I have a such index vector which maps the cells.
For example (1 -> 310),(2 -> 910),....(1024 -> 67)
It maps the 8*8 cell(eg:1st cell) of image to 8*8 cell(eg:310th cell).
I have 1024 8*8 cells and I want to reverse this process to get their first indices.
I mean, after this mapping It reorder to the basic order.
(eg: 310th cell comes to 1th and 910th cell comes to 2).
Anyone can help?
yeah sure ,I write my program, for example I have these image blocks for an image.
% dividing of main image to 8*8 blocks:
U = floor(M1/8);
V = floor(M2/8);
I_div = cell(U,V);
K = 1; L = 1;
for u = 1:U
for v = 1:V
I_div{u,v} = I2(K:K+7,L:L+7);
I_div_DCT{u,v} = dct2(I_div{u,v}); %%% the main blocks
L = L + 8;
end
K = K + 8;
L = 1;
end
a=reshape(I_div_DCT',[],1)';
b=cell(1,1024);
for i=1:1024
b(i)=a(index(i));
end
%%% index is the 1*1024 array, something like this:
%%% 1 2 3 4 5 6 7 8 ....
%%% 367 910 274 308 275 429 403 276 ....
if the first image block be a(1,1)=[8*8 double]; after implementing b(i)=a(index(i)) it goes to the 367th [8*8]block and for example The 787th [8*8]block comes to the first block,(just a simple mapping).
I want to bring back this blocks into their original indeices after implementing my changes!
Suppose you have a forward map
fMap = [367, 910, 274, 308, 275, 429, 403, 276, ...
Such that index k is mapped to fMap( k ).
Now you war the "inverse" map that will map fMap(k) back to k.
Or in other words k == revMap( fMap( k ) ).
For simplicity I'll assume fMap is a permutation. then
revMap( fMap ) = 1:n
A small example:
>> fMap
fMap =
1 4 5 3 2
>> rMap( fMap ) = 1:5
rMap =
1 5 4 2 3
Related
How to generate a sequence of random pairs without repeating pairs?
The following code already generates the pairs, but does not avoid repetitions:
for k=1:8
Comb=[randi([-15,15]) ; randi([-15,15])];
T{1,k}=Comb;
end
When running I got:
T= [-3;10] [5;2] [1;-5] [10;9] [-4;-9] [-5;-9] [3;1] [-3;10]
The pair [-3,10] is repeated, which cannot happen.
PS : The entries can be positive or negative.
Is there any built in function for this? Any sugestion to solve this?
If you have the Statistics Toolbox, you can use randsample to sample 8 numbers from 1 to 31^2 (where 31 is the population size), without replacement, and then "unpack" each obtained number into the two components of a pair:
s = -15:15; % population
M = 8; % desired number of samples
N = numel(s); % population size
y = randsample(N^2, M); % sample without replacement
result = s([ceil(y/N) mod(y-1, N)+1]); % unpack pair and index into population
Example run:
result =
14 1
-5 7
13 -8
15 4
-6 -7
-6 15
2 3
9 6
You can use ind2sub:
n = 15;
m = 8;
[x y]=ind2sub([n n],randperm(n*n,m));
Two possibilities:
1.
M = nchoosek(1:15, 2);
T = datasample(M, 8, 'replace', false);
2.
T = zeros(8,2);
k = 1;
while (k <= 8)
t = randi(15, [1,2]);
b1 = (T(:,1) == t(1));
b2 = (T(:,2) == t(2));
if ~any(b1 & b2)
T(k,:) = t;
k = k + 1;
end
end
The first method is probably faster but takes up more memory and may not be practicable for very large numbers (ex: if instead of 15, the max was 50000), in which case you have to go with 2.
I have a matrix D of distances between 3 places and 4 persons
example D(2,3) = 10 means person 3 is far away from place 2 of 10 units.
