I have an array:
v = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20];
I want to extract multiple arrays of length 3 equally spaced by 6 elements, starting at the fifth element of v, and then combine them together:
v1 = [5 6 7];
v2 = [11 12 13];
v3 = [17 18 19];
v_combined = [5 6 7 11 12 13 17 18 19];
Are there any simple ways to do this without using a for loop?
You can do it using logical indexing. You need to create an index mask like this
idx = [0 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0]
which you can create like this:
idx = false(size(v))
k = 5
idx(k:end) = ~mod(floor((0:numel(v)-k)/3),2)
And finally
v_combined = v(idx)
In general for m elements spaces by n elements starting from k you can use
k=5;
m=3;
n=6;
I=1:numel(v);
v_combined = v((I>=k) & mod(I-k,n)<m)
Related
I have managed to transfer the 2d array int a 1d array to make it simpler. So given:
local table = {0,0,0,0,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,0,30,31,32,33, 0,34,35,36,37, 0 }
How can I extract a sub-array as below?
local sub = {2,3,4,7,8,9,12,13,14,17,18,19,22,23,24}
Here's one way...
> t = {}
> for i = 7, 30, 5 do
>> for j = i, i + 2 do
>> t[#t+1] = ble[j]
>> end
>> end
> for k,v in ipairs(t) do print(k,v) end
1 2
2 3
3 4
4 7
5 8
6 9
7 12
8 13
9 14
10 17
11 18
12 19
13 22
14 23
15 24
>
I renamed your table to ble so it doesn't shadow the table library.
This question already has an answer here:
Vector as column index in matrix
(1 answer)
Closed 7 years ago.
While coding in GNU Octave/MATLAB I came through this simple problem I couldn't figure out by myself: I'm trying to select some elements of a matrix by using some indexes stored in an array. Let me put it clear with an example:
Given:
A = zeros(5, 3)
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
I would like to select some elements in A matrix row-wise, by using the values in the auxiliary array B as subindices.
Ie. the following B array
B = [ 1 3 2 1 3 ]'
1
3
2
1
3
should be read as:
1 -> index '1' on first row (element [1, 1])
3 -> index '3' on second row (element [2, 3])
2 -> index '2' on third row (element [3, 2])
1 -> index '1' on fourth row (element [4, 1])
3 -> index '3' on fifth row (element [5, 3])
Therefore, if we assign value '1' to the elements selected using the aforementioned criteria, the resulting matrix would be:
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
I believe this is a simple operation and I'm convinced that there must be a way to achieve the described behaviour without having to loop across the rows in matrix A.
Thank you.
Edit: Rewrite question so that it is (hopefully) less confusing.
Your question is a bit confusing. You're saying you want to select the elements in A by using the values in the vector B as column indexes, but your example sets (not gets) new values in matrix A. I'm explaining both cases.
Consider this matrix
A = magic(5)
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
Say you want to get/set the diagonal elements of A.
Index pairs in that case are [1,1], [2,2], [3,3], [4,4] and [5,5].
To access elements as a vector, run this
A(sub2ind([5,5], (1:5)',(1:5)'))
17
5
13
21
9
To set elements run this
A(sub2ind([5,5], (1:5)',(1:5)')) = 0
0 24 1 8 15
23 0 7 14 16
4 6 0 20 22
10 12 19 0 3
11 18 25 2 0
These commands can be written as
r = 1:5
c = 1:5
A(sub2ind([max(r),max(c)], r',c'))
# to assign values
A(sub2ind([max(r),max(c)], r',c')) = 0
# and to assign different value to each index pair
A(sub2ind([max(r),max(c)], r',c')) = [20 10 50 12 99]
In your example,
r = 1:5
c = B'
A(sub2ind([max(r),max(c)], r',c')) = 1
# or simply A(sub2ind([max(r),max(B)], r',B)) = 1
1 0 0
0 0 1
0 1 0
1 0 0
0 0 1
You can read how sub2ind works here.
I have a matrix of unsorted numbers and I want to keep the n largest (not necessarily unique) values per column and set the rest to zero.
I figured out how to do it with a loop:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4]
k = 2
for n = 1:4
[y, ind] = sort(a(:,n), 'descend');
a(ind(k+1:end),n) = 0;
end
a
which gives me:
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
k =
2
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
However when I try to eliminate the loop, I can't seem to get the indexing right, because this:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4]
k = 2
[y, ind] = sort(a, 'descend');
b = ind(k+1:end,:)
a(b) = 0
which gives me this: (which is not what I wanted to do)
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
k =
2
b =
4 3 3 1
1 2 2 4
a =
0 8 12 5
0 2 6 18
0 3 9 7
0 9 12 4
Am I indexing this wrong? Do I have to use the loop?
I referenced this question to get started but it wasn't exactly what I was trying to do: How to find n largest elements in an array and make the other elements zero in matlab?
You're very close. ind in the sort function gives you the row locations for each column where that particular value would appear in the sorted output. You need to do some additional work if you want to index into the matrix properly and eliminate the entries. You know that for each column of I, that tells you that we need to eliminate those entries from that particular column. Therefore, what I would do is generate column-major linear indices using each column of I to be the rows we need to eliminate.
Try doing this:
a = [4 8 12 5; 9 2 6 18; 11 3 9 7; 8 9 12 4];
k = 2;
[y, ind] = sort(a, 'descend');
%// Change here
b = sub2ind(size(a), ind(k+1:end,:), repmat(1:size(a,2), size(a,1)-k, 1));
a(b) = 0;
We use sub2ind to help us generate our column major indices where the rows are denoted by the values in ind after the kth element and the columns we need are for each column in this matrix. There are size(a,1)-k rows remaining after you truncate out the k values after sorting, and so we generate column values that go from 1 up to as many columns as we have in a and as many rows as there are remaining.
