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How to create matrix in python of repeating number?


By : Kath Reinarz
Date : October 16 2020, 08:10 PM
it should still fix some issue The python itertools module product() function can do this:
code :
for code in itertools.product( range(5), repeat=3 ):
    print(code)
(0, 0, 0)
(0, 0, 1)
(0, 0, 2)
(0, 0, 3)
...
(4, 4, 2)
(4, 4, 3)
(4, 4, 4)
import itertools
matrix = []
for code in itertools.product( range(5), repeat=3 ):
     matrix.append( list(code) )


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Python create list from smallest number of repeating values in a nested list


By : user3648538
Date : March 29 2020, 07:55 AM
To fix this issue I attacked it by building a tree of all the potential additions, and then finding the cheapest option among those. Here is a working (albeit ugly and unoptimized) example:
https://gist.github.com/1288835

Matlab: create band matrix with repeating and alternating values


By : Carl
Date : March 29 2020, 07:55 AM
I hope this helps you . You can use spdiags to specify the upper diagonals in a sparse matrix and then add the transpose of the strictly upper triangular part for exact symmetry:
code :
>> n  = 6;
>> a = 1;b = 2;c = 3;d = 4;e = 5;f = 6;g = 7;
>> n  = 6;
>> A = spdiags(repmat([[a;e] , [f;b] , [c;g] , [0;d]],n/2,1),0:3,n,n);
>> A = A + triu(A,1).';
>> issymmetric(A)
ans =
     1
>> full(A)
ans =
     1     2     3     4     0     0
     2     5     6     7     0     0
     3     6     1     2     3     4
     4     7     2     5     6     7
     0     0     3     6     1     2
     0     0     4     7     2     5
>> m = 13;
>> n = 6;
>> p = max([m,n]);
>> p = p + mod(p,2); % added to make p even.
>> A = spdiags(repmat([[a;e] , [f;b] , [c;g] , [0;d]],p/2,1),0:3,p,p);
>> A = A + triu(A,1).';
>> A = A(1:m,1:n);
>> full(A)
ans =
     1     2     3     4     0     0
     2     5     6     7     0     0
     3     6     1     2     3     4
     4     7     2     5     6     7
     0     0     3     6     1     2
     0     0     4     7     2     5
     0     0     0     0     3     6
     0     0     0     0     4     7
     0     0     0     0     0     0
     0     0     0     0     0     0
     0     0     0     0     0     0
     0     0     0     0     0     0
     0     0     0     0     0     0

How to make repeating matrix with increasing number without loop


By : мишель жюльен
Date : March 29 2020, 07:55 AM
This might help you Using a combination of repmat and reshape, you can achieve it as follows:
code :
A = [1 2];
n = 5;
D = 10;
B = reshape(repmat(A(:), 1, n) + D*(0:(n-1)), 1, [])
B = reshape(bsxfun(@plus, repmat(A(:), 1, n), D*(0:(n-1))), 1, [])

Create a 2d array/list in python 3 based on only one number to create a blank matrix


By : pSPS
Date : March 29 2020, 07:55 AM
This might help you I am attempting to create a 2D array/list that can dynamically change based on one number, so say I input the number 3, it would create a 2d array like this, with three rows and three columns. inputting 10 would create an array with 10 columns and 10 rows, etc. , If the arrays need to be filled with 0's, this should work:
code :
def createTestingEnv(numberOfCells):
   holder = []
   env = []
   for i in range(numberOfCells):
      holder.append(0)
   for i in range(numberOfCells):
      env.append(holder)

   print(env)

Create matrix of some repeating numbers


By : user3605987
Date : March 29 2020, 07:55 AM
hope this fix your issue You can use implicit expansion implicitly (introduced in MATLAB R2016b) or explicitly by using MATLAB's bsxfun to create that matrix:
code :
% Parameters
EN = 10;
n = 6;
step = 3;

% Matrix (implict expansion, MATLAB >= R2016b)
etable = (0:step:EN*step-1).' + (1:n)

% Matrix (bsxfun, MATLAB < R2016b)
etable_legacy = bsxfun(@plus, (0:step:EN*step-1).', (1:n))
etable =
    1    2    3    4    5    6
    4    5    6    7    8    9
    7    8    9   10   11   12
   10   11   12   13   14   15
   13   14   15   16   17   18
   16   17   18   19   20   21
   19   20   21   22   23   24
   22   23   24   25   26   27
   25   26   27   28   29   30
   28   29   30   31   32   33

etable_legacy =
    1    2    3    4    5    6
    4    5    6    7    8    9
    7    8    9   10   11   12
   10   11   12   13   14   15
   13   14   15   16   17   18
   16   17   18   19   20   21
   19   20   21   22   23   24
   22   23   24   25   26   27
   25   26   27   28   29   30
   28   29   30   31   32   33
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