Given a two-dimensional array T of size NxN, filled with various natural numbers (They do not have to be sorted in any way as in the example below.). My task is to write a program that transforms the array in such a way that all elements lying above the main diagonal are larger than each element lying on the diagonal and all elements lying below the main diagonal are to be smaller than each element on the diagonal.
For example:
T looks like this:
[2,3,5][7,11,13][17,19,23] and one of the possible solutions is:
[13,19,23][3,7,17][5,2,11]
I have no clue how to do this. Would anyone have an idea what algorithm should be used here?
Let's say the matrix is NxN.
Put all N² values inside an array.
Sort the array with whatever method you prefer (ascending order).
In your final array, the (N²-N)/2 first values go below the diagonal, the following N values go to the diagonal, and the final (N²-N)/2 values go above the diagonal.
The following pseudo-code should do the job:
mat <- array[N][N] // To be initialized.
vec <- array[N*N]
for i : 0 to (N-1)
for j : 0 to (N-1)
vec[i*N+j]=mat[i][j]
next j
next i
sort(vec)
p_below <- 0
p_diag <- (N*N-N)/2
p_above <- (N*N+N)/2
for i : 0 to (N-1)
for j : 0 to (N-1)
if (i>j)
mat[i][j] = vec[p_above]
p_above <- p_above + 1
endif
if (i<j)
mat[i][j] = vec[p_below]
p_below <- p_below + 1
endif
if (i=j)
mat[i][j] = vec[p_diag]
p_diag <- p_diag + 1
endif
next j
next i
Code can be heavily optimized by sorting directly the matrix, using a (quite complex) custom sort operator, so it can be sorted "in place". Technically, you'll do a bijection between the matrix indices to a partitioned set of indices representing "below diagonal", "diagonal" and "above diagonal" indices.
But I'm unsure that it can be considered as an algorithm in itself, because it will be highly dependent on the language used AND on how you stored, internally, your matrix (and how iterators/indices are used). I could write one in C++, but I lack knownledge to give you such an operator in Python.
Obviously, if you can't use a standard sorting function (because it can't work on anything else but an array), then you can write your own with the tricky comparison builtin the algorithm.
For such small matrixes, even a bubble-sort can work properly, but obviously implementing at least a quicksort would be better.
Elements about optimizing:
First, we speak about the trivial bijection from matrix coordinate [x][y] to [i]: i=x+y*N. The invert is obviously x=floor(i/N) & y=i mod N. Then, you can parse the matrix as a vector.
This is already what I do in the first part initializing vec, BTW.
With matrix coordinates, it's easy:
Diagonal is all cells where x=y.
The "below" partition is everywhere x<y.
The "above" partition is everywhere x>y.
Look at coordinates in the below 3x3 matrix, it's quite evident when you know it.
0,0 1,0 2,0
0,1 1,1 2,1
0,2 1,2 2,2
We already know that the ordered vector will be composed of three parts: first the "below" partition, then the "diagonal" partition, then the "above" partition.
The next bijection is way more tricky, since it requires either a piecewise linear function OR a look-up table. The first requires no additional memory but will use more CPU power, the second use as much memory as the matrix but will require less CPU power.
As always, optimization for speed often cost memory. If memory is scarse because you use huge matrixes, then you'll prefer a function.
In order to shorten a bit, I'll explain only for "below" partition. In the vector, the (N-1) first elements will be the ones belonging to the first column. Then, we'll have (N-2) elements for the 2nd column, (N-3) for the third, until we had only 1 element for the (N-1)th column. You see the scheme, sum of the number of elements and the column (zero-based index) is always (N-1).
I won't write the function, because it's quite complex and, honestly, it won't help so much to understand. Simply know that converting from matrix indices to vector is "quite easy".
The opposite is more tricky and CPU-intensive, and it SHOULD use a (N-1) element vector to store where each column starts within the vector to GREATLY speed up the process. Thanks, this vector can also be used (from end to begin) for the "above" partition, so it won't burn too much memory.
Now, you can sort your "vector" normally, simply by chaining the two bijection together with the vector index, and you'll get a matrix cell instead. As long as the sorting algorithm is stable (that's usually the case), it will works and will sort your matrix "in place", at the expense of a lot of mathematical computing to "route" the linear indexes to matrix indexes.
Please note that, despite we speak about bijections, we need ONLY the "vector to matrix" formulas. The "matrix to vector" are important - it MUST be a bijection! - but you won't use them, since you'll sort directly the (virtual) vector from 0 to N²-1.