We get this output:
>> a
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
Here's one using bsxfun -
%// Get descending sorting indices per column
[~, ind] = sort(a,1, 'descend')
%// Get linear indices that are to be set to zeros and set those in a to 0s
rem_ind = bsxfun(#plus,ind(n+1:end,:),[0:size(a,2)-1]*size(a,1))
a(rem_ind) = 0
Sample run -
a =
4 8 12 5
9 2 6 18
11 3 9 7
8 9 12 4
n =
2
ind =
3 4 1 2
2 1 4 3
4 3 3 1
1 2 2 4
rem_ind =
4 7 11 13
1 6 10 16
a =
0 8 12 0
9 0 0 18
11 0 0 7
0 9 12 0
How do I combine all the vectors in small subsets of vectors in Matlab?
a= [5 6 7] b = [8 9 10] c=[11 12 13] d=[14 15 16] e=[17 18 19]
a combine with b and c:
Outcome:
M1= [ 5 6 7 8 9 10 11 12 13]
a with b and d:
M2 = [5 6 7 8 9 10 14 15 16]
and so on .....
This answer covers the case for an arbitrary number of vectors. The vectors are assumed to be row vectors of equal length.
Let your example data be defined as
a = [5 6 7]; b = [8 9 10]; c = [11 12 13]; d = [14 15 16]; e = [17 18 19];
vectors = {a, b, c, d, e}; %// cell array with any number of row vectors of equal size
n = 3; %// desired subset size
Then: generate all combinations of indices, use that to index into vectors, concatenate into one big row vector, and reshape that to obtain the desired result:
combs = nchoosek(1:numel(vectors), n);
result = reshape([vectors{combs.'}], numel(vectors{1})*n, []).';
This gives a matrix whose first row is your M1, second row is M2 etc:
result =
5 6 7 8 9 10 11 12 13
5 6 7 8 9 10 14 15 16
5 6 7 8 9 10 17 18 19
5 6 7 11 12 13 14 15 16
5 6 7 11 12 13 17 18 19
5 6 7 14 15 16 17 18 19
8 9 10 11 12 13 14 15 16
8 9 10 11 12 13 17 18 19
8 9 10 14 15 16 17 18 19
11 12 13 14 15 16 17 18 19
You can use the cat function:
res = cat(2,a,b,c);
or simply the [] syntax:
res = [a b c];
In both cases, res will contain [5 6 7 8 9 10 11 12 13].
Best,
Use nchoosek.
a = [5 6 7]; b = [8 9]; c = [11 12 13]; d = [14 15 16]; e = [17 18 19];
N = 3;
x = {a,b,c,d,e};
y = nchoosek(x,N);
And you have all the combination of your arrays taken N at a time in a cell array. Each row i of the cell x is a combination, so to have it back as a row vector just do
horzcat(y{i,:})
Or, if you want to get them all and put them in a cell array of size n_combs
n_combs = size(y,1);
out = cell(0,n_combs);
for i = 1 : n_combs
out{i} = horzcat(1, y{i,:});
end
There is no constraint on the size of the arrays that you want to combine, e.g., you can combine
a = [5 7]; b = [8 9]; c = [11]; d = [20 14 15 16]; e = [17 18 19];
However, if you must put together all the combination in a matrix, then the arrays have to be of the same size. In this case Luis Mendo's answer does the job.
Finally, if repetitions are allowed use nmultichoosek instead of nchoosek.
I am considering an easy algorithm to rank my 2D array and mark their rank in the same size of the 2D array.
For example, I have a matrix in below:
[0 2 15 34;
0 15 21 24;
0 3 5 8;
1 14 23 29]
The output should be as follow:
[1 5 10 16;
1 10 12 14;
1 6 7 8;
4 9 13 15]
I am kind of new to matlab, I not sure if the matlab have the functionality to directly do it. Or it would be even better if you could provide some ideas for implementing the algorithm. Thank you very much!
If I understand correctly, you want to replace each element by its rank. I offer three ways to do it; the third seems to be what you want.
Let your example data be defined as
data = [0 2 15 34;
0 15 21 24;
0 3 5 8;
1 14 23 29];
This assigns equal ranks to equal data values (as in your example), but doesn't skip ranks in that case (your example seems to do so):
[~, ~, vv] = unique(data(:));
result = reshape(vv, size(data));
With your example data, this gives
result =
1 3 8 13
1 8 9 11
1 4 5 6
2 7 10 12
This assigns different ranks to equal data values (so skipping ranks is out of the question):
[~, vv] = sort(data(:));
[~, vv] = sort(vv);
result = reshape(vv, size(data));
With your example data,
result =
1 5 11 16
2 10 12 14
3 6 7 8
4 9 13 15
This assigns equal ranks to equal data values, and in that case it skips ranks:
[~, vv] = sort(data(:));
[~, vv] = sort(vv);
[~, jj, kk] = unique(data(:), 'first');
result = reshape(vv(jj(kk)), size(data));
With your example data,
result =
1 5 10 16
1 10 12 14
1 6 7 8
4 9 13 15
Another approach, single-line: for each entry, find how many other entries are smaller, and add 1:
result = reshape(sum(bsxfun(#lt,data(:),data(:).'))+1, size(data));