How to create a square matrix with consecutive numbers on each row?
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Ming
am 20 Aug. 2015
Bearbeitet: David Alejandro Ramirez Cajigas
am 18 Aug. 2021
Hi everyone,
Given a vector i.e. n=[1 12 25 78], is there any way to create a matrix A, such that
A=[ 1 2 3 4; 11 12 13 14; 23 24 25 26; 75 76 77 78]?
without FOR LOOP?
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Sebastian Castro
am 20 Aug. 2015
Yeah, for sure.
I'm sure there are more efficient ways to do this, but this one will show you a few examples of the "repmat" function to string together vectors and matrices (either row-wise or column-wise).
I first avoided hard-coding parameters by using a variable "nCols" for number of columns, which should be the same as number of rows (or numel(n)). Note that I had to transpose n to n' to meet your desired solution.
>> nCols = numel(n);
>> baseMatrix = repmat(n',[1 nCols])
baseMatrix =
1 1 1 1
12 12 12 12
25 25 25 25
78 78 78 78
Next, you have to make the pretty complicated matrix to add to that matrix above. I would copy-paste both of the terms below into MATLAB to see what each of those does. Basically, I create a column pattern and then a row pattern, and subtract them.
>> addMatrix = repmat(0:nCols-1,[nCols 1]) - repmat((0:nCols-1)',[1 nCols])
addMatrix =
0 1 2 3
-1 0 1 2
-2 -1 0 1
-3 -2 -1 0
Finally, add 'em up!
>> A = baseMatrix + addMatrix
A =
1 2 3 4
11 12 13 14
23 24 25 26
75 76 77 78
- Sebastian
4 Kommentare
Walter Roberson
am 18 Aug. 2021
N = 22;
v = [0:N];
M = toeplitz([v(1) fliplr(v(2:end))], v)
result = mod(tril(-tril(M)) + triu(M), N+1)
David Alejandro Ramirez Cajigas
am 18 Aug. 2021
Bearbeitet: David Alejandro Ramirez Cajigas
am 18 Aug. 2021
Bingo!
The answer is:
N=22
Top1=N
Top12=repmat(0:Top1-1,[Top1 1]) - repmat((0:Top1-1)',[1 Top1]); %genera matriz 0 hasta n
Top17=(tril(Top12,-1)*-1);
Top18=Top17+Top12;
Top19=Top17+Top18
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