In the MATLAB if:
x = rand(a,b,c);
What does the third input to rand, c, mean?
a: numbers of rows
b: numbers of columns
c: ??
In MATLAB, matrices are not limited to 2 dimensions (i.e. only rows and columns), but can have many higher dimensions.
The third input to the rand function is just telling rand the size of the third dimension that you want. By default it is 1, but in your case it will instead be c.
In general, the nth input to rand will be the size of the nth dimension of the matrix of random numbers it produces.
See the docs:
X = rand(sz1,...,szN) returns an sz1-by-...-by-szN array of random
numbers where sz1,...,szN indicate the size of each dimension. For
example, rand(3,4) returns a 3-by-4 matrix.
Related
I need obtain a vector with n numbers disposed in random order in MatLab. How can I do it?
If I use, for example, randi(10,1,10), I will obtain random values, but I have no garantee that all the numbers will be in the sequence.
For example:
I need a vector with all the numbers 1,2,3,4,5,6,7,8,9,10, disposed in a random order, like in 2,5,3,6,8,10,4,7,9,1.
It appears that you are looking for the randperm function. From the docs:
p = randperm(n) returns a row vector containing a random permutation of the integers from 1 to n inclusive.
Your example would be randperm(10).
There is an optional second argument that you could pass in to determine how many elements will be chosen. For example, randperm(10, 5) would choose five random numbers from 1 to 10.
You can also use the results of randperm to shuffle or select from an arbitrary vector. Say you wanted the numbers 101 to 110 in random order instead of 1 to 10:
nums = 101:110;
nums = nums(randperm(numel(nums)));
i have a char matrix (in matlab) 18x6692 and i want this to be a matrix with 1 row and 6692x18=120450 column.
I'm not able to do this, can you help me?
I also tried with a smaller matrix: from 2x4 to 1x8 with no results.
thank you
Simply use the colon operator and transpose the vector:
A = A(:).'
You can use the reshape function:
B = reshape(A,1,[]);
where A is the input matrix, 1 is the number of rows and [] is to indicate that the number of columns is to be calculated from the number of elements in A.
Note that this stacks all columns of A. If you want to concatenate along the rows, you can do this by transposing A first
B = reshape(A.',1,[]);
I am trying to find the largest vector inside a matrix compound by vectors with MATLAB, however I am having some difficulties, so I would be very thankful if someone help me. I have this:
The matrix paths (solution of a Dijkstra function), which is a 1000x1000 matrix, whose values are vectors of 1 row and different number of columns (when the columns are bigger than 10, the values appear as "1x11 double, 1x12 double, etc"). The matrix paths have this form:
1 2 3 ....
1 1 <1x20 double> <1x16 double>
2 <1x20 double> 2 [2,870,183,492,641,863,611,3]
3 <1x16 double> [3,611,863,641,492,183,870,2] 3
4 <1x25 double> <1x12 double> <1x14 double>
.
.
.
At first I thought in just finding the largest vector in the matrix by
B = max(length(paths))
However MATLAB returns B = 1000, value which is feasible, but not likely. When trying to find out the position of the vector by using:
[row,column] = find(length(paths) == B)
MATLAB returns row = 1, column = 1, which for sure is wrong... I have thought that maybe is a problem of how MATLAB takes the data. It is like it doesn't consider the entries of the matrix as vectors, because when I enter:
length(paths(3,2))
It returns me 1, but it should return 8 as I understand, also when introducing:
paths(3,2)
It returns [1x8 double] but I expect to see the whole vector. I don't know what to do, maybe a "for" loop, I really do not know if MATLAB takes the data of the matrix as vectors or as simple double values.
The cell with the largest vector can be found using cellfun and numel to get the number of elements in each numeric matrix stored in the cells of paths:
vecLens = cellfun(#numel,paths);
[maxLen,im] = max(vecLens(:));
[rowMax,colMax] = ind2sub(size(vecLens),im)
This gets a 1000x1000 numeric matrix vecLens containing the sizes, max gets the linear index of the largest element, and ind2sub translates that to row,column indexes.
A note on length: It gives you the size of the largest dimension. The size of paths is 1000x1000, so length(paths) is 1000. My advice is, Don't ever use length. Use size, specifying the dimension you want.
If multiple vectors are the same length, you get the first one with the above approach. To get all of them (starting after the max command):
maxMask = vecLens==maxLen;
if nnz(maxMask)>1,
[rowMax,colMax] = find(maxMask);
else
[rowMax,colMax] = ind2sub(size(vecLens),im)
end
or just
[rowMax,colMax] = find(vecLens==maxLen);
Suppose in MATLAB I have a real matrix A which is n x m and a binary matrix B of the same size. The latter matrix defines the optimization set (all indices for which the element of B equals one): over this set I would like to find the maximal element of A. How can I do this?
