I have some problems using reshape. I want to reshape a 4-dimensional matrix, and the fourth dimension has to become a column.
so if I have:
A(:,:,1,1) =
1 4
2 5
A(:,:,2,1) =
2 5
3 6
A(:,:,1,2) =
10 14
12 15
A(:,:,2,2) =
12 15
13 16
My reshape should be:
Columns 1 through 5
1 4 2 5 2
10 14 12 15 12
Columns 6 through 8
5 3 6
15 13 16
This should work:
reshape(permute(A,[1 4 2 3]),[2 8])
To understand this, you can go step by step. First do the following:
reshape(A,[2 8])
ans =
1 2 2 3 10 12 12 13
4 5 5 6 14 15 15 16
You observe that the columns of the reshaped matrix are taken by sliding across the second dimension in the original matrix. After second dimension is over, you slide over to third dimension and the re-iterate over second dimension (here first dimension is rows, second is column, so on...).
What you want to do is, iterate over 4th dimension (as if its a second dimension). You also want (4,14) after (1,10). You can see that corresponding elements vary across second dimension (but reshape is going to slide over third dimension, no matter what. So swap 2nd and 3rd dimensions).
Finally, you get, reshape(permute(A,[1 4 2 3]),[2 8]).
I have always had a hard time explaining permute to somebody. I hope I didn't confuse you more.
You're going to have to do a permutation first to put the dimensions in the appropriate order. Try this:
reshape(permute(A,[4,2,1,3]),[2,8])
or break it up into two separate permutations, one to switch dimensions 1 and 2 in the original array, then reshape to 8x2, then take the transpose:
reshape(permute(A,[2,1,3,4]),[8,2])'
Related
I am trying to achieve multiple matrices that will cover the full set of numbers. For example say I want to generate 5 matrices of length 10 that cover all the numbers from 1-20.
So matrix one will contain half the numbers say
m1 = [1 2 3 4 5 6 7 8 9 10];
while matrix two contains
m2 = [11 12 13 14 15 16 17 18 19 20];
Although this satisfies my condition with only two matrices not 5, I preferably need to generate all matrices randomly. Other than randomly generating the matrices and checking all values are generated is there a more efficient way to do this?
You can do it like that:
>> l=[1:20,randi(20,1,30)];
>> vec=l(randperm(length(l)));
>> v=reshape(vec,5,10);
The first line generates an array of 50 numbers from 1 to 20. It guarantees that each such number appears at least once. The second line randomizes the order of the numbers. The third line reshapes the vector into an array of arrays (that is, a 2D matrix, where each row is one of the arrays).
I have a matrix A
A =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
i have another matrix to specify the indices
index =
1 2
3 4
Now, i have got third matrix C from A
C = A(index)
C =
1 6
11 16
Problem: I am unable to understand how come i have received such a matrixC. I mean, what is logi behind it?
The logic behind it is linear indexing: When you provide a single index, Matlab moves along columns first, then along rows, then along further dimensions (according to their order).
So in your case (4 x 5 matrix) the entries of A are being accessed in the following order (each number here represents order, not the value of the entry):
1 5 9 13 17
2 6 10 14 18
3 7 11 15 19
4 8 12 16 20
Once you get used to it, you'll see linear indexing is a very powerful tool.
As an example: to obtain the maximum value in A you could just use max(A(1:20)). This could be further simplified to max(A(1:end)) or max(A(:)). Note that "A(:)" is a common Matlab idiom, used to turn any array into a column vector; which is sometimes called linearizing the array.
See also ind2sub and sub2ind, which are used to convert from linear index to standard indices and vice versa.
all.
I have a 15 element array = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15];.
I was wondering if there was a command such that it would step through iterations of the array without repeating itself. In other words, since there is a chance that randperm() will create the same matrix twice, I want to step through each permutation only once and perform a calculation.
I concede that there are factorial(15) permutations, but for my purposes, these two vectors (and similar) are identical and don't need to be counted twice:
[1 2 3 4 5 6 7 8 9 10 11 12 13 14 15]
[15 14 13 12 11 10 9 8 7 6 5 4 3 2 1]
Thus, is there any way to step through this?
Thanks.
I think what you are looking for is perms. randperm returns a single random permutation, you want all the permutations.