D=[23 54 67 32
32 5 10 2
3 11 13 5]
another matrix A with the same number of rows (3 places) and where A(i,:) correspond to the persons that picked place i
example for place 1, persons 1 and 3 picked it
no one picked place 2
and persons 2 and 4 picked place 3
A=[1 3 0
0 0 0
2 4 0]
I want to reorder each row of A by the persons who are closest to the place it represents.
In this example, for place 1, person 1 is closer to it than person 3 based on D so nothing to do.
nothing to do for place 2
and there is a change for place 3 since person 4 is closer than 2 to place 3 D(3,2)>D(3,4)
The result should be
A=[1 3
0 0
4 2 ]
each row(place) in A can have 0 or many non zeros elements in it (persons that picked it)
Basically, I want to reorder elements in each row of A based on the rows of D (the closest to the location comes first), something like this but here A and D are not of the same size (number of columns).
[SortedD,Ind] = sort(D,2)
for r = 1:size(A,1)
A(r,:) = A(r,Ind(r,:));
end
There is another Matlab function sortrows(C,colummn_index) that can do the trick. It can sort rows based on the elements in a particular column. So if you transpose your matrix A (C = A') and extend the result by adding to the end the proper column, according to which you want to sort a required row, then you will get what you want.
To be more specific, you can do something like this:
clear all
D=[23 54 67 32;
32 5 10 2;
3 11 13 5];
A=[1 0;
3 0;
4 2 ];
% Sort elements in each row of the matrix A,
% because indices of elements in each row of the matrix D are always
% ascending.
A_sorted = sort(A,2);
% shifting all zeros in each row to the end
for i = 1:length(A_sorted(:,1))
num_zeros = sum(A_sorted(i,:)==0);
if num_zeros < length(A_sorted(i,:))
z = zeros(1,num_zeros);
A_sorted(i,:) = [A_sorted(i,num_zeros+1:length(A_sorted(i,:))) z];
end;
end;
% Prelocate in memory an associated array of the corresponding elements in
% D. The matrix Dr is just a reduced derivation from the matrix D.
Dr = zeros(length(A_sorted(:,1)),length(A_sorted(1,:)));
% Create a matrix Dr of elements in D corresponding to the matrix A_sorted.
for i = 1:length(A_sorted(:,1)) % i = 1:3
for j = 1:length(A_sorted(1,:)) % j = 1:2
if A_sorted(i,j) == 0
Dr(i,j) = 0;
else
Dr(i,j) = D(i,A_sorted(i,j));
end;
end;
end;
% We don't need the matrix A_sorted anymore
clear A_sorted
% In order to use the function SORTROWS, we need to transpose matrices
A = A';
Dr = Dr';
% The actual sorting procedure starts here.
for i = 1:length(A(1,:)) % i = 1:3
C = zeros(length(A(:,1)),2); % buffer matrix
C(:,1) = A(:,i);
C(:,2) = Dr(:,i);
C = sortrows(C,2);
A(:,i) = C(:,1);
% shifting all zeros in each column to the end
num_zeros = sum(A(:,i)==0);
if num_zeros < length(A(:,i))
z = zeros(1,num_zeros);
A(:,i) = [A(num_zeros+1:length(A(:,i)),i) z]';
end;
end;
% Transpose the matrix A back
A = A';
clear C Dr z
I am given the following matrices A of size 3x1 and B of size 5x1
A = B=
1 A
2 B
3 C
D
E
I want to convert matrix C in a 15x2 matrix
C =
1 A
1 B
1 C
1 D
1 E
2 A
.
.
.
3 E
How can I make it?
Can be done with repmat
D = repmat(A',size(B,1),1);
C = [D(:),repmat(B,size(A,1),1)]
Here's a different alternative based on code for generating truth tables from Generate All Possible combinations of a Matrix in Matlab
ind = dec2base(0:power(5,2)-1,5)-47;
C = [A(ind(1:15,1) + 48, B(ind(1:15,2)];
And if you want to generalize it
m = max(size(A,1),size(B,1));
n = size(A,1)*size(B,1);
col = 2;
ind = dec2base(0:power(n,col)-1,n)-47;
ind = ind(1:n,:);
C = [A(ind(:,1) + 48, B(ind(:,2)];
The + 48 is just to convert your A matrix from a numerical matrix to a char matrix so that C can hold both number and letters. You can leave it out if A was already a char matrix.
