Can someone tell me if there is a subtle reason why the jacobian
function in matrices.py is restricted to only work when the Jacobian
is square?
The example in the documentation is
from sympy import symbols, sin, cos
rho, phi = symbols("rho phi")
X = Matrix([rho*cos(phi), rho*sin(phi)])
Y = Matrix([rho, phi])
X.jacobian(Y)
which correctly yields the 2x2 matrix
[cos(phi), -rho*sin(phi)]
[sin(phi), rho*cos(phi)]
Suppose instead that
X = Matrix([rho*cos(phi), rho*sin(phi), rho**2])
I would hope that
X.jacobian(Y)
would now yield the 3x2 matrix
[cos(phi), -rho*sin(phi)]
[sin(phi), rho*cos(phi)]
[ 2*rho, 0]
but the code is restricted to the square Jacobian case so it instead
yields an assertion
---------------------------------------------------------------------------
AssertionError Traceback (most recent call
last)
/usr/lib/pymodules/python2.5/sympy/matrices/<ipython console> in
<module>()
/usr/lib/pymodules/python2.5/sympy/matrices/matrices.pyc in jacobian
(self, X)
943 assert X.shape[1] == n
944 else:
--> 945 assert X.shape[0] == n
946
947 # n is the dimension of the matrix, computing the
Jacobian is now easy:
AssertionError:
Would it break something I don't know about if the jacobian function
handled the nonsquare case? It appears as if the following modified
function yields the behavior I expected in the 3x2 case without
causing any tests to fail.
if not isinstance(X, Matrix):
X = Matrix(X)
# Both X and self can be a row or a column matrix, so we need
to make
# sure all valid combinations work, but everything else fails:
assert len(self.shape) == 2
assert len(X.shape) == 2
if self.shape[0] == 1:
m = self.shape[1]
else:
m = self.shape[0]
if X.shape[0] == 1:
n = X.shape[1]
else:
n = X.shape[0]
# m is the number of functions and n is the number of
variables
# computing the Jacobian is now easy:
return Matrix(m, n, lambda j, i: self[j].diff(X[i]))
Thank you,
Ben
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