Can anybody help me to find out the method to calculate the elements of different sized matrix in Matlab ?
Let say that I have 2 matrices with numbers.
Example:
A=[1 2 3;
4 5 6;
7 8 9]
B=[10 20 30;
40 50 60]
At first,we need to find maximum number in each column.
In this case, Ans=[40 50 60].
And then,we need to find ****coefficient** (k).
Coefficient(k) is equal to 1 divided by quantity of column of matrix A.
In this case, **coefficient (k)=1/3=0.33.
I wanna create matrix C filling with calculation.
Example in MS Excel.
H4 = ABS((C2-C6)/C9)*0.33+ABS((D2-D6)/D9)*0.33+ABS((E2-E6)/E9)*0.33
I4 = ABS((C3-C6)/C9)*0.33+ABS((D3-D6)/D9)*0.33+ABS((E3-E6)/E9)*0.33
J4 = ABS((C4-C6)/C9)*0.33+ABS((D4-D6)/D9)*0.33+ABS((E4-E6)/E9)*0.33
And then (Like above)
H5 = ABS((C2-C7)/C9)*0.33+ABS((D2-D7)/D9)*0.33+ABS((E2-E7)/E9)*0.33
I5 = ABS((C3-C7)/C9)*0.33+ABS((D3-D7)/D9)*0.33+ABS((E3-E7)/E9)*0.33
J5 = ABS((C4-C7)/C9)*0.33+ABS((D4-D7)/D9)*0.33+ABS((E4-E7)/E9)*0.33
C =
0.34 =|(1-10)|/40*0.33+|(2-20)|/50*0.33+|(3-30)|/60*0.33
0.28 =|(4-10)|/40*0.33+|(5-20)|/50*0.33+|(6-30)|/60*0.33
0.22 =|(7-10)|/40*0.33+|(8-20)|/50*0.33+|(9-30)|/60*0.33
0.95 =|(1-40)|/40*0.33+|(2-50)|/50*0.33+|(3-60)|/60*0.33
0.89 =|(4-40)|/40*0.33+|(5-50)|/50*0.33+|(6-60)|/60*0.33
0.83 =|(7-40)|/40*0.33+|(8-50)|/50*0.33+|(9-60)|/60*0.33
Actually A is a 15x4 matrix and B is a 5x4 matrix.
Perhaps,the matrices dimensions are more than this matrices (variables).
How can i write this in Matlab?
Thanks you!
You can do it like so. Let's assume that A and B are defined as you did before:
A = vec2mat(1:9, 3)
B = vec2mat(10:10:60, 3)
A =
1 2 3
4 5 6
7 8 9
B =
10 20 30
40 50 60
vec2mat will transform a vector into a matrix. You simply specify how many columns you want, and it will automatically determine the right amount of rows to transform the vector into a correctly shaped matrix (thanks #LuisMendo!). Let's also define more things based on your post:
maxCol = max(B); %// Finds maximum of each column in B
coefK = 1 / size(A,2); %// 1 divided by number of columns in A
I am going to assuming that coefK is multiplied by every element in A. You would thus compute your desired matrix as so:
cellMat = arrayfun(#(x) sum(coefK*(bsxfun(#rdivide, ...
abs(bsxfun(#minus, A, B(x,:))), maxCol)), 2), 1:size(B,1), ...
'UniformOutput', false);
outputMatrix = cell2mat(cellMat).'
You thus get:
outputMatrix =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
Seems like a bit much to chew right? Let's go through this slowly.
Let's start with the bsxfun(#minus, A, B(x,:)) call. What we are doing is taking the A matrix and subtracting with a particular row in B called x. In our case, x is either 1 or 2. This is equal to the number of rows we have in B. What is cool about bsxfun is that this will subtract every row in A by this row called by B(x,:).
Next, what we need to do is divide every single number in this result by the corresponding columns found in our maximum column, defined as maxCol. As such, we will call another bsxfun that will divide every element in the matrix outputted in the first step by their corresponding column elements in maxCol.
Once we do this, we weight all of the values of each row by coefK (or actually every value in the matrix). In our case, this is 1/3.
