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Generate a matrix containing all combinations of elements taken from n vectors
(4 answers)
Closed 5 years ago.
I want to create all pattern combination which possible occur
For example, I have three ball and I want to pick 3 times. All possible is 27 events but I want to create all possible event in array like this
[1 1 1; 1 1 2; 1 1 3; 1 2 1 ;....]
Dose anyone can help me to write m-file in matlab program, please?
This can be done very easily with base conversion.
If the number of balls does not exceed 10
M = 3; % Number of balls. Must not exceed 10
N = 3; % Number of draws
result = dec2base(0:M^N-1, M)-'0'+1;
Note that dec2base ouputs chars, not numbers, and hence the -'0' part. Characters in arithmetic operations behave like their corresponding ASCII codes. So subtracting '0' transforms characters '0', '1', ..., '9' into the corresponding numbers.
With this approach M cannot exceed 10 because then dec2bin would output '0', '1', ..., '9', 'A', 'B' ... and the character arithmetic would not give the correct result for 'A', 'B', ... But this can be easily solved as follows.
For number of balls up to 36
M = 12; % Number of balls. Must not exceed 36
N = 2; % Number of draws
result = dec2base(0:M^N-1, M)-'0'+1; % same as before
result = result - ('A'-'9'-1)*(result>10); % correction
The new line simply corrects the results for 'A', 'B', ... by subtracting 'A'-'9'-1, to compensate the fact that '9' and 'A' do not have consecutive ASCII codes.
With this approach M cannot exceed 36 because of dec2base restrictions.
You can create the result like this for the specific case:
firstCol = [ones(9,1);2*ones(9,1);3*ones(9,1)];
secondCol = repeat([ones(3,1);2*ones(3,1);3*ones(3,1)],1,3);
thirdCol = repeat([1;2;3],1,9);
result = [firstCol secondCol thirdCol];
First, repeat 9 times 1,2, and 3 for first column. then repeat each of them 3 times for the second column and choose the third column once for each item. Indeed, this generate the all possible choices for each location.
How? If you suppose the first element is 1, you have 3 choice for the second place, and 3 choice for the third place. Hence, you have 9 possible option when the first place is 1. Also, fix the second place, and analyze this. You can generalize this for 2 and 3. The above code, try to generate possibilities based on this explanation.
In the above, ones generate a matrix which all elements are 1 with the specified size and repeat function repeats the specified matrix in the specified size and dimension. You can check the documentation to know more about them.
Hence, You can generalize it for n like the following:
n = 10;
result = zeros(3^n,3);
for idx = 1:n
result(:,idx) = repeat([ones(3^(n-idx),1);2*ones(3^(n-idx),1);3*ones(3^(n-idx),1)],1,3^(idx-1));
end
Related
I have tried different repeating functions which tell me the frequency of the numbers, but I want to know the order of the repeating numbers.
For example I have an array which has numbers
a=[ 1,1,1,1,1,2,2,2,2,2,1,1,1,1,4,4,4,5,5,5,7,7,2,2,2,2]
I want my function to give me the answer, i.e. 1,2,1,4,5,7,2.
You can use the diff, you're interested in the first element of a and then every subsequent index where the difference between elements is non-zero
>> a = [ 1,1,1,1,1,2,2,2,2,2,1,1,1,1,4,4,4,5,5,5,7,7,2,2,2,2];
>> b = a( [true, diff(a)~=0] );
b =
1 2 1 4 5 7 2
I need to find all possible combinations of numbers 1:8 such that sum of all elements is equal to 8
The combinations need to be arranged in an ascending order.
Eg
1 7
2 2 4
1 3 5
1 2 2 3
1 1 1 1 1 1 1 1
A number can repeat itself. But a combination must not..
i.e 1 2 2 3 and 2 1 2 3
I need the the solution in ascending order So there will be only one possibility of every combination
I tried a few codes online suggested on Find vector elements that sum up to specific number in MATLAB
VEC = [1:8];
NUM = 8;
n = length(VEC);
finans = zeros(2^n-1,NUM);
for i = 1:(2^n - 1)
ndx = dec2bin(i,n) == '1';
if sum(VEC(ndx)) == NUM
l = length(VEC(ndx));
VEC(ndx)
end
end
but they dont include the possibilities where the numbers repeat.
I found a better approach through recursion and it's more elegant (I like elegant) and faster than my previous attempt (0.00399705213 seconds on my computer).
EDIT: You will need my custom function stretchmat.m that stretches a vector to fit the size of another matrix. Kinda like repmat but stretching the first parameter (see help for details). Very useful!
script.m
% Define funciton to prepend a cell x with a variable i
cellprepend = #(x,i) {[i x]};
% Execute and time function
tic;
a = allcomb(cellprepend,1,8); % Solution in a
toc;
allcomb.m
function a = allcomb( cellprepend, m, n )
% Add entire block as a combination
a{1} = n;
% Exit recursion if block size 1
if n == 1
return;
end
% Recurse cutting blocks at different segments
for i = m:n/2
b = allcomb(cellprepend,i,n-i);
a = [a cellfun( cellprepend, b, num2cell( stretchmat( i, b ) ) )];
end
end
So the idea is simple, for solutions that add to 8 is exhaustive. If you look for only valid answers, you can do a depth first search by breaking up the problem into 2 blocks. This can be written recursively as I did above and is kinda similar to Merge Sort. The allcomb call takes the block size (n) and finds all the ways of breaking it up into smaller pieces.
