I have a matrix with following shape:
A = [1 2 3;
4 5 6;
7 8 9]
Now I want starting with the last column to multiply the column with a number and then decrease the number and move to the next column.
So if we start with the number 1 and use for step 0.2 to modify all columns:
Anew = [1*0.6 2*0.8 3*1;
4*0.6 5*0.8 6*1;
7*0.6 8*0.8 9*1]
Or for second example we start with 0.9 with 0.1 as step and modify 3 columns:
B = [1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16]
And to get:
Bnew = [1 2*0.7 3*0.8 4*0.9;
5 6*0.7 7*0.8 8*0.9;
9 10*0.7 11*0.8 12*0.9;
13 14*0.7 15*0.8 16*0.9]
The matrices might vary in their amount of columns, and I would like to set starting number, ending number, step number and the amount of columns I want to modify.
What you are describing can be achieved with broadcasted element-wise multiplication in matlab R2016b and beyond.
Let's say your inputs are the matrix A, start value start, step size step, and number n. You can start by constructing the factors you want to multiply by. I am going to assume that when n > size(A, 2), you want to just use the first n steps rather than error out:
k = size(A, 2);
n = min(n, k);
factors = ones(1, k);
factors(1 + k - n:end) = linspace(start - (n - 1) * step, start, n);
Now you can just multiply your matrix:
result = A .* factors;
This solution has the advantage of being extremely simple and fully vectorized.
If you have an older version of MATLAB, do the following instead:
result = A .* repmat(factors, size(A, 1), 1);
Or use Tony's trick:
result = A .* factors(ones(3, 1), :)
I just found the solution:
count = 0;
A = randi([-10,10],4,4);
Anew = [];
for i=0.9:-0.1:0
number_columns = 3;
if count == number_columns
rest = existing_columns - count;
for i=rest:-1:1
Anew = [(A(:,i)) Anew];
end
break
end
existing_columns = size(A,1);
Anew = [(A(:,existing_columns-count)*i) Anew];
count = count + 1;
end
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 as follows:
id value
=============
1 0.5
2 0.5
3 0.8
4 0.3
5 0.2
From this array, I wish to find all the possible combinations that have a sum less than or equal to 1. That is,
result
======
1 2
1 4 5
2 4 5
3 5
1 5
1 4
2 4
2 5
...
In order to get the above result, my idea has been to initially compute all the possibilities of finding sum of elements in the array, like so:
for ii = 1 : length(a) % compute number of possibilities
no_of_possibilities = no_of_possibilities + nchoosek(length(a),ii);
end
Once this is done, then loop through all possible combinations.
I would like to know if there's an easier way of doing this.
data = [0.5, 0.5, 0.8, 0.3, 0.2];
required = cell(1, length(data));
subsets = cell(1, length(data));
for k = 2:length(data)-1 % removes trivial cases (all numbers or one number at a time)
% generate all possible k-pairs (if k = 3, then all possible triplets
% will be generated)
combination = nchoosek(1:length(data), k);
% for every triplet generated, this function sums the corresponding
% values and then decides whether then sum is less than equal to 1 or
% not
findRequired = #(x) sum(data(1, combination(x, :))) <= 1;
% generate a logical vector for all possible combinations like [0 1 0]
% which denotes that the 2nd combination satisfies the condition while
% the others do not
required{k} = arrayfun(findRequired, 1:size(combination, 1));
% access the corresponding combinations from the entire set
subsets{k} = combination(required{k}, :);
end
This produces the following subsets:
1 2
1 4
1 5
2 4
2 5
3 5
4 5
1 4 5
2 4 5
It is not in easy way, however is a faster way, as I removed the combination which its subsets are not passed the condition.
bitNo = length(A); % number of bits
setNo = 2 ^ bitNo - 1; % number of sets
subsets = logical(dec2bin(0:setNo, bitNo) - '0'); % all subsets
subsets = subsets(2:end,:); % all subsets minus empty set!
subsetCounter = 1;
resultCounter = 1;
result = {};
while(1)
if( subsetCounter >= size(subsets,1))
break;
end
if(sum(A(subsets(subsetCounter,:).',2)) <= 1)
result{resultCounter} = A(subsets(subsetCounter,:).',1).';
resultCounter = resultCounter + 1;
subsetCounter = subsetCounter + 1;
else
% remove all bad cases related to the current subset
subsets = subsets(sum((subsets & subsets(subsetCounter,:)) - subsets(subsetCounter,:),2) ~= 0,:);
end
end
Generate the subsets using this method. After that, check the condition for each subset. If the subset does not pass the condition, all its supersets are removed from the subsets. To do this, using sum((subsets & subsets(i,:)) - subsets(i,:),2) ~= 0 which mean get some rows from subsets which has not the same elements of the not passed subset. By doing this, we able to not to consider some bad cases anymore. Although, theoretically, this code is Θ(2^n).
Here is potential solution, using inefficient steps, but borrowing efficient code from various SO answers. Credit goes to those original peeps.
data = [0.5, 0.5, 0.8, 0.3, 0.2];
First get all combinations of indices, not necessarily using all values.
combs = bsxfun(#minus, nchoosek(1:numel(data)+numel(data)-1,numel(data)), 0:numel(data)-1);
Then get rid of repeated indices in each combination, regardless of index order
[ii, ~, vv] = find(sort(combs,2));
uniq = accumarray(ii(:), vv(:), [], #(x){unique(x.')});
Next get unique combinations, regardless of index order... NOTE: You can do this step much more efficiently by restructuring the steps, but it'll do.