I have two scipy sparse csr matrices with the exact same shape but potentially different data values and nnz value. I now want to get the top 10 elements of one matrix and increase the value on the same indices on the other matrix. My current approach is as follows:
idx = a.data.argpartition(-10)[-10:]
i, j = matrix.nonzero()
i_idx = i[idx]
j_idx = j[idx]
b[i_idx, j_idx] += 1
The reason I have to go this way is that a.data and b.data do not necessarily have the same number of elements and hence the indices would differ.
My question now is whether I can improve this in some way. As far as I know the nonzero procedure is not elegant as I have to allocate two new arrays and I am very tough on memory already. I can get the j_indices via csr_matrix.indices but what about the i_indices? Can I use the indptr in a nice way for that?
Happy for any hints.
I'm not sure what the "top 10 elements" means. I assume that if you have matrices A and B you want to set B[i, j] += 1 if A[i, j] is within the first 10 nonzero entries of A (in CSR format).
I also assume that B[i, j] could be zero, which is the worst case performance wise, since you need to modify the sparsity structure of your matrix.
CSR is not an ideal format to use for changing sparsity structure. This is because every new insertion/deletion is O(nnzs) complexity (assuming the CSR storage is backed by an array - and it usually is).
You could use the DOK format for your second matrix (the one you are modifying), which provides O(1) access to elements.
I haven't benchmarked this but I suppose your option is 10 * O(nnzs) at worst, when you are adding 10 new nonzero values, whereas the DOK version should need O(nnzs) to build the matrix, then O(1) for each insertion and, finally, O(nnzs) to convert it back to CSR (assuming this is needed).
This may be more of an 'approach' or conceptual question.
Basically, I have a python a multi-dimensional list like so:
my_list = [[0,1,1,1,0,1], [1,1,1,0,0,1], [1,1,0,0,0,1], [1,1,1,1,1,1]]
What I have to do is iterate through the array and compare each element with those directly surrounding it as though the list was layed out as a matrix.
For instance, given the first element of the first row, my_list[0][0], I need to know know the value of my_list[0][1], my_list[1][0] and my_list[1][1]. The value of the 'surrounding' elements will determine how the current element should be operated on. Of course for an element in the heart of the array, 8 comparisons will be necessary.
Now I know I could simply iterate through the array and compare with the indexed values, as above. I was curious as to whether there was a more efficient way which limited the amount of iteration required? Should I iterate through the array as is, or iterate and compare only values to either side and then transpose the array and run it again. This, however would ignore those values to the diagonal. And should I store results of the element lookups, so I don't keep determining the value of the same element multiple times?
I suspect this may have a fundamental approach in Computer Science, and I am eager to get feedback on the best approach using Python as opposed to looking for a specific answer to my problem.
You may get faster, and possibly even simpler, code by using numpy, or other alternatives (see below for details). But from a theoretical point of view, in terms of algorithmic complexity, the best you can get is O(N*M), and you can do that with your design (if I understand it correctly). For example:
def neighbors(matrix, row, col):
for i in row-1, row, row+1:
if i < 0 or i == len(matrix): continue
for j in col-1, col, col+1:
if j < 0 or j == len(matrix[i]): continue
if i == row and j == col: continue
yield matrix[i][j]
matrix = [[0,1,1,1,0,1], [1,1,1,0,0,1], [1,1,0,0,0,1], [1,1,1,1,1,1]]
for i, row in enumerate(matrix):
for j, cell in enumerate(cell):
for neighbor in neighbors(matrix, i, j):
do_stuff(cell, neighbor)
This has takes N * M * 8 steps (actually, a bit less than that, because many cells will have fewer than 8 neighbors). And algorithmically, there's no way you can do better than O(N * M). So, you're done.
(In some cases, you can make things simpler—with no significant change either way in performance—by thinking in terms of iterator transformations. For example, you can easily create a grouper over adjacent triplets from a list a by properly zipping a, a[1:], and a[2:], and you can extend this to adjacent 2-dimensional nonets. But I think in this case, it would just make your code more complicated that writing an explicit neighbors iterator and explicit for loops over the matrix.)
However, practically, you can get a whole lot faster, in various ways. For example:
Using numpy, you may get an order of magnitude or so faster. When you're iterating a tight loop and doing simple arithmetic, that's one of the things that Python is particularly slow at, and numpy can do it in C (or Fortran) instead.
Using your favorite GPGPU library, you can explicitly vectorize your operations.
Using multiprocessing, you can break the matrix up into pieces and perform multiple pieces in parallel on separate cores (or even separate machines).
Of course for a single 4x6 matrix, none of these are worth doing… except possibly for numpy, which may make your code simpler as well as faster, as long as you can express your operations naturally in matrix/broadcast terms.
In fact, even if you can't easily express things that way, just using numpy to store the matrix may make things a little simpler (and save some memory, if that matters). For example, numpy can let you access a single column from a matrix naturally, while in pure Python, you need to write something like [row[col] for row in matrix].