We want non-zero pieces so we loop it from 1:n-1. It then prepends the first block to all the combinations of the second block. By only doing all comb on one of the blocks, we can ensure that all solutions are unique.
As for the sorting, I'm not quite sure what you mean by ascending. From what I see, you appear to be sorting from the last number in ascending order. Can you confirm? Any sort can be appended to the end of script.m.
EDIT 2/3 Notes
For the permutatively unique case, the code can be found here
Thanks to #Simon for helping me QA the code multiple times
EDIT: Look at my second more efficient answer!
The Naive approach! Where the cartprod.m function can be found here.
% Create all permutations
p(1:8) = {0:8};
M = fliplr( cartprod( p{:} ) );
% Check sums
r = sum( M, 2 ) == 8;
M = M(sum( M, 2 ) == 8,:); % Solution here
There are definitely more efficient solutions than this but if you just need a quick and dirty solution for small permutations, this will work. Please note that this made Matlab take 3.5 GB of RAM to temporarily store the permutations.
First save all combinations with repetitions in a cell array. In order to do that, just use nmultichoosek.
v = 1 : 8;
combs = cell(length(v),0);
for i = v
combs{i} = nmultichoosek(v,i);
end
In this way, each element of combs contains a matrix where each row is a combination. For instance, the i-th row of combs{4} is a combination of four numbers.
Now you need to check the sum. In order to do that to all the combinations, use cellfun
sums = cellfun(#(x)sum(x,2),combs,'UniformOutput',false);
sums contains the vectors with the sum of all combinations. For
instance, sums{4} has the sum of the number in combination combs{4}.
The next step is check for the fixed sum.
fixed_sum = 10;
indices = cellfun(#(x)x==fixed_sum,sums,'UniformOutput',false);
indices contains arrays of logical values, telling if the combination satisfies the fixed sum. For instance, indices{4}(1) tells you if the first combination with 4 numbers sums to fixed_sum.
Finally, retrieve all valid combinations in a new cell array, sorting them at the same time.
valid_combs = cell(length(v),0);
for i = v
idx = indices{i};
c = combs{i};
valid_combs{i} = sortrows(c(idx,:));
end
valid_combs is a cell similar to combs, but with only combinations that sum up to your desired value, and sorted by the number of numbers used: valid_combs{1} has all valid combinations with 1 number, valid_combs{2} with 2 numbers, and so on. Also, thanks to sortrows, combinations with the same amount of numbers are also sorted. For instance, if fixed_sum = 10 then valid_combs{8} is
1 1 1 1 1 1 1 3
1 1 1 1 1 1 2 2
This code is quite efficient, on my very old laptop I am able to run it in 0.016947 seconds.
I am new to matlab and I was wondering what it meant to use logical indexing/masking to extract data from a matrix.
I am trying to write a function that accepts a matrix and a user-inputted value to compute and display the total number of values in column 2 of the matrix that match with the user input.
The function itself should have no return value and will be called on later in another loop.
But besides all that hubbub, someone suggested that I use logical indexing/masking in this situation but never told me exactly what it was or how I could use it in my particular situation.
EDIT: since you updated the question, I am updating this answer a little.
Logical indexing is explained really well in this and this. In general, I doubt, if I can do a better job, given available time. However, I would try to connect your problem and logical indexing.
Lets declare an array A which has 2 columns. First column is index (as 1,2,3,...) and second column is its corresponding value, a random number.
A(:,1)=1:10;
A(:,2)=randi(5,[10 1]); //declares a 10x1 array and puts it into second column of A
userInputtedValue=3; //self-explanatory
You want to check what values in second column of A are equal to 3. Imagine as if you are making a query and MATLAB is giving you binary response, YES (1) or NO (0).
q=A(:,2)==3 //the query, what values in second column of A equal 3?
Now, for the indices where answer is YES, you want to extract the numbers in the first column of A. Then do some processing.
values=A(q,2); //only those elements will be extracted: 1. which lie in the
//second column of A AND where q takes value 1.
Now, if you want to count total number of values, just do:
numValues=length(values);
I hope now logical indexing is clear to you. However, do read the Mathworks posts which I have mentioned earlier.
I over simplified the code, and wrote more code than required in order to explain things. It can be achieved in a single-liner:
sum(mat(:,2)==userInputtedValue)
I'll give you an example that may illustrate what logical indexing is about:
array = [1 2 3 0 4 2];
array > 2
ans: [0 0 1 0 1 0]
using logical indexing you could filter elements that fullfil a certain condition
array(array>2) will give: [3 4]
you could also perform alterations to only those elements:
array(array>2) = 100;
array(array<=2) = 0;
will result in "array" equal to
[0 0 100 0 100 0]
Logical indexing means to have a logical / Boolean matrix that is the same size as the matrix that you are considering. You would use this as input into the matrix you're considering, and any locations that are true would be part of the output. Any locations that are false are not part of the output. To perform logical indexing, you would need to use logical / Boolean operators or conditions to facilitate the selection of elements in your matrix.