After, we then sum over all of the columns to give us our corresponding elements for each column of the output matrix for row x.
As we wish to do this for all of the rows, going from 1, 2, 3, ... up to as many rows as we have in B, we apply arrayfun that will substitute values of x going from 1, 2, 3... up to as many rows in B. For each value of x, we will get a numCol x 1 vector where numCol is the total number of columns shared by A and B. This code will only work if A and B share the same number of columns. I have not placed any error checking here. In this case, we have 3 columns shared between both matrices. We need to use UniformOutput and we set this to false because the output of arrayfun is not a single number, but a vector.
After we do this, this returns each row of the output matrix in a cell array. We need to use cell2mat to transform these cell array elements into a single matrix.
You'll notice that this is the result we want, but it is transposed due to summing along the columns in the second step. As such, simply transpose the result and we get our final answer.
Good luck!
Dedication
This post is dedicated to Luis Mendo and Divakar - The bsxfun masters.
Assuming by maximum number in each column, you mean columnwise maximum after vertically concatenating A and B, you can try this one-liner -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(vertcat(A,B)),[1 3 2]))),3)./size(A,2)
Output -
ans =
0.3450 0.2833 0.2217
0.9617 0.9000 0.8383
If by maximum number in each column, you mean columnwise maximum of B, you can try -
sum(abs(bsxfun(#rdivide,bsxfun(#minus,permute(A,[3 1 2]),permute(B,[1 3 2])),permute(max(B),[1 3 2]))),3)./size(A,2)
The output for this case stays the same as the previous case, owing to the values of A and B.
Related
I'm looking for a way to test if all elements of a matrix are greater than or equal to their corresponding indexes values in another matrix, and if not to stop evaluating. This is part of an elseif statement for setting a lower bound for values, thus as simplified example:
Lower Bound matrix: A = [4 5 6 7]
New values matrix: B = [7 8 9 10]
Is B>=A then yes, accept and proceed
whereas
Lower Bound matrix: A = [4 5 6 7]
New values matrix: C = [5 3 5 8]
Is C>=A? then no, all elements of C are not greater than A, reject and stop
My solution so far is a bit hackneyed:
Is sum(C>=A) < length(C)? no, then reject and stop
This gives the sum of the true/false values in the comparison C>=A, which if all values of C were greater would equal the length of C, else the sum would be less than the length of C. I keep thinking there's a simple and more elegant solution to this that I'm overlooking and would be grateful for any thoughts. Thanks!
MATLAB has a built-in function for performing this action called all. You can use this on a logical matrix to determine if all values are true. In your case you would pass the logical matrix: B >= A.
A = [4,5,6,7];
B = [7,8,9,10];
all(B(:) >= A(:))
1
Notice that I have used (:) above which ensures that both A and B are column vectors prior to performing the comparison. This way, they can be of any arbitrary dimension and the result of all will always be a scalar.
While you're at it, you may also look into any.
You must indeed rely on logical indexing. To check whether a given matrix B has elements greater than or equal to their corresponding indexes values in another given matrix A you can do something like:
if (sum(sum(B>=A))==numel(A))
%enter if body here
end
The statement B>=A will return a logical matrix with 1 in position (i,j) if B(i,j)>=A(i,j). You then sum all the 1s inside such matrix and then check if such sum is equal to the number of elements (numel()) in A (or B).
In your example. Given
A=[4 5 6 7];
B=[7 8 9 10];
the statement B>=A will return
ans =
1 1 1 1
because all elements in B are greater then elements in A. Sum this vector, you'll get 4. Is the sum (4) equal to the number of elements (4)? Yes. Then all elements in B are greater than or equal to the elements in A.
Note: the double sum() makes your code more robust. It will indeed work also with matrices and not just with vectors. That is because if you do sum() on a matrix, Matlab by default does not return the sum of all its elements, but the sum along the first dimension whose size is different from 1. So if our matrix is:
C =
8 1 6
3 5 7
4 9 2
sum(C) will return
ans =
15 15 15
(Matlab took the sum of every column).
By taking also the sum of such vector we'll get the sum of all the elements.
