Well I do not know if I used the exact term. I tried to find an answer on the net.
Here is what i need:
I have a matix
a = 1 4 7
2 5 8
3 6 9
If I do a(4) the value is 4. So it is reading first column top to buttom then continuing to next .... I don't know why. However,
What I need is to call it using two indices. As row and column:
a(1,2)= 4
or even better if i can call it in the following way:
a{1}(2)=4
What is this process really called (want to learn) and how to perform in matlab.
I thought of a loop. Is there a built in function
Thanks a lot
Check this:
a =
18 18 16 18 18 18 16 0 0 0
16 16 18 0 18 16 0 18 18 16
18 0 18 18 0 16 0 0 0 18
18 0 18 18 16 0 16 0 18 18
>> a(4)
ans =
18
>> a(5)
ans =
18
>> a(10)
ans =
18
I tried reshape. it is reshaping not converting into 2 indeces
If you've already got a matrix, you already can access it with two indices:
if you've got
a = 1 4 7
2 5 8
3 6 9
you can access it as
a(3,2) = 6
However, the indexing goes from the top left, row first then column. If you want to get at the "4" in the matrix then do:
a(1,2)
To reshape a vector/matrix/array, use reshape().
Or you could leave it as a one dimensional array and just use
((Column - 1) * 3) + Row - 1) as the index. 3 because there are three columns.
NB a(4) = 4 because of the way you've arranged columns and rows in the one dimensional array, yours is "loaded" as
R1C1,R2C1,R3C1, R1C2 etc wher R is row and C is column
If that's inconvenient then you just need to get whatever fills the array row then column so the above mapping would be
((Row - 1) * 3) + Column - 1)
Don't do Matlab so above code assumes array starts at 0, if not just add 1 to it.
Related
If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
I have a problem on a part of a program and I would appreciate some help.
My main objective is to use all possible pairs in two arrays. With some help i managed to get this
A = nchoosek(0:15, 2)
arr1 = A(:,1);
arr2 = A(:,2);
Result = arr1.*arr2 + arr1.^2 + arr2.^2;
I want to use all the combinations in arr1 and arr2 to solve the result equation and print out the result like this:
arr1 arr2 Result
0 0 0
1 1 3
2 0 4
and so on.. but not all the combinations are used when I try this approach. What should I do to get all the possible combinations?
Matlab has meshgrid function to eliminate loops for this purpose, for example
>> a1=[1:4];
>> a2=[0:3];
>> [x1,x2]=meshgrid(a1,a2);
>> r=x1.*x2+x1.^2+x2.^2;
or to use square once
>> r1=(x1+x2).^2-x1.*x2;
UPDATE: for your case you use 0:15 values, using them will result with
>> a1=[0:15];a2=[0:15];
>> [x1,x2]=meshgrid(a1,a2);
>> r=-x1.*x2+(x1+x2).^2;
>> size(r)
ans =
16 16
UPDATE 2 Note that your method doesn't create all pairs, for example (0,0) or (1,1) won't be there also only of one of the (x,y) (y,x) pairs will be there for x!=y values. Other than double loops the preferred approach is what I proposed. You can gather the results in a matrix in the form you want easily as well
>> n=size(r,1);
>> R=[reshape(x1,1,n*n); reshape(x2,1,n*n); reshape(r,1,n*n)]'
R =
0 0 0
0 1 1
0 2 4
0 3 9
0 4 16
0 5 25
0 6 36
0 7 49
...
15 6 351
15 7 379
15 8 409
15 9 441
15 10 475
15 11 511
15 12 549
15 13 589
15 14 631
15 15 675
Say I have a matrix A
A =
0 1 2
2 1 1
3 1 2
and another matrix B
B =
0 42
1 24
2 32
3 12
I want to replace each value in A by the one associated to it in B.
I would obtain
A =
42 24 32
32 24 24
12 24 32
How can I do that without loops?
There are several ways to accomplish this, but here is an short one:
[~,ind]=ismember(A,B(:,1));
Anew = reshape(B(ind,2),size(A))
If you can assume that the first column of B is always 0:size(B,1)-1, then it is easier, becoming just reshape(B(A+1,2),size(A)).
arrayfun(#(x)(B(find((x)==B(:,1)),2)),A)
I have a matrix of 2d lets assume the values of the matrix
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
17 24 1 8 15
11 18 25 2 9
This matrix is going to be divided into three different matrices randomly let say
b =
17 24 1 8 15
23 5 7 14 16
c =
4 6 13 20 22
11 18 25 2 9
d =
10 12 19 21 3
17 24 1 8 15
How can i know the index of the vectors in matrix d for example in the original matrix a,note that the values of the matrix can be duplicated.
for example if i want to know the index of {10 12 19 21 3} in matrix a?
or the index of {17 24 1 8 15} in matrix a,but for this one should return only on index value?
I would appreciate it so much if you can help me with this. Thank you in advance
You can use ismember with the 'rows' option. For example:
tf = ismember(a, c, 'rows')
Should produce:
tf =
0
0
1
0
0
1
To get the indices of the rows, you can apply find on the result of ismember (note that it's redundant if you're planning to use this vector for matrix indexing). Here find(tf) return the vector [3; 6].
If you want to know the number of the row in matrix a that matches a single vector, you either use the method explained and apply find, or use the second output parameter of ismember. For example:
[tf, loc] = ismember(a, [10 12 19 21 3], 'rows')
returns loc = 4 for your example. Note that here a is the second parameter, so that the output variable loc would hold a meaningful result.
Handling floating-point numbers
If your data contains floating point numbers, The ismember approach is going to fail because floating-point comparisons are inaccurate. Here's a shorter variant of Amro's solution:
x = reshape(c', size(c, 2), 1, []);
tf = any(all(abs(bsxfun(#minus, a', x)) < eps), 3)';
Essentially this is a one-liner, but I've split it into two commands for clarity:
x is the target rows to be searched, concatenated along the third dimension.
bsxfun subtracts each row in turn from all rows of a, and the magnitude of the result is compared to some small threshold value (e.g eps). If all elements in a row fall below it, mark this row as "1".
It depends on how you build those divided matrices. For example:
a = magic(5);
d = a([2 1 2 3],:);
then the matching rows are obviously: 2 1 2 3
EDIT:
Let me expand on the idea of using ismember shown by #EitanT to handle floating-point comparisons:
tf = any(cell2mat(arrayfun(#(i) all(abs(bsxfun(#minus, a, d(i,:)))<1e-9,2), ...
1:size(d,1), 'UniformOutput',false)), 2)
not pretty but works :) This would be necessary for comparisons such as: 0.1*3 == 0.3
(basically it compares each row of d against all rows of a using an absolute difference)
Suppose I have:
>> X = magic(5)
X =
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
How do I get i'th element from the second column?
I already figured that indices in (some?) collections in Octave are one-based, but I'm not sure if that holds for matrices, too.
See the index expressions section of the manual. To get the i'th element from second column:
X(i,2) # element 'i' from column 2
X(1:end,2) # the whole 2nd column
X(:,2) # same thing but shorter
x(:, [2 3]) # whole 2nd and 3rd column
Note that Octave is a language where array elements are in column-major order.