So, how would you tackle this with numpy?
First, you should read over numpy.matrix and ufunc (or, better, some higher-level tutorial, but I don't have one to recommend) before going too much further.
Anyway, it depends on what you're doing with each set of neighbors, but there are three basic ideas.
First, if you can convert your operation into simple matrix math, that's always easiest.
If not, you can create 8 "neighbor matrices" just by shifting the matrix in each direction, then perform simple operations against each neighbor. For some cases, it may be easier to start with an N+2 x N+2 matrix with suitable "empty" values (usually 0 or nan) in the outer rim. Alternatively, you can shift the matrix over and fill in empty values. Or, for some operations, you don't need an identical-sized matrix, so you can just crop the matrix to create a neighbor. It really depends on what operations you want to do.
For example, taking your input as a fixed 6x4 board for the Game of Life:
def neighbors(matrix):
for i in -1, 0, 1:
for j in -1, 0, 1:
if i == 0 and j == 0: continue
yield np.roll(np.roll(matrix, i, 0), j, 1)
matrix = np.matrix([[0,0,0,0,0,0,0,0],
[0,0,1,1,1,0,1,0],
[0,1,1,1,0,0,1,0],
[0,1,1,0,0,0,1,0],
[0,1,1,1,1,1,1,0],
[0,0,0,0,0,0,0,0]])
while True:
livecount = sum(neighbors(matrix))
matrix = (matrix & (livecount==2)) | (livecount==3)
(Note that this isn't the best way to solve this problem, but I think it's relatively easy to understand, and likely to illuminate whatever your actual problem is.)
I am using Scipy to construct a large, sparse (250k X 250k) co-occurrence matrix using scipy.sparse.lil_matrix. Co-occurrence matrices are triangular; that is, M[i,j] == M[j,i]. Since it would be highly inefficient (and in my case, impossible) to store all the data twice, I'm currently storing data at the coordinate (i,j) where i is always smaller than j. So in other words, I have a value stored at (2,3) and no value stored at (3,2), even though (3,2) in my model should be equal to (2,3). (See the matrix below for an example)
My problem is that I need to be able to randomly extract the data corresponding to a given index, but, at least the way, I'm currently doing it, half the data is in the row and half is in the column, like so:
M =
[1 2 3 4
0 5 6 7
0 0 8 9
0 0 0 10]
So, given the above matrix, I want to be able to do a query like M[1], and get back [2,5,6,7]. I have two questions:
1) Is there a more efficient (preferably built-in) way to do this than first querying the row, and then the column, and then concatenating the two? This is bad because whether I use CSC (column-based) or CSR (row-based) internal representation, one of the two queries is highly inefficient.
2) Am I even using the right part of Scipy? I have seen a few functions in the Scipy library that mention triangular matrices, but they seem to revolve around getting triangular matrices from a full matrix. In my case, (I think) I already have a triangular matrix, and want to manipulate it.
Many thanks.
I would say that you can't have the cake and eat it too: if you want efficient storage, you cannot store full rows (as you say); if you want efficient row access, I'd say that you have to store full rows.
While real performances depend on your application, you could check whether the following approach works for you:
You use Scipy's sparse matrices for efficient storage.
You automatically symmetrize your matrix (there is a small recipe on StackOverflow, that works at least on regular matrices).
You can then access its rows (or columns); whether this is efficient depends on the implementation of sparse matrices…
i want to create a matrix of size 1234*5678 with it being filled with 1 to 5678 in row major order?>..!!
I think you will need to use numpy to hold such a big matrix efficiently , not just computation. You have ~5e6 items of 4/8 bytes means 20/40 Mb in pure C already, several times of that in python without an efficient data structure (a list of rows, each row a list).
Now, concerning your question:
import numpy as np
a = np.empty((1234, 5678), dtype=np.int)
a[:] = np.linspace(1, 5678, 5678)
You first create an array of the requested size, with type int (I assume you know you want 4 bytes integer, which is what np.int will give you on most platforms). The 3rd line uses broadcasting so that each row (a[0], a[1], ... a[1233]) is assigned the values of the np.linspace line (which gives you an array of [1, ....., 5678]). If you want F storage, that is column major:
a = np.empty((1234, 4567), dtype=np.int, order='F')
...
The matrix a will takes only a tiny amount of memory more than an array in C, and for computation at least, the indexing capabilities of arrays are much better than python lists.
A nitpick: numeric is the name of the old numerical package for python - the recommended name is numpy.
Or just use Numerical Python if you want to do some mathematical stuff on matrix too (like multiplication, ...). If they use row major order for the matrix layout in memory I can't tell you but it gets coverd in their documentation
Here's a forum post that has some code examples of what you are trying to achieve.