Let's concentrate on vectors as it's the easiest to deal with. Let's say we had the following vector:
>> A = 1:9
A =
1 2 3 4 5 6 7 8 9
Let's say I wanted to retrieve all values that are 5 or more. The logical condition for this would be A >= 5. We want to retrieve all values in A that are greater than or equal to 5. Therefore, if we did A >= 5, we get a logical vector which tells us which values in A satisfy the above condition:
>> A >= 5
ans =
0 0 0 0 1 1 1 1 1
This certainly tells us where in A the condition is satisfied. The last step would be to use this as input into A:
>> B = A(A >= 5)
B =
5 6 7 8 9
Cool! As you can see, there isn't a need for a for loop to help us select out elements that satisfy a condition. Let's go a step further. What if I want to find all even values of A? This would mean that if we divide by 2, the remainder would be zero, or mod(A,2) == 0. Let's extract out those elements:
>> C = A(mod(A,2) == 0)
C =
2 4 6 8
Nice! So let's go back to your question. Given your matrix A, let's extract out column 2.
>> col = A(:,2)
Now, we want to check to see if any of column #2 is equal to a certain value. Well we can generate a logical indexing array for that. Let's try with the value of 3:
>> ind = col == 3;
Now you'll have a logical vector that tells you which locations are equal to 3. If you want to determine how many are equal to 3, you just have to sum up the values:
>> s = sum(ind);
That's it! s contains how many values were equal to 3. Now, if you wanted to write a function that only displayed how many values were equal to some user defined input and displayed this event, you can do something like this:
function checkVal(A, val)
disp(sum(A(:,2) == val));
end
Quite simply, we extract the second column of A and see how many values are equal to val. This produces a logical array, and we simply sum up how many 1s there are. This would give you the total number of elements that are equal to val.
Troy Haskin pointed you to a very nice link that talks about logical indexing in more detail: http://www.mathworks.com/help/matlab/math/matrix-indexing.html?refresh=true#bq7eg38. Read that for more details on how to master logical indexing.
Good luck!
%% M is your Matrix
M = randi(10,4)
%% Val is the value that you are seeking to find
Val = 6
%% Col is the value of the matrix column that you wish to find it in
Col = 2
%% r is a vector that has zeros in all positions except when the Matrix value equals the user input it equals 1
r = M(:,Col)==Val
%% We can now sum all the non-zero values in r to get the number of matches
n = sum(r)
M =
4 2 2 5
3 6 7 1
4 4 1 6
5 8 7 8
Val =
6
Col =
2
r =
0
1
0
0
n =
1
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
I have a matrix S something like:
1 4 7
2 5 8
3 6 9
Then I make a=complex(S{2},S{3}) and wanted to find the abs(a);. This is not possible in MATLAB as a is not an integer - it is a matrix. How can I get the magnitude of each row of matrix a?
PS: the matrix is read from a text file using textscan() as S = textscan(fileID,'%d %d %d', 'delimiter','\t');.
Second question:
Assuming again hav the following S matrix.
1 4 7 2 1
2 5 8 3 4
3 6 9 6 8
Now I wanted to arrange them in such way that column 2,3 and 4,5 alternate like this:
4 2
7 1
5 3
8 4
6 6
9 8
How can I do that without using a loop?
Thanks.
Going with my assumption in the comments, I'm going to assume that the second column consists of your real component of your matrix while the third column consists of your imaginary components. Your matrix S is actually a cell array of elements. You don't need to use complex then abs. You can simply take each of the columns, individually square them, add them together and take the square root. What I would do is convert the cell array into a 2D matrix, cast it to double to allow for floating point precision when finding the magnitude, and do what I just did above. This is necessary because abs and sqrt will only work for floating-point numbers. Your elements in S are already int32 due to the %d delimiter from textread. In other words:
Smat = double(cell2mat(S));
realComp = Smat(:,2);
imagComp = Smat(:,3);
mag = sqrt(realComp.^2 + imagComp.^2);
mag will thus return the magnitude of each row for you, assuming that the second column is the real component and the third component is the imaginary component as we specified.
However, if you're dead set on using complex and abs, you can do it like so:
Smat = double(cell2mat(S));
imagNumbers = complex(Smat(:,2), Smat(:,3));
mag = abs(imagNumbers);
This should still give you the same results as we talked about above.
Edit
Seeing your edit in your post above, we can achieve that quite easily by subsetting the matrix, then applying reshape to each part of the matrix you want. In other words:
Smat = double(cell2mat(S));
realMat = Smat(:,2:3); %// Grab second and third columns
imagMat = Smat(:,4:5); %// Grab fourth and fifth columns
realCol = reshape(realMat.', [], 1); % // Form the columns like you specified
imagCol = reshape(imagMat.', [], 1);
finalMatrix = [realCol imagCol];
finalMatrix should contain those two columns that you specified above in a single matrix.