CODEWITHSUNDEEP
pythonperformancematrixsympy
Zorg
Zorg

Reputation: 201

speeding up sympy matrix operations

I have written a code, which uses sympy to set up a matrix and a vector. The elements of these two are sympy symbols. Then I invert the matrix and multiply the inverted matrix and the vector. This should be a generic solver for linear equation systems with n variables. I am interested in the symbolic solution of these linear equations. The problem is that my code is too slow. For instance, for n=4 it takes roughly 30 sec but for n=7 I haven't been able to solve it so far, the code ran all night (8h) and hasn't finished in the morning. This is my code.

from sympy import * 
import pprint 

MM = Matrix(niso,1, lambda i,j:var('MM_%s' % (i+1) ))
MA = Matrix (niso,1, lambda i,j:var('m_%s%s' % ('A', chr(66+i)) ) )
MX = Matrix (niso,1, lambda i,j:var('m_%s%s'% (chr(66+i), 'A')))
RB = Matrix(niso,1, lambda i,j:var('R_%s%s' % ('A'+chr(66+i),i+2)))
R = Matrix (niso, niso-1, lambda i,j: var('R_%s%d' % (chr(65+i) , j+2 )))
K= Matrix(niso-1,1, lambda i,j:var('K_%d' % (i+2) ) )
C= Matrix(niso-1,1, lambda i,j:var('A_%d' % i))

A = Matrix(niso-1,niso-1, lambda i,j:var('A_%d' % i))
b = Matrix(niso-1,1, lambda i,j:var('A_%d' % i))

for i in range(0,niso-1):
        for j in range(0,niso-1):
            A[i,j]=MM[j+1,0]*(Add(Mul(R[0,j],1/MA[i,0]/(RB[i,0]-R[0,i])))+R[i+1,j]/MX[i,0]/(-RB[i,0]+R[0,i]))

for i in range(0,niso-1):
    b[i,0]=MM[0,0]*(Add(Mul(1,1/MA[i,0]/(RB[i,0]-R[0,i])))+1/MX[i,0]/(-RB[i,0]+R[0,i]))


A_in = Inverse(A)
if niso <= 4:
    X =simplify(A_in*b)
if niso > 4:
    X = A_in*b
pprint(X)

Is there a way to speed it up?

Upvotes: 2

Views: 1338

Answers (1)

ptb
ptb

Reputation: 2148

Don't invert! With n=4

%timeit soln = A.LUsolve(b)
697 µs ± 12.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

With n=10

%timeit soln = A.LUsolve(b)
431 ms ± 13 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Upvotes: 3

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