What's useful about this technique is that by changing col, this generalizes to combing more than just 2 vectors in a similar fashion
I had a question in Matlab. It is so, I try to take average of the different number of values in a column. For example, if we have the column below,
X = [1 1 2 3 4 3 8 2 1 3 5 6 7 7 5]
first I want to start by taking the average of 5 values and plot them. In the case above, I should receive three averages that I could plot. Then take 10 values at a time and so on.
I wonder if you have to write custom code to fix it.
The fastest way is probably to rearrange your initial vector X into some matrix, with each column storing the required values to average:
A = reshape(X, N, []);
where N is the desired number of rows in the new matrix, and the empty brackets ([]) tell MATLAB to calculate the number of columns automatically. Then you can average each column using mean:
X_avg = mean(A);
Vector X_avg stores the result. This can be done in one line like so:
X_avg = mean(reshape(X, N, []));
Note that the number of elements in X has to be divisible by N, otherwise you'll have to either pad it first (e.g with zeroes), or handle the "leftover" tail elements separately:
tail = mod(numel(X), N);
X_avg = mean(reshape(X(1:numel(X) - tail), N, [])); %// Compute average values
X_avg(end + 1) = mean(X(end - tail + 1:end)); %// Handle leftover elements
Later on you can put this code in a loop, computing and plotting the average values for a different value of N in each iteration.
Example #1
X = [1 1 2 3 4 3 8 2 1 3 5 6 7 7 5];
N = 5;
tail = mod(numel(X), N);
X_avg = mean(reshape(X(1:numel(X) - tail), N, []))
X_avg(end + 1) = mean(X(end - tail + 1:end))
The result is:
X_avg =
2.2000 3.4000 6.0000
Example #2
Here's another example (this time the length of X is not divisible by N):
X = [1 1 2 3 4 3 8 2 1 3 5 6 7 7 5];
N = 10;
tail = mod(numel(X), N);
X_avg = mean(reshape(X(1:numel(X) - tail), N, []))
X_avg(end + 1) = mean(X(end - tail + 1:end))
The result is:
X_avg =
2.8000 6.0000
This should do the trick:
For a selected N (the number of values you want to take the average of):
N = 5;
mean_vals = arrayfun(#(n) mean(X(n-1+(1:N))),1:N:length(X))
Note: This does not check if Index exceeds matrix dimensions.
If you want to skip the last numbers, this should work:
mean_vals = arrayfun(#(n) mean(X(n-1+(1:N))),1:N:(length(X)-mod(length(X),N)));
To add the remaining values:
if mod(length(X),N) ~= 0
mean_vals(end+1) = mean(X(numel(X)+1-mod(length(X),N):end))
end
UPDATE: This is a modification of Eitan's first answer (before it was edited). It uses nanmean(), which takes the mean of all values that are not NaN. So, instead of filling the remaining rows with zeros, fill them with NaN, and just take the mean.
X = [X(:); NaN(mod(N - numel(X), N), 1)];
X_avg = nanmean(reshape(X, N, []));
It would be helpful if you posted some code and point out exactly what is not working.
As a first pointer. If
X = [1 1 2 3 4 3 8 2 1 3 5 6 7 7 5]
the three means in blocks of 5 you are interested in are
mean(X(1:5))
mean(X(6:10))
mean(X(11:15))
You will have to come up with a for loop or maybe some other way to iterate through the indices.
