Related
Let's say we have the following matrix
A=magic(4)
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
and we want to extract 3 submatrices, identified by the indexes for top left and bottom right corners. The indexes for a submatrix are contained in a row of the matrix i; columns 1 and 2 of i are the row indexes of the corners, columns 3 and 4 of i are the column indexes of the corners.
i.e.
i =
1 1 1 3
2 4 1 2
3 4 3 4
>> A(i(1,1):i(1,2),i(1,3):i(1,4))
ans =
16 2 3
>> A(i(2,1):i(2,2),i(2,3):i(2,4))
ans =
5 11
9 7
4 14
>> A(i(3,1):i(3,2),i(3,3):i(3,4))
ans =
6 12
15 1
The command A(i(,):i(,),i(,):i(,)) which I used to extract the submatrices is not very convenient, so I wonder is there a better way to do the job ?
If you don't want to type it all out then why not write a wrapper function?
A = magic(4);
S = #(r) A(i(r,1):i(r,2),i(r,3):i(r,4));
S(1)
S(2)
S(3)
If A may change after the definition of S then you would need to make it a parameter to the function.
S = #(A,r) A(i(r,1):i(r,2),i(r,3):i(r,4));
A = magic(4)
S(A,1)
S(A,2)
S(A,3)
Similarly if i may change then you would need to make it a parameter as well.
Edit
Unfortunately, contrary to my comment, if you want to perform assignment then A(I(r)) won't work exactly the same as what you've posted since this always returns an array instead of a matrix. One possible workaround is to use cell arrays in place of comma-separated-lists, but this isn't as elegant as the read only option. For example
S = #(r) {i(r,1):i(r,2) , i(r,3):i(r,4)};
s = S(1); A(s{:})
s = S(2); A(s{:})
s = S(3); A(s{:})
Following the same principle you could pre-define a cell array from i to make access one line.
s = arrayfun(#(r) {i(r,1):i(r,2),i(r,3):i(r,4)}, 1:size(i,1), 'UniformOutput', false);
A(s{1}{:})
A(s{2}{:})
A(s{3}{:})
I got a 4-by-n matrix, like
A =
1 5 9
3 0 6
2 3 10
7 8 4
What I want to do with A is getting each half column of A as
Line1Point1 = [1 3]
Line1Point2 = [2 7]
Line2Point1 = [5 0]
Line2Point2 = [3 8]
Line3Point1 = [9 6]
Line3Point2 = [10 4]
How could I do that? I’m pretty new to matlab coding.. Any help is really appreciated..
Cheers
Use reshape function, for example:
>> A = [1 5 9;
3 0 6;
2 3 10;
7 8 4];
>> reshape(A,2,6)
ans =
1 2 5 3 9 10
3 7 0 8 6 4
Storing such information as many variables is generally a bad idea
Some options for storing and accessing are
Cell array
Line=mat2cell(A,[2,2],ones(1,size(A,2))).'
access with
Line{2,1}
ans =
5
0
Indexing
as other answers
Anonymous Function
Line=#(l,p)A(2*p-1:2*p,l)
access with
Line(2,1)
ans =
5
0
Structure
Not really a useful solution, more for interests sake
for ii=1:size(A,2);for jj=1:2;Line(ii).Point(jj).Value=A(2*jj-1:2*jj,ii);end;end
access with
Line(2).Point(1).Value
ans =
5
0
A(1:2,1) will give you first half of the first column.
A(3:4,1) will give you second half of the first column.
A(1:2,2) will give you first half of the second column.
A(3:4,2) will give you second half of the second column.
A(1:2,3) will give you first half of the third column.
A(3:4,3) will give you second half of the third column.
You can create the variables with the eval function, which executes the input string. Using eval is commonly regarded as bad practice since it is horrible to debug.
Nevertheless, here's the code:
A = [1 5 9; 3 0 6; 2 3 10; 7 8 4];
for ii = 1:length(A(1,:))
eval(['Line' num2str(ii) 'Point1 = A(1:2, ii)' ]);
eval(['Line' num2str(ii) 'Point2 = A(3:4, ii)' ]);
end
% Now all variables are created - for example: Line2Point1
A more elegant solution could be to store the vectors in a cell array. You can acces the first vectors for example by typing: c{1,1}
c = cell(length(A(1,:)),2)
for ii = 1:length(A(1,:))
c{ii,1} = A(1:2, ii);
c{ii,2} = A(3:4, ii);
end
I would suggest using 3D arrays to store and then access those values.
Code
N = size(A,1)/2;
LinePoint = permute(reshape(A,N,size(A,1)/N,[]),[1 3 2])
Here,
2nd dimension indices (columns) would represent Line IDs
3rd dimension indices would represent Point IDs.
Thus, the representative 3D array would be - LinePoint(:,LineID,PointID).
Example run
For your given A, we would have LinePoint as -
LinePoint(:,:,1) =
1 5 9
3 0 6
LinePoint(:,:,2) =
2 3 10
7 8 4
Thus,
Line1Point1 would be denoted by LinePoint(:,1,1)
Line1Point2 would be denoted by LinePoint(:,1,2)
Line2Point1 would be denoted by LinePoint(:,2,1)
Line2Point2 would be denoted by LinePoint(:,2,2)
Line3Point1 would be denoted by LinePoint(:,3,1)
Line3Point2 would be denoted by LinePoint(:,3,2)
Lets say I have matrix such that A(:,:,1)=[1,2,3;2,3,4], A(:,:,2)=[3,4,5;4,5,6].
