I have a 10x5 double matrix mat. I also have a 1x5 row vector start_rows. In mat I would like to replace all numbers from specific rows onwards using start_rows. I could use a loop and replace all the numbers column by column. However, I'm sure there is some vectorized solution.
mat = nan(10, 5);
start_rows = [3,5,1,7,2];
% How to avoid that loop
for idx = 1 : numel(start_rows)
mat(start_rows(idx):end, idx) = 1;
end
This can be solved comparing an array of the following form to your start_rows vector:
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
6 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
10 10 10 10 10
which will return a logical array for when the condition is met (this uses broadcasting AKA implicit expansion).
If mat always contains zeros, and you're replacing with ones:
(1:size(mat,1)).'+ mat >= start_rows;
If mat is nonzero:
(1:size(mat,1)).'+ 0*mat >= start_rows; % option 1
(1:size(mat,1)).'+ zeros(size(mat)) >= start_rows; % option 2
If replacing with values other than 1 (or true):
((1:size(mat,1)).'+ 0*mat >= start_rows) * newVal;
Related
Is there any command to find mean of first 5 values then next 5 values from a total of 25 values present in a vector in MATLAB. If the dataset is X.
If anyone can help me to provide a code where I can get mean at every 5th value.
X=[4 5 6 7 2 5 7 4 2 6 7 3 2 1 5 7 8 3 4 6 8 4 2 6 8];
You can for instance reshape the vector in an array with reshape and then apply the mean function:
M = mean(reshape(X, [5, numel(X)/5]),1);
or simply
M = mean(reshape(X, 5, []),1);
But there as stated in the comments there are many other ways.
Here is one simple way to do it. Rearrange the vector into a matrix loop over the columns and take the mean of all values in each column. Store the results in a new vector.
X=[4 5 6 7 2 5 7 4 2 6 7 3 2 1 5 7 8 3 4 6 8 4 2 6 8];
Xr = reshape(X,5,5)
cols = size(Xr)(2)
avgs=zeros(1,cols)
for i= 1:cols
avgs(i) = mean(Xr(:,i))
end
So i have this data:
A=
2
4
8
9
4
6
1
3
And 3 interval
B=
1 4
5 8
9 12
How to make an output like this
Output=
1
1
2
3
1
2
1
1
The output is based on the interval
you can solve it in several ways. for example, with arrayfun:
A = [2 4 8 9 4 6 1 3].';
B = [1 4;
5 8;
9 12];
res = arrayfun(#(x) find((x >= B(:,1)) & (x <= B(:,2))),A);
If the interval always has the same length, as in your case 4, you can solve it as follows:
Output=ceil(A/4);
If it is not the case, and if not all numbers necessarily fall between any of the intervals, you can compute it as follows. A zero is outputted if a number does not fall within any of the intervals.
% example entry
A=[2 3 4 8 9 4 6 1 3]';
B=[1 4;5 7;9 12]';
Arep=A(:,ones(size(B,2),1)); % replicate array (alternatively use repmat)
Alog=Arep>=B(1,:)&Arep<=B(2,:); % conditional statements, make logical array
Output=Alog*(1:size(B,2))'; % matrix product with natural array to obtain indices
I have a vector of 13 entities in Matlab.
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]
I want to append values [1 2 3 4 5] at regular intervals at position 1 5 9 13 & 17.
The final value of a looks like this.
a=[1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2].
The values with italics show the appended values.
How can I do it?
Since you are looking for regular intervals, you can take advantage of the reshape and cat function:
a = [3 4 6 8 1 5 8 9 3 7 3 6 2];
v = [1 2 3 4 5];
l = [1 5 9 13 17];
interval = l(2)-l(1)-1; %computes the interval between inserts
amax = ceil(size(a,2)/interval) * interval; %calculating maximum size for zero padding
a(amax) = 0; %zero padding to allow `reshape`
b = reshape (a,[interval,size(v,2)]); %reshape into matrix
result = reshape(vertcat (v,b), [1,(size(b,1)+1)*size(b,2)]); %insert the values into the right position and convert back into vector
%remove padded zeros
final = result(result ~= 0) %remove the zero padding.