So use
my_permuations = perms([1:15]);
If forward-backward is the same as backward-foward then you can use the top half of the list only...
my_permutation_to_use = my_permutations(1:length(my_permutations)/2, :);
You may compare all permutations, but this would require to store all past permutations. Instead a local decision is better. I recommend this simple rule:
A permutation is valid, if the first element is smaller than the last element.
A permutation is redundant, if the first element is larger than the last element.
For small sizes, this could simply be done with this code:
%generate all permutations
x=perms(1:10)
%select only the valid lines, remove all redundant lines
x(x(:,1)<x(:,end),:)
Remains the problem, that generating x for 1:15 breaks all memory limits and would require about 100h.
Edit for clarity:
I have two matrices, p.valor 2x1000 and p.clase 1x1000. p.valor consists of random numbers spanning from -6 to 6. p.clase contains, in order, 200 1:s, 200 2:s and 600 3:s. What I wan´t to do is
Print p.valor using a diferent color/prompt for each clase determined in p.clase, as in following figure.
I first wrote this, in order to find out which locations in p.valor represented where the 1,2 respective 3 where in p.clase
%identify the locations of all 1,2 respective 3 in p.clase
f1=find(p.clase==1);
f2=find(p.clase==2);
f3=find(p.clase==3);
%define vectors in p.valor representing the locations of 1,2,3 in p.clase
x1=p.valor(f1);
x2=p.valor(f2);
x3=p.valor(f3);
There is 200 ones (1) in p.valor, thus, is x1=(1:200). The problem is that each number one(1) (and, respectively 2 and 3) represents TWO elements in p.valor, since p.valor has 2 rows. So even though p.clase and thus x1 now only have one row, I need to include the elements in the same colums as all locations in f1.
So the different alternatives I have tried have not yet been succesfull. Examples:
plot(x1(:,1), x1(:,2),'ro')
hold on
plot(x2(:,1),x2(:,2),'k.')
hold on
plot(x3(:,1),x3(:,2),'b+')
and
y1=p.valor(201:400);
y2=p.valor(601:800);
y3=p.valor(1401:2000);
scatter(x1,y1,'k+')
hold on
scatter(x2,y1,'b.')
hold on
scatter(x3,y1,'ro')
and
y1=p.valor(201:400);
y2=p.valor(601:800);
y3=p.valor(1401:2000);
plot(x1,y1,'k+')
hold on
plot(x2,y2,'b.')
hold on
plot(x3,y3,'ro')
My figures have the axisies right, but the plotted values does not match the correct figure provided (see top of the question).
Ergo, my question is: how do I include tha values on the second row in p.valor in my plotted figure?
I hope this is clearer!
Values from both rows simultaneously can be accessed using this syntax:
X=p.value(:,findX)
In this case, resulting X matrix will be a matrix having 2 rows and length(findX) columns.
M = magic(5)
M =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
M2 = M(1:2, :)
M2 =
17 24 1 8 15
23 5 7 14 16
Matlab uses column major indexing. So to get to the next row, you actually just have to add 1. Adding 2 to an index on M2 here gets you to the next column, or adding 5 to an index on M
e.g. M2(3) is 24. To get to the next row you just add one i.e. M2(4) returns 5.To get to the next column add the number of rows so M2(2 + 2) gets you 1. If you add the number of columns like you suggested you just get gibberish.
So your method is very wrong. Freude's method is 100% correct, it's much easier to use subscript indexing than linear indexing for this. But I just wanted to explain why what you were trying doesn't work in Matlab. (aside from the fact that X=p.value(findX findX+1000) gives you a syntax error, I assume you meant X=p.value([findX findX+1000]))
I can easily flatten an entire matrix into one row using reshape(M,1,[]). However, this time I want to flatten every n rows into one row. Thus, if we start with 100 rows and n=10, we will end up with 10 rows.
e.g.
1 2 3
4 5 6
7 8 9
10 11 12
with n=2 changes into
1 2 3 4 5 6
7 8 9 10 11 12
Is there a simple way to do this?
Suppose your original matrix is m, then:
reshape(m',[6 2])'
produces the required output. I'll leave it to you to generalise to other cases; comment or post again if that causes you problems.
This should work.
reshape(M',l/n,n)'
Where n is what you've defined and l is the total elements in M.
EDIT: Made it one-liner