I think you want something like this (I didn't use Matlab in a while, I hope the syntax is right):
X = [1 1 2 3 4 3 8 2 1 3 5 6 7 7 5],
currentAmount=5,
block=0,
while(numel(X)<=currentAmount)
while(numel(X)<=currentAmount+block*currentAmount)
mean(X(block*currentAmount+1:block*currentAmount+currentAmount));
block =block+1;
end;
currentAmount = currentAmount+5;
block=0;
end
This code will first loop through all elements calculating means of 5 elements at a time. Then, it will expand to 10 elements. Then to 15, and so on, until the number of elements from which you want to make the mean is bigger than the number of elements in the column.
If you are looking to average K random samples in your N-dimensional vector, then you could use:
N = length(X);
K = 20; % or 10, or 30, or any integer less than or equal to N
indices = randperm(N, K); % gives you K random indices from the range 1:N
result = mean(X(indices)); % averages the values of X at the K random
% indices from above
A slightly more compact form would be:
K = 20;
result = mean(X(randperm(length(X), K)));
If you are just looking to take every K consecutive samples from the list and average them then I am sure one of the previous answers will give you what you want.
If you need to do this operation a lot, it might be worth writing your own function for it. I would recommend using #EitanT's basic idea: pad the data, reshape, take mean of each column. However, rather than including the zero-padded numbers at the end, I recommend taking the average of the "straggling" data points separately:
function m = meanOfN(x, N)
% function m = meanOfN(x, N)
% create groups of N elements of vector x
% and return their mean
% if numel(x) is not a multiple of N, the last value returned
% will be for fewer than N elements
Nf = N * floor( numel( x ) / N ); % largest multiple of N <= length of x
xr = reshape( x( 1:Nf ), N, []);
m = mean(xr);
if Nf < N
m = [m mean( x( Nf + 1:end ) )];
end
This function will return exactly what you were asking for: in the case of a 15 element vector with N=5, it returns 3 values. When the size of the input vector is not a multiple of N, the last value returned will be the "mean of what is left".
Often when you need to take the mean of a set of numbers, it is the "running average" that is of interest. So rather than getting [mean(x(1:5)) mean(x(6:10)) mean(11:15))], you might want
m(1) = mean(x(1:N));
m(2) = mean(x(2:N+1));
m(3) = mean(x(3:N+2));
...etc
That could be achieved using a simple convolution of your data with a vector of ones; for completeness, here is a possible way of coding that:
function m = meansOfN(x, n)
% function m = meansOfN(x, n)
% taking the running mean of the values in x
% over n samples. Returns a row vector of size (sizeof(x) - n + 1)
% if numel(x) < n, this returns an empty matrix
mv = ones(N,1) / N; % vector of ones, normalized
m = convn(x(:), mv, 'valid'); % perform 1D convolution
With these two functions in your path (save them in a file called meanOfN.m and meansOfN.m respectively), you can do anything you want. In any program you will be able to write
myMeans = meanOfN(1:30, 5);
myMeans2 = meansOfN(1:30, 6);
etc. Matlab will find the function, perform the calculation, return the result. Writing your custom functions for specific operations like this can be very helpful - not only does it keep your code clean, but you only have to test the function once...
I have a matrix sorted in ascending order.
S = 25;
RT = zeros(S,2);
for i = 1:S;
for j = 1:i;
R = i *j;
T = R + j;
RT(j,:) = [R T];
end
end
sortRT = sortrows(RT, [1 2]);
disp(sortRT);
I want to find the sortRT elements which values is lower than 500 (for R) and 490 (for T) per column and place these values inside a matrix. Is it possible?
Just use find:
idx = find(sortRT(:,1)<500 & sortRT(:,2)<490)
idx' =
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
These are the rows where both R<500 and T<490. You can of course separate these two:
idxR500 = find(sortRT(:,1)<500);
idxT490 = find(sortRT(:,2)<490);
If you're just going to copy elements or rows, then find isn't even necessary and you can use logical indexing:
R500 = sortRT(find(sortRT(:,1)<500) , 1);
is the same as
R500 = sortRT(sortRT(:,1)<500 , 1);
This copies elements of the first column, if you want to copy the whole row, use the colon operator:
R500 = sortRT(sortRT(:,1)<500 , :);