The first idea I had is that I consider C = A.*B and look for the maximal element of C. This works fine for all matrices A which have at least one positive element, however it does not work for matrices with all negative elements.
You can do
C = A(B==1);
to give you an array of just the values of A corresponding to a value of 1 in B. And
max( C )
will give you the maximum value of A where B is 1
With this method you don't run into a problem when all values of A are negative as the zeros don't appear in C.
Obviously you can condense this to
desiredValue = max(A(B(:)==1));
I am using the colon operator to make sure that the result of A(B(:)==1) is a column vector - if B is all ones I am not sure if Matlab would return a vector or a nxm matrix (and I can't confirm right now).
update to get the index of the value, you can do:
f = find(B==1);
[m mi] = max(A(f));
maxIndex = f(mi);
And to get that back to the 2D elements:
[i j] = ind2sub(size(A), maxIndex);
I am currently trying to code up a function to assign probabilities to a collection of vectors using a histogram count. This is essentially a counting exercise, but requires some finesse to be able to achieve efficiently. I will illustrate with an example:
Say that I have a matrix X = [x1, x2....xM] with N rows and M columns. Here, X represents a collection of M, N-dimensional vectors. IN other words, each of the columns of X is an N-dimensional vector.
As an example, we can generate such an X for M = 10000 vectors and N = 5 dimensions using:
X = randint(5,10000)
This will produce a 5 x 10000 matrix of 0s and 1s, where each column is represents a 5 dimensional vector of 1s and 0s.
I would like to assign a probability to each of these vectors through a basic histogram count. The steps are simple: first find the unique columns of X; second, count the number of times each unique column occurs. The probability of a particular occurrence is then the #of times this column was in X / total number of columns in X.
Returning to the example above, I can do the first step using the unique function in MATLAB as follows:
UniqueXs = unique(X','rows')'
The code above will return UniqueXs, a matrix with N rows that only contains the unique columns of X. Note that the transposes are due to weird MATLAB input requirements.
However, I am unable to find a good way to count the number of times each of the columns in UniqueX is in X. So I'm wondering if anyone has any suggestions?
Broadly speaking, I can think of two ways of achieving the counting step. The first way would be to use the find function, though I think this may be slow since find is an elementwise operation. The second way would be to call unique recursively as it can also provide the index of one of the unique columns in X. This should allow us to remove that column from X and redo unique on the resulting X and keep counting.
Ideally, I think that unique might already be doing some counting so the most efficient way would probably be to work without the built-in functions.
Here are two solutions, one assumes all values are either 0's or 1's (just like the example in your description), the other does not. Both codes should be very fast (more so the one with binary values), even on large data.
1) only zeros and ones
%# random vectors of 0's and 1's
x = randi([0 1], [5 10000]); %# RANDINT is deprecated, use RANDI instead
%# convert each column to a binary string
str = num2str(x', repmat('%d',[1 size(x,1)])); %'
%# convert binary representation to decimal number
num = (str-'0') * (2.^(size(s,2)-1:-1:0))'; %'# num = bin2dec(str);
%# count frequency of how many each number occurs
count = accumarray(num+1,1); %# num+1 since it starts at zero
%# assign probability based on count
prob = count(num+1)./sum(count);
2) any positive integer
%# random vectors with values 0:MAX_NUM
x = randi([0 999], [5 10000]);
%# format vectors as strings (zero-filled to a constant length)
nDigits = ceil(log10( max(x(:)) ));
frmt = repmat(['%0' num2str(nDigits) 'd'], [1 size(x,1)]);
str = cellstr(num2str(x',frmt)); %'
%# find unique strings, and convert them to group indices
[G,GN] = grp2idx(str);
%# count frequency of occurrence
count = accumarray(G,1);
%# assign probability based on count
prob = count(G)./sum(count);
Now we can see for example how many times each "unique vector" occurred:
>> table = sortrows([GN num2cell(count)])
table =
'000064850843749' [1] # original vector is: [0 64 850 843 749]
'000130170550598' [1] # and so on..
'000181606710020' [1]
'000220492735249' [1]
'000275871573376' [1]
'000525617682120' [1]
'000572482660558' [1]
'000601910301952' [1]
...
Note that in my example with random data, the vector space becomes very sparse (as you increase the maximum possible value), thus I wouldn't be surprised if all counts were equal to 1...