This ends the quick explanation regarding the nested sum().
I assume by an elegant solution you mean a solution which is more efficient.
Lets create test data:
A = zeros(100000,100000); B = zeros(100000,100000);
Linear Loop
for i = 1:numel(A)
if A(i) < B(i)
display('different')
break
end
end
Logical indexing
if (sum(sum(B>=A))~=numel(A))
display('different')
end
The loop is better when its comes to the best case (first element is smaller):
Elapsed time is 0.000236 seconds. Elapsed time is 0.131802 seconds.
and when its the average case:
Elapsed time is 0.001993 seconds. Elapsed time is 0.196228 seconds.
But we only care about the worst case:
B(end) = 1;
Elapsed time is 8.181473 seconds. Elapsed time is 0.108002 seconds.
I'm optimizing my codes. Now I have an MxN matrix, and I want to generate a mean MxN matrix which is the mean of other rows
for example: if i have matrix A:
1 2 3
3 4 5
2 3 2
In the new matrix B, I want each one is the mean of other rows.
mean(row2,row3)
mean(row1,row3)
mean(row1,row2)
I've think of many ways, but can't avoid a loop
for row=1:3
temp = A;
temp(row,:) = [];
B(row,:) = mean(temp);
end
any thoughts?
Simple with bsxfun -
B = (bsxfun(#minus,sum(A,1),A))./(size(A,1)-1)
The trick is to subtract the current row from the sum all rows with bsxfun in a vectorized manner, thus giving us the sum of all rows except the current one. Finally, get the average values by dividing them by the number of rows minus 1.
I'm trying to represent a simple matrix m*n (let's assume it has only one row!) such that m1n1 = m1n1^1, m1n2 = m1n1^2, m1n3 = m1n1^3, m1n3 = m1n1^4, ... m1ni = m1n1^i.
In other words, I am trying to iterate over a matrix columns n times to add a new vector(column) at the end such that each of the indices has the same value as the the first vector but raised to the power of its column number n.
This is the original vector:
v =
1.2421
2.3348
0.1326
2.3470
6.7389
and this is v after the third iteration:
v =
1.2421 1.5429 1.9165
2.3348 5.4513 12.7277
0.1326 0.0176 0.0023
2.3470 5.5084 12.9282
6.7389 45.4128 306.0329
now given that I'm a total noob in Matlab, I really underestimated the difficulty of such a seemingly easy task, that took my almost a day of debugging and surfing the web to find any clue. Here's what I have come up with:
rows = 5;
columns = 3;
v = x(1:rows,1);
k = v;
Ncol = ones(rows,1);
extraK = ones(rows,1);
disp(v)
for c = 1:columns
Ncol = k(:,length(k(1,:))).^c; % a verbose way of selecting the last column only.
extraK = cat(2,extraK,Ncol);
end
k = cat(2,k,extraK);
disp(extraK(:,2:columns+1)) % to cut off the first column
now this code (for some weird reason) work only if rows = 6 or less, and columns = 3 or less.
when rows = 7, this is the output:
v = 1.0e+03 *
0.0012 0.0015 0.0019
0.0023 0.0055 0.0127
0.0001 0.0000 0.0000
0.0023 0.0055 0.0129
0.0067 0.0454 0.3060
0.0037 0.0138 0.0510
0.0119 0.1405 1.6654
How could I get it to run on any number of rows and columns?
Thanks!
I have found a couple of things wrong with your code:
I'm not sure as to why you are defining d = 3;. This is just nitpicking, but you can remove that from your code safely.
You are not doing the power operation properly. Specifically, look at this statement:
Ncol = k(:,length(k(1,:))).^c; % a verbose way of selecting the last column only.
You are selectively choosing the last column, which is great, but you are not applying the power operation properly. If I understand your statement, you wish to take the original vector, and perform a power operation to the power of n, where n is the current iteration. Therefore, you really just need to do this:
Ncol = k.^c;
Once you replace Ncol with the above line, the code should now work. I also noticed that you crop out the first column of your result. The reason why you are getting duplicate columns is because your for loop starts from c = 1. Since you have already computed v.^1 = v, you can just start your loop at c = 2. Change your loop starting point to c = 2, and you can get rid of the removal of the first column.