>>final =
Columns 1 through 16
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6
Columns 17 through 18
5 2
Here's an approach using boolean-indexing -
% Inputs
a = [3 4 6 8 1 5 8 9 3 7 3 6 2]
append_vals = [1 2 3 4 5]
append_interval = 4 % Starting at 1st index
% Find out indices of regular intervals where new elements are to be inserted.
% This should create that array [1,5,9,13,17]
N_total = numel(a) + numel(append_vals)
append_idx = find(rem(0:N_total-1,append_interval)==0)
% Get boolean array with 1s at inserting indices, 0s elsewhere
append_mask = ismember(1:N_total,append_idx)
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(~append_mask) = a
out(append_mask) = append_vals
Alternatively, we can also use linear-indexing and avoid creating append_mask, like so -
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(append_idx) = append_vals
out(setdiff(1:numel(out),append_idx)) = a
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]; % // Your original values
pos = [1 5 9 13 17]; % // The position of the values you want to insert
b=[1 2 3 4 5]; % // The values you want to insert
% // Pre-allocate a vector with the total size to hold the resulting values
r = zeros(size(a,2)+size(pos,2),1);
r(pos) = b % // Insert the appended values into the resulting vector first
r3 = r.' <1 % // Find the indices of the original values. These will be zero in the variable r but 1 in r3
ans =
0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1
ind= find(r3==1) % // Find the indices of the original values
ind =
2 3 4 6 7 8 10 11 12 14 15 16 18
r(ind) = a; % // Insert those into the resulting vector.
r.'
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
You can use this function to append a bunch of values to an existing vector, given their positions in the new vector:
function r=append_interval(a,v,p)
% a - vector with initial values
% v - vector containing values to be inserted
% p - positions for values in v
lv=numel(v); % number of elements in v vector
la=numel(a); % number of elements in a vector
column_a=iscolumn(a); % check if a is a column- or row- wise vector
tot_elements=la+lv;
% size of r is tha max between the total number of elements in the two vectors and the higher positin in vector p (in this case missing positions in a are filled with zeros)
lr=max([max(p) tot_elements]);
% initialize r as nan vector
r=zeros(column_a*(lr-1)+1,~column_a*(lr-1)+1)/0;
% set elements in p position to the corresponding values in v
r(p)=v;
% copy values in a in the remaining positions and fill with zeros missing entries (if any)
tot_missing_values=lr-tot_elements;
if(tot_missing_values)
remaining_values=cat(2-iscolumn(a),a,zeros(column_a*(tot_missing_values-1)+1,~column_a*(tot_missing_values-1)+1));
else
remaining_values=a;
end
% insert values
r(isnan(r))=remaining_values;
You can use row-wise or column-wise vectors; the orientation of r will be the same of that of a.
Input:
a =
3 4 6 8 1 5 8 9 3 7 3 6 2
v =
1 2 3 4 5
p =
1 5 9 13 17
Output:
>> append_interval(a,v,p)
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
Every sequence of positive positions is allowed and the function will pad for you with zeros the final vector, in case you indicate a position exceding the sum of the original vector and added items.
For example, if:
v3 =
1 2 3 4 5 6 90
p3 =
1 5 9 13 17 30 33
you get:
append_interval(a,v3,p3)
ans =
Columns 1 through 19
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2 0
Columns 20 through 33
0 0 0 0 0 0 0 0 0 0 6 0 0 90
Hope this will help.
I have two matrix A and B. Suppose I would like to find in each row of matrix A the smallest number, and for the same cell that this number is in Matrix A, do find the corresponding number of the same cell in matrix B. For example the number in matrix A will be in the position A(1,3), A(2,9)...and I want the corresponding number in B(1,3), B(2,9)... Is it possible to do it, or I am asking something hard for matlab. Hope someone will help me.
