsum(dRL(1:end-1,:));
It only works along columns since it uses linear indexing. Since you want do something similar along rows, you need to transpose back and forth, so you could use it for your case like so:
>> original = [1 1 1;2 2 3; 1 2 3];
>> original = original.'; % transpose, since runlength works along columns
>> output = runlength(original);
>> output = output.'; % transpose back
>> output(output == 1) = 0; % see hitzg's comment
>> output
output =
3 3 3
2 2 0
0 0 0
I have a non-fixed dimensional matrix M, from which I want to access a single element.
The element's indices are contained in a vector J.
So for example:
M = rand(6,4,8,2);
J = [5 2 7 1];
output = M(5,2,7,1)
This time M has 4 dimensions, but this is not known in advance. This is dependent on the setup of the algorithm I'm writing. It could likewise be that
M = rand(6,4);
J = [3 1];
output = M(3,1)
so I can't simply use
output=M(J(1),J(2))
I was thinking of using sub2ind, but this also needs its variables comma separated..
#gnovice
this works, but I intend to use this kind of element extraction from the matrix M quite a lot. So if I have to create a temporary variable cellJ every time I access M, wouldn't this tremendously slow down the computation??
I could also write a separate function
function x= getM(M,J)
x=M(J(1),J(2));
% M doesn't change in this function, so no mem copy needed = passed by reference
end
and adapt this for different configurations of the algorithm. This is of course a speed vs flexibility consideration which I hadn't included in my question..
BUT: this is only available for getting the element, for setting there is no other way than actually using the indices (and preferably the linear index). I still think sub2ind is an option. The final result I had intended was something like:
function idx = getLinearIdx(J, size_M)
idx = ...
end
RESULTS:
function lin_idx = Lidx_ml( J, M )%#eml
%LIDX_ML converts an array of indices J for a multidimensional array M to
%linear indices, directly useable on M
%
% INPUT
% J NxP matrix containing P sets of N indices
% M A example matrix, with same size as on which the indices in J
% will be applicable.
%
% OUTPUT
% lin_idx Px1 array of linear indices
%
% method 1
%lin_idx = zeros(size(J,2),1);
%for ii = 1:size(J,2)
% cellJ = num2cell(J(:,ii));
% lin_idx(ii) = sub2ind(size(M),cellJ{:});
%end
% method 2
sizeM = size(M);
J(2:end,:) = J(2:end,:)-1;
lin_idx = cumprod([1 sizeM(1:end-1)])*J;
end
method 2 is 20 (small number of index sets (=P) to convert) to 80 (large number of index sets (=P)) times faster than method 1. easy choice
For the general case where J can be any length (which I assume always matches the number of dimensions in M), there are a couple options you have:
You can place each entry of J in a cell of a cell array using the num2cell function, then create a comma-separated list from this cell array using the colon operator:
cellJ = num2cell(J);
output = M(cellJ{:});
You can sidestep the sub2ind function and compute the linear index yourself with a little bit of math:
sizeM = size(M);
index = cumprod([1 sizeM(1:end-1)]) * (J(:) - [0; ones(numel(J)-1, 1)]);
output = M(index);
Here is a version of gnovices option 2) which allows to process a whole matrix of subscripts, where each row contains one subscript. E.g for 3 subscripts:
J = [5 2 7 1
1 5 2 7
4 3 9 2];
sizeM = size(M);
idx = cumprod([1 sizeX(1:end-1)])*(J - [zeros(size(J,1),1) ones(size(J,1),size(J,2)-1)]).';