B = cellfun(#mat2str,uniq,'uniformoutput',false);
[~,ia] = unique(B);
uniq=uniq(ia);
Now sum all values in data based on cell array (uniq) of index combinations
idx = cumsum(cellfun('length', uniq));
x = diff(bsxfun(#ge, [0; idx(:)], 1:max(idx)));
x = sum(bsxfun(#times, x', 1:numel(uniq)), 2); %'// Produce subscripts
y = data([uniq{:}]); % // Obtain values
sums_data = accumarray(x, y);
And finally only keep the index combinations that sum to <= 1
allCombLessThanVal = uniq(sums_data<=1)
I have a matrix [1 2 3 4] and I want to shuffle it with randperm in few times but I want to obtain different matrices. For example
for i=1:4
m(i,:)=randperm(4);
end
will give me 4 rows with 4 columns but I want every row to be different from every other one; e.g. like this:
m(1,:)=[1 3 4 2]
m(2,:)=[2 3 1 4]
m(3,:)=[2 1 4 3]
m(4,:)=[4 3 2 3]
You can just check the existing rows to see if the current permutation already exists
m = zeros(4, 4);
counter = 1;
while counter < 4
new = randperm(4);
if ~ismember(new, m, 'rows')
m(counter, :) = new;
counter = counter + 1;
end
end
Another (memory intensive) approach would be to generate all permutations and then randomly select N of them
allperms = perms(1:4);
N = 4;
m = allperms(randsample(size(allperms,1), N), :);
You can easily use the MATLAB function ismember to check if the random permutation that you just created is already contained in your matrix.
So you can just try something like that:
for i=1:4
temp = randperm(4);
while ismember(m,temp,'rows')
temp = randperm(4);
end
m(i,:) = temp;
end
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 very large matrix (216 rows, 31286 cols) of doubles. For reasons specific to the data, I want to average every 9 rows to produce one new row. So, the new matrix will have 216/9=24 rows.
I am a Matlab beginner so I was wondering if this solution I came up with can be improved upon. Basically, it loops over every group, sums up the rows, and then divides the new row by 9. Here's a simplified version of what I wrote:
matrix_avg = []
for group = 1:216/9
new_row = zeros(1, 31286);
idx_low = (group - 1) * 9 + 1;
idx_high = idx_low + 9 - 1;
% Add the 9 rows to new_row
for j = idx_low:idx_high
new_row = new_row + M(j,:);
end
% Compute the mean
new_row = new_row ./ 9
matrix_avg = [matrix_avg; new_row];
end
You can reshape your big matrix from 216 x 31286 to 9 x (216/9 * 31286).
Then you can use mean, which operates on each column. Since your matrix only has 9 rows per column, this takes the 9-row average.
Then you can just reshape your matrix back.
% generate big matrix
M = rand([216 31286]);
n = 9 % want 9-row average.
% reshape
tmp = reshape(M, [n prod(size(M))/n]);
% mean column-wise (and only 9 rows per col)
tmp = mean(tmp);
% reshape back
matrix_avg = reshape(tmp, [ size(M,1)/n size(M,2) ]);
In a one-liner (but why would you?):
matrix_avg = reshape(mean(reshape(M,[n prod(size(M))/n])), [size(M,1)/n size(M,2)]);
Note - this will have problems if the number of rows in M isn't exactly divisible by 9, but so will your original code.
I measured the 4 solutions and here are the results:
reshape: Elapsed time is 0.017242 seconds.
blockproc [9 31286]: Elapsed time is 0.242044 seconds.
blockproc [9 1]: Elapsed time is 44.477094 seconds.
accumarray: Elapsed time is 103.274071 seconds.
This is the code I used:
M = rand(216,31286);
fprintf('reshape: ');
tic;
n = 9;
matrix_avg1 = reshape(mean(reshape(M,[n prod(size(M))/n])), [size(M,1)/n size(M,2)]);
toc
fprintf('blockproc [9 31286]: ');
tic;
fun = #(block_struct) mean(block_struct.data);
matrix_avg2 = blockproc(M,[9 31286],fun);
toc
fprintf('blockproc [9 1]: ');
tic;
fun = #(block_struct) mean(block_struct.data);
matrix_avg3 = blockproc(M,[9 1],fun);
toc
fprintf('accumarray: ');
tic;
[nR,nC] = size(M);
n2average = 9;
[xx,yy] = ndgrid(1:nR,1:nC);
x = ceil(xx/n2average); %# makes xx 1 1 1 1 2 2 2 2 etc
matrix_avg4 = accumarray([xx(:),yy(:)],M(:),[],#mean);
toc
Here's an alternative based on accumarray. You create an array with row and column indices into matrix_avg that tells you which element in matrix_avg a given element in M contributes to, then you use accumarray to average the elements that contribute to the same element in matrix_avg. This solution works even if the number of rows in M is not divisible by 9.
M = rand(216,31286);
[nR,nC] = size(M);
n2average = 9;
[xx,yy] = ndgrid(1:nR,1:nC);
x = ceil(xx/n2average); %# makes xx 1 1 1 1 2 2 2 2 etc
matrix_avg = accumarray([xx(:),yy(:)],M(:),[],#mean);