What you can do is use min and find the minimum across all of the rows for each column. You would actually use the second output in order to find the location of each column per row that you want to find. Once you locate these, simply use sub2ind to access the corresponding values in B. As such, try something like this:
[~,ind] = min(A,[],2);
val = B(sub2ind(size(A), (1:size(A,1)).', ind));
val would contain the output values in the matrix B which correspond to the same positions as the minimum values of each row in A. This is also assuming that A and B are the same size. As an illustration, here's an example. Let's set A and B to be a random 4 x 4 array of integers each.
rng(123);
A = randi(10, 4, 4)
B = randi(10, 4, 4)
A =
7 8 5 5
3 5 4 1
3 10 4 4
6 7 8 8
B =
2 7 8 3
2 9 4 7
6 8 4 1
6 7 3 5
By running the first line of code, we get this:
[~,ind] = min(A,[],2)
ind =
3
4
1
1
This tells us that the minimum value of the first row is the third column, the minimum value of the next row is the 4th column, and so on and so forth. Once we have these column numbers, let's access what the corresponding values are in B, so we would want row and columns (1,3), (2,4), etc. Therefore, after running the second statement, we get:
val = B(sub2ind(size(A), (1:size(A,1)).', ind))
val =
8
7
6
6
If you quickly double check the accessed positions in B in comparison to A, we have found exactly those spots in B that correspond to A.
A = randi(9,[5 5]);
B = randi(9,[5 5]);
[C,I] = min(A');
B.*(A == repmat(C',1,size(A,2)))
example,
A =
2 1 6 9 1
2 4 4 4 2
5 6 5 5 5
9 3 9 3 6
4 5 6 8 3
B =
3 5 6 8 1
9 2 9 7 1
5 6 6 5 6
4 6 1 4 5
5 3 7 1 9
ans =
0 5 0 0 1
9 0 0 0 1
5 0 6 5 6
0 6 0 4 0
0 0 0 0 9
You can use it like,
B(A == repmat(C',1,5))
ans =
9
5
5
6
6
5
4
1
1
6
9
How to generate random matrix without repetition in rows and cols with specific range
example (3x3): range 1 to 3
2 1 3
3 2 1
1 3 2
example (4x4): range 1 to 4
4 1 3 2
1 3 2 4
3 2 4 1
2 4 1 3
A way of approaching this problem is to generate a circular matrix and shuffle it.
mat_size = 4
A = gallery('circul', 1:mat_size); % circular matrix
B = A( randperm(length(A)) , randperm(length(A)) ); % shuffle rows and columns with randperm
It gives
A =
1 2 3 4
4 1 2 3
3 4 1 2
2 3 4 1
B =
3 4 1 2
2 3 4 1
4 1 2 3
1 2 3 4
This method should be fast. An 11 size problem is computed in 0.047021 seconds.
This algorithm will do the trick, assuming you want to contain all elements between 1 and n
%// Elements to be contained, but no zero allowed
a = [1 2 3 4];
%// all possible permutations and its size
n = numel(a);
%// initialization
output = zeros(1,n);
ii = 1;
while ii <= n;
%// random permuation of input vector
b = a(randperm(n));
%// concatenate with already found values
temp = [output; b];
%// check if the row chosen in this iteration already exists
if ~any( arrayfun(#(x) numel(unique(temp(:,x))) < ii+1, 1:n) )
%// if not, append
output = temp;
%// increase counter
ii = ii+1;
end
end
output = output(2:end,:) %// delete first row with zeros
It definitely won't be the fastest implementation. I would be curios to see others.
The computation time increases exponentially. But everything up to 7x7 is bearable.