However, I'm going to do this in an alternative way in one line of code. Before we do this, let's go through the theory of what you're trying to do.
Given a vector v that is m elements long stored in a m x 1 vector, what you want is to have a matrix of size m x n, where n is the desired number of columns, and for each column starting from left to right, you wish to take v to the nth power.
Therefore, given your example from your third "iteration", the first column represents v, the second column represents v.^2, and the third column represents v.^3.
I'm going to introduce you to the power of bsxfun. bsxfun stands for Binary Singleton EXpansion function. What bsxfun does is that if you have two inputs where either or both inputs has a singleton dimension, or if either of both inputs has only one dimension which has value of 1, each input is replicated in their singleton dimensions to match the size of the other input, and then an element-wise operation is applied to these inputs together to produce your output.
For example, if we had two vectors like so:
A = [1 2 3]
B = [1
2
3]
Note that one of them is a row vector, and the other is a column vector. bsxfun would see that A and B both have singleton dimensions, where A has a singleton dimension being the number of rows being 1, and B having a singleton dimension which is the number of columns being 1. Therefore, we would duplicate B as many columns as there are in A and duplicate A for as many rows as there are in B, and we actually get:
A = [1 2 3
1 2 3
1 2 3]
B = [1 1 1
2 2 2
3 3 3]
Once we have these two matrices, you can apply any element wise operations to these matrices to get your output. For example, you could add, subtract, take the power or do an element wise multiplication or division.
Now, how this scenario applies to your problem is the following. What you are doing is you have a vector v, and you will have a matrix of powers like so:
M = [1 2 3 ... n
1 2 3 ... n
...........
...........
1 2 3 ... n]
Essentially, we will have a column of 1s, followed by a column of 2s, up to as many columns as you want n. We would apply bsxfun on the vector v which is a column vector, and another vector that is only a single row of values from 1 up to n. You would apply the power operation to achieve your result. Therefore, you can conveniently calculate your output by doing:
columns = 3;
out = bsxfun(#power, v, 1:columns);
Let's try a few examples given your vector v:
>> v = [1.2421; 2.3348; 0.1326; 2.3470; 6.7389];
>> columns = 3;
>> out = bsxfun(#power, v, 1:columns)
out =
1.2421 1.5428 1.9163
2.3348 5.4513 12.7277
0.1326 0.0176 0.0023
2.3470 5.5084 12.9282
6.7389 45.4128 306.0321
>> columns = 7;
>> format bank
>> out = bsxfun(#power, v, 1:columns)
out =
Columns 1 through 5
1.24 1.54 1.92 2.38 2.96
2.33 5.45 12.73 29.72 69.38
0.13 0.02 0.00 0.00 0.00
2.35 5.51 12.93 30.34 71.21
6.74 45.41 306.03 2062.32 13897.77
Columns 6 through 7
3.67 4.56
161.99 378.22
0.00 0.00
167.14 392.28
93655.67 631136.19
Note that for setting the columns to 3, we get what we see in your post. For pushing the columns up to 7, I had to change the way the numbers were presented so you can see the numbers clearly. Not doing this would put this into exponential form, and there were a lot of zeroes that followed the significant digits.
Good luck!
When computing cumulative powers you can reuse previous results: for scalar x and n, x.^n is equal to x * x.^(n-1), where x.^(n-1) has been already obtained. This may be more efficient than computing each power independently, because multiplication is faster than power.
Let N be the maximum exponent desired. To use the described approach, the column vector v is repeated N times horizontally (repmat), and then a cumulative product is applied along each row (cumprod):
v =[1.2421; 2.3348; 0.1326; 2.3470; 6.7389]; %// data. Column vector
N = 3; %// maximum exponent
result = cumprod(repmat(v, 1, N), 2);
For my experiment I have 20 categories which contain 9 pictures each. I want to show these pictures in a pseudo-random sequence where the only constraint to randomness is that one image may not be followed directly by one of the same category.