How is the easiest way of accessing and plotting the vectors (1,2,3),(2,3,4),(3,4,5),(4,5,6). I tried creating B=[A(:,:,1);A(:,:,2)], but i need a procedure to arbitrary number of A's.
Hope this isn't trivial and I've formulated myself satisfactory.
You should think 'vertically'. This will allow you to use colon indexing:
>> A(:,:,1) = [1,2,3;2,3,4].'; %'// NOTE: transpose of your original
>> A(:,:,2) = [3,4,5;4,5,6].'; %'// NOTE: transpose of your original
>> A(:,:)
ans =
1 2 3 4
2 3 4 5
3 4 5 6
The colon indexing with two colons works for any dimension A:
>> A(:,:,:,:,1,1) = [1 2 3; 2 3 4].'; %'
>> A(:,:,:,:,2,1) = [3 4 5; 4 5 6].'; %'
>> A(:,:,:,:,1,2) = [5 6 7; 6 7 8].'; %'
>> A(:,:,:,:,2,2) = [7 8 9; 8 9 0].'; %'
>> A(:,:)
ans =
1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 4 5 6 7 8 9 0
Colon indexing in MATLAB is quite interesting and really powerful once you master it. For example, if you use fewer colons than there are dimensions in the array (like above), MATLAB will automatically concatenate the remainder of the data along the dimension equal to the colon count.
So, if A has 48 dimensions, but you index with just 2 colons: you'll get a 2D array, that is the concatenation of the remaining 46 dimensions along the 2nd dimension.
In general: if A has N dimensions, but you index with just M ≤ N colons: you'll get an M-D array, that is the concatenation of the remaining N-M dimensions along the Mth dimension.
So as long as you are free to define your A to contain vectors on the columns rather than the rows (you should advise everyone to do this, as virtually everything in MATLAB is a bit faster that way), I think this is the fastest and most elegant way to do what you want.
If not, well, then just reshape like Dan :)
Assuming the order does not matter, here is how you can do it for vectors of length 3:
B = reshape(shiftdim(A,2), [], 3)
plot(B')
For vectors of arbitrary dimensions, replace 3 by size(A,2)
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
Here is an array i need to extract the exact values where the increasing and decreasing trend starts.
the output for the array a will be [2(first element) 2 6 9]
a=[2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2].
^ ^ ^ ^
| | | |
Kindly help me to get the result in MATLAB for any similar type of array..
You just have to find where the sign of the difference between consecutive numbers changes.
With some common sense and the functions diff, sign and find, you get this solution:
a = [2 3 6 7 2 1 0.01 6 8 10 12 15 18 9 6 5 4 2];
sda = sign(diff(a));
idx = [1 find(sda(1:end-1)~=sda(2:end))+2 ];
result = a(idx);
EDIT:
The sign function messes things up when there are two consecutive numbers which are the same, because sign(0) = 0, which is falsely identified as a trend change. You'd have to filter these out. You can do this by first removing the consecutive duplicates from the original data. Since you only want the values where the trend change starts, and not the position where it actually starts, this is easiest:
a(diff(a)==0) = [];
This is a great place to use the diff function.
Your first step will be to do the following:
B = [0 diff(a)]
The reason we add the 0 there is to keep the matrix the same length because of the way the diff function works. It will start with the first element in the matrix and then report the difference between that and the next element. There's no leading element before the first one so is just truncates the matrix by one element. We add a zero because there is no change there as it's the starting element.
If you look at the results in B now it is quite obvious where the inflection points are (where you go from positive to negative numbers).
To pull this out programatically there are a number of things you can do. I tend to use a little multiplication and the find command.
Result = find(B(1:end-1).*B(2:end)<0)
This will return the index where you are on the cusp of the inflection. In this case it will be:
ans =
4 7 13
I want to get a vector of elements of a Matlab matrix at predefined locations. For example, I have the following
>> i = [1,2,3];
>> j = [1,3,4];
>> A = [1,2,3,4; 5,6,7,8; 9,10,11,12; 13,14,15,16]
A =
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
I want a vector that will give me the values of A at the locations correspongin to i,j. I tried
A(i,j)
ans =
1 3 4
5 7 8
9 11 12
but this is not what I wanted. I want to get the following
>> [A(i(1),j(1)); A(i(2),j(2));A(i(3),j(3))]
ans =
1
7
12
What is the matlab syntax for that? Please, avoid suggesting for loops or anything that is not in a vectorized form, as I need this to be done fast. Hopefully there will be some built-in function.
to get it in the fastest way, use linear indexing:
A((j-1)*size(A,1)+i)
remember that MATLAB uses a column-major order.
A(sub2ind(size(A),i,j))
If you really crave speed, you might try making your own copy of sub2ind.m that strips out all the input-checking that that function does.
To understand how to do this, it is best to understand how matlab stores its arrays. In the matrix:
i = [1,2,3];
j = [1,3,4];
A = [1,2,3,4; 5,6,7,8; 9,10,11,12; 13,14,15,16]
matlab stores the elements going DOWN the columns. So they actually reside in memory in the order:
{1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16}
You can actually index a 2-d array using a SINGLE index. This is why the sub2ind trick works. Since the elements that you want to get are the 1st, 10th and 15th elements, as stored in a column-wise ordering,
singleIndex = [1 10 15];
A(singleIndex)
ans =
1 7 12
To confirm that sub2ind gave that index list, try it...
ind = sub2ind([4 4],i,j)
ind =
1 10 15
For now I'm using this:
>> diag(A(i,j))
ans =
1
7
12