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By : user3044423
Date : December 01 2020, 04:47 PM
will be helpful for those in need I expect that by "cuboid", you mean that it must be the same size in all 3 dimensions.
In that case, the size of the largest cuboid with maximal point (x,y,z) can be calculated from the sizes of the largest cuboids with maximal points (x-1,y,z), (x,y-1,z), (x,y,z-1), (x-1,y-1,z), (x-1,y,z-1), (x,y-1,z-1), and (x-1, y-1, z-1). code : ## How do I get the index of the largest list inside a list of lists using Python?

By : user2102577
Date : March 29 2020, 07:55 AM
Any of those help I am storing animation key frames from Cinema4D(using the awesome py4D) into a lists of lists:
code :
``````max(enumerate(props), key = lambda tup: len(tup))
`````` ## Writing a script for finding the largest and second largest eigenvectors of a symmetric matrix

By : venkatasubramaniam
Date : March 29 2020, 07:55 AM
I wish this helpful for you You have a solution for checking for a symmetric matrix.
For the eigenvectors, see the documentation for eig as suggested by Luis Mendo, but also the documentation for eigs, which allows you to request k eigenvectors according to sigma:
code :
``````eigs(A,k,sigma)
`````` ## Finding the largest vector inside a matrix

By : S M Azeem Dogar
Date : March 29 2020, 07:55 AM
I wish this help you The cell with the largest vector can be found using cellfun and numel to get the number of elements in each numeric matrix stored in the cells of paths:
code :
``````vecLens = cellfun(@numel,paths);
[maxLen,im] = max(vecLens(:));
[rowMax,colMax] = ind2sub(size(vecLens),im)
``````
``````maxMask = vecLens==maxLen;
else
[rowMax,colMax] = ind2sub(size(vecLens),im)
end
``````
``````[rowMax,colMax] = find(vecLens==maxLen);
`````` ## Packing many small cuboids into given large cuboids

By : Treasey Rodriguez
Date : March 29 2020, 07:55 AM
this one helps. Assuming length is greater than the breadth for both the rectangles (smaller and larger ones), following are the possibilities while you try to pack smaller rectangles on the larger one. Let the length of larger rectangle be L, its breadth be B and length and breadth of smaller rectangles be l and b respectively.
Case 1: Pack the smaller rectangles such that their lengths are parallel to the breadth of the larger rectangle until you fall short of space. Then try the other way round (Length of larger rectangle parallel to Lengths of smaller one) on the available space. ## Get five largest values from a column of a Dataframe inside a List

``````lapply(your_list, function(d) head(d\$sum_of_weights[order(-d\$sum_of_weights)], 5))
``````lapply(your_list, function(d) head(d["sum_of_weights"][order(-d\$sum_of_weights)], 5))
``````lapply(your_list, function(d) head(d[order(-d\$sum_of_weights), ], 5)) 