matrix
Why does m[1] – m[0] return 3 where m is a 3×3 matrix?
In your code: m[1] – m[0] denotes a pointer subtraction which gives you the difference of the two pointers based on the type. In this case, both the pointers are differentiated by 3 elements, so the result is 3. To quote C11 standard, chapter §6.5.6 When two pointers are subtracted, both shall point to elements … Read more
How to extract all columns but one from an array (or matrix) in python?
Since for the general case you are going to be returning a copy anyway, you may find yourself producing more readable code by using np.delete: >>> a = np.arange(12).reshape(3, 4) >>> np.delete(a, 2, axis=1) array([[ 0, 1, 3], [ 4, 5, 7], [ 8, 9, 11]])
How to get row and column from index?
For a zero-based index, the two operations are, where width is the width of the structure: row = index / width column = index % width Those are for C using integers, where the division rounds down and the % modulo operator gives the remainder. If you’re not using C, you may have to translate … Read more
Swap rows with columns (transposition) of a matrix in javascript [duplicate]
DuckDucking turned up this by Ken. Surprisingly, it’s even more concise and complete than Nikita’s answer. It retrieves column and row lengths implicitly within the guts of map(). function transpose(a) { return Object.keys(a[0]).map(function(c) { return a.map(function(r) { return r[c]; }); }); } console.log(transpose([ [1,2,3], [4,5,6], [7,8,9] ]));
Organizing felt tip pens: optimizing the arrangement of items in a 2D grid by similarity of adjacent items, using JS [updated]
I managed to find a solution with objective value 1861.54 by stapling a couple ideas together. Form unordered color clusters of size 8 by finding a min-cost matching and joining matched subclusters, repeated three times. We use d(C1, C2) = ∑c1 in C1 ∑c2 in C2 d(c1, c2) as the distance function for subclusters C1 … Read more
Replace sub part of matrix by another small matrix in numpy
Here is how you can do it: >>> A[3:5, 3:5] = B >>> A array([[ 1. , 1. , 1. , 1. , 1. ], [ 1. , 1. , 1. , 1. , 1. ], [ 1. , 1. , 1. , 1. , 1. ], [ 1. , 1. , 1. , 0.1, … Read more
Selecting only a specific number of rows fulfilling a condition
Try: A = [ 1 11 22 33 44 13 12 33 1 14 33 44 ]; idx = ( A(:,4)==33 ); A_new = A(idx,:) This is using logical indexing