I wrote another code (interesting to compare timings and, if possible, to make it parallel). Also had problem with perms (needed to restart Matlab to be able to generate for 11 elements, I have x64 and 16GB of memory). Than I decided to keep characters instead of the numbers, reducing the memory occupied by the matrix. It, of course, generates all permutations, and I shuffle them in the beginning, selecting in the loop in a new random order. It runs faster this way and 'eats' less memory. Time for 11 x 11 (of course it differs from run to run) is shown in results.
clear all;
t = cputime;
sze = 11;
variations = perms(char(1 : sze)); % permutations
varN = length(variations);
variations = variations(randperm(varN)', :); % shuffle
sudoku = zeros(sze, sze);
sudoku(1, :) = variations(1, :); % set the first row
indx = 2;
for ii = 2 : varN
% take a random index
rowVal = variations(ii, :);
% check that row numbers do not present in table at
% corresponding columns
if (~isempty(find(repmat(rowVal, sze, 1) - sudoku == 0, 1)))
continue;
end;
sudoku(indx, :) = rowVal;
disp(['Found row ' num2str(indx)]);
indx = indx + 1;
if indx > sze, break; end;
end;
disp(cputime - t);
disp(sudoku);
Result
252.9712 seconds
7 11 3 9 6 2 4 1 8 10 5
1 9 6 3 10 7 11 5 2 4 8
9 6 11 8 2 10 1 7 4 5 3
4 10 7 11 1 8 5 2 6 3 9
2 5 9 1 3 6 8 4 10 7 11
10 3 5 6 7 4 2 9 11 8 1
6 4 2 10 8 5 3 11 9 1 7
3 8 10 4 11 1 7 6 5 9 2
11 1 8 5 4 9 6 3 7 2 10
5 2 4 7 9 3 10 8 1 11 6
8 7 1 2 5 11 9 10 3 6 4
Here's a memory-efficient approach. The time it takes is random, but not very large. All possible output matrices are equally likely.
This works by randomly filling the matrix until no more positions are available or until the whole matrix has been filled. The code is commented so it should be obvious how it works.
For size 11 this takes of the order of a few thousands or tens of thousands attempts. On my old laptop that means a (random) running time from a few seconds to tens of seconds.
It could perhaps be sped up using uint8 values instead of double. I don't think that brings a large gain, though.
The code:
clear all
n = 11; %// matrix size
[ ii jj ] = ndgrid(1:n); %// rows and columns of S
ii = ii(:);
jj = jj(:);
success = 0; %// ...for now
attempt = 0; %// attempt count (not really needed)
while ~success
attempt = attempt + 1;
S = NaN(n, n); %// initiallize result. NaN means position not filled yet
t = 1; %// number t is being placed within S ...
u = 1; %// ... for the u-th time
mask = true(1, numel(ii)); %// initiallize mask of available positions
while any(mask) %// while there are available positions
available = find(mask); %// find available positions
r = randi(numel(available), 1); %// pick one available position
itu = ii(available(r)); %// row of t, u-th time
jtu = jj(available(r)); %// col of t, u-th time
S(itu, jtu) = t; %// store t at that position
remove = (ii==itu) | (jj==jtu);
mask(remove) = false; %// update mask of positions available for t
u = u+1; %// next u
if u > n %// we are done with number t
t = t+1; %// let's go with new t
u = 1; %// initiallize u
mask = isnan(S(:)); %// initiallize mask for this t
end
if t > n %// we are done with all numbers
success = 1; %// exit outer loop (inner will be exited too)
end
end
end
disp(attempt) %// display number of attempts
disp(S) %// show result
An example result:
10 11 8 9 7 2 3 4 1 6 5
8 4 2 1 10 11 6 5 7 9 3
2 3 5 6 11 8 1 10 4 7 9
9 8 7 4 6 10 11 3 5 1 2
3 5 9 8 2 1 4 7 6 11 10
11 9 4 5 3 6 2 1 8 10 7
1 2 6 3 8 7 5 9 10 4 11
7 1 11 10 5 4 9 8 2 3 6
4 7 1 2 9 3 10 6 11 5 8
6 10 3 11 1 5 7 2 9 8 4
5 6 10 7 4 9 8 11 3 2 1