So I need something similar to
r = randi([1 20],1,180);
just with an added constraint of two numbers not directly following each other. E.g.
14 8 15 15 7 16 6 4 1 8 is not legitimate, whereas
14 8 15 7 15 16 6 4 1 8 would be.
An alternative way I was thinking of was naming the categories A,B,C,...T, have them repeat 9 times and then shuffle the bunch. But there you run into the same problem I think?
I am an absolute Matlab beginner, so any guidance will be welcome.
The following uses modulo operations to make sure each value is different from the previous one:
m = 20; %// number of categories
n = 180; %// desired number of samples
x = [randi(m)-1 randi(m-1, [1 n-1])];
x = mod(cumsum(x), m) + 1;
How the code works
In the third line, the first entry of x is a random value between 0 and m-1. Each subsequent entry represents the change that, modulo m, will give the next value (this is done in the fourth line).
The key is to choose that change between 1 and m-1 (not between 0 and m-1), to assure consecutive values will be different. In other words, given a value, there are m-1 (not m) choices for the next value.
After the modulo operation, 1 is added to to transform the range of resulting values from 0,...,m-1 to 1,...,m.
Test
Take all (n-1) pairs of consecutive entries in the generated x vector and count occurrences of all (m^2) possible combinations of values:
count = accumarray([x(1:end-1); x(2:end)].', 1, [m m]);
imagesc(count)
axis square
colorbar
The following image has been obtained for m=20; n=1e6;. It is seen that all combinations are (more or less) equally likely, except for pairs with repeated values, which never occur.
You could look for the repetitions in an iterative manner and put new set of integers from the same group [1 20] only into those places where repetitions have occurred. We continue to do so until there are no repetitions left -
interval = [1 20]; %// interval from where the random integers are to be chosen
r = randi(interval,1,180); %// create the first batch of numbers
idx = diff(r)==0; %// logical array, where 1s denote repetitions for first batch
while nnz(idx)~=0
idx = diff(r)==0; %// logical array, where 1s denote repetitions for
%// subsequent batches
rN = randi(interval,1,nnz(idx)); %// new set of random integers to be placed
%// at the positions where repetitions have occured
r(find(idx)+1) = rN; %// place ramdom integers at their respective positions
end
In MATLAB have a large matrix with transition probabilities transition_probs, and an adjacency matrix adj_mat. I want to compute the cumulative sum of the transition matrix along the columns and then element wise multiply it against the adjacency matrix which acts as a mask in this way:
cumsumTransitionMat = cumsum(transition_probs,2) .* adj_mat;
I get a MEMORY error because with the cumsum all the entries of the matrix are then non-zero.
I would like to avoid this problem by only having the cumulative sum entries where there are non zero entries in the first place. How can this be done without the use of a for loop?
when CUMSUM is applied on rows, for each row it will go and fill with values starting with the first nonzero column it finds up until the last column, thats what it does by definition.
The worst case in terms of storage is when the sparse matrix contains values at the first column, the best case is when all nonzero values occur at the last column. Example:
% worst case
>> M = sparse([ones(5,1) zeros(5,4)]);
>> MM = cumsum(M,2); % completely dense matrix
>> nnz(MM)
ans =
25
% best case
>> MM = cumsum(fliplr(M),2);
If the resulting matrix does not fit in memory, I dont see what else you can do, except maybe use a for-loop over the rows, and process the matrix is smaller batches...
Note that you cannot apply the masking operation before computing the cumulative sum, since this will alter the results. So you cant say cumsum(transition_probs .* adj_mat, 2).
You can apply cumsum on the non-zero elements only. Here is some code:
A = sparse(round(rand(100,1))); %some sparse data
A_cum = A; %instantiate A_cum by copy A
idx_A = find(A); %find non-zeros
A_cum(idx_A) = cumsum(A(idx_A)); %cumsum on non-zeros elements only
You can check the output with
B = cumsum(A);
A_cum B
1 1
0 1
0 1
2 2
3 3
4 4
5 5
0 5
0 5
6 6
and isequal(A_cum(find(A_cum)), B(find(A_cum))) gives 1.