N-dim array¶
N-dim array module for SymPy.
Four classes are provided to handle N-dim arrays, given by the combinations dense/sparse (i.e. whether to store all elements or only the non-zero ones in memory) and mutable/immutable (immutable classes are SymPy objects, but cannot change after they have been created).
Examples¶
The following examples show the usage of Array. This is an abbreviation for
ImmutableDenseNDimArray, that is an immutable and dense N-dim array, the
other classes are analogous. For mutable classes it is also possible to change
element values after the object has been constructed.
Array construction can detect the shape of nested lists and tuples:
>>> from sympy.tensor.array import Array
>>> a1 = Array([[1, 2], [3, 4], [5, 6]])
>>> a1
[[1, 2], [3, 4], [5, 6]]
>>> a1.shape
(3, 2)
>>> a1.rank()
2
>>> from sympy.abc import x, y, z
>>> a2 = Array([[[x, y], [z, x*z]], [[1, x*y], [1/x, x/y]]])
>>> a2
[[[x, y], [z, x*z]], [[1, x*y], [1/x, x/y]]]
>>> a2.shape
(2, 2, 2)
>>> a2.rank()
3
Otherwise one could pass a 1-dim array followed by a shape tuple:
>>> m1 = Array(range(12), (3, 4))
>>> m1
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]
>>> m2 = Array(range(12), (3, 2, 2))
>>> m2
[[[0, 1], [2, 3]], [[4, 5], [6, 7]], [[8, 9], [10, 11]]]
>>> m2[1,1,1]
7
>>> m2.reshape(4, 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]]
Slice support:
>>> m2[:, 1, 1]
[3, 7, 11]
Elementwise derivative:
>>> from sympy.abc import x, y, z
>>> m3 = Array([x**3, x*y, z])
>>> m3.diff(x)
[3*x**2, y, 0]
>>> m3.diff(z)
[0, 0, 1]
Multiplication with other SymPy expressions is applied elementwisely:
>>> (1+x)*m3
[x**3*(x + 1), x*y*(x + 1), z*(x + 1)]
To apply a function to each element of the N-dim array, use applyfunc:
>>> m3.applyfunc(lambda x: x/2)
[x**3/2, x*y/2, z/2]
N-dim arrays can be converted to nested lists by the tolist() method:
>>> m2.tolist()
[[[0, 1], [2, 3]], [[4, 5], [6, 7]], [[8, 9], [10, 11]]]
>>> isinstance(m2.tolist(), list)
True
If the rank is 2, it is possible to convert them to matrices with tomatrix():
>>> m1.tomatrix()
Matrix([
[0, 1, 2, 3],
[4, 5, 6, 7],
[8, 9, 10, 11]])
Products and contractions¶
Tensor product between arrays \(A_{i_1,\ldots,i_n}\) and \(B_{j_1,\ldots,j_m}\) creates the combined array \(P = A \otimes B\) defined as
\(P_{i_1,\ldots,i_n,j_1,\ldots,j_m} := A_{i_1,\ldots,i_n}\cdot B_{j_1,\ldots,j_m}.\)
It is available through tensorproduct(...):
>>> from sympy.tensor.array import Array, tensorproduct
>>> from sympy.abc import x,y,z,t
>>> A = Array([x, y, z, t])
>>> B = Array([1, 2, 3, 4])
>>> tensorproduct(A, B)
[[x, 2*x, 3*x, 4*x], [y, 2*y, 3*y, 4*y], [z, 2*z, 3*z, 4*z], [t, 2*t, 3*t, 4*t]]
Tensor product between a rank-1 array and a matrix creates a rank-3 array:
>>> from sympy import eye
>>> p1 = tensorproduct(A, eye(4))
>>> p1
[[[x, 0, 0, 0], [0, x, 0, 0], [0, 0, x, 0], [0, 0, 0, x]], [[y, 0, 0, 0], [0, y, 0, 0], [0, 0, y, 0], [0, 0, 0, y]], [[z, 0, 0, 0], [0, z, 0, 0], [0, 0, z, 0], [0, 0, 0, z]], [[t, 0, 0, 0], [0, t, 0, 0], [0, 0, t, 0], [0, 0, 0, t]]]
Now, to get back \(A_0 \otimes \mathbf{1}\) one can access \(p_{0,m,n}\) by slicing:
>>> p1[0,:,:]
[[x, 0, 0, 0], [0, x, 0, 0], [0, 0, x, 0], [0, 0, 0, x]]
Tensor contraction sums over the specified axes, for example contracting positions \(a\) and \(b\) means
\(A_{i_1,\ldots,i_a,\ldots,i_b,\ldots,i_n} \implies \sum_k A_{i_1,\ldots,k,\ldots,k,\ldots,i_n}\)
Remember that Python indexing is zero starting, to contract the a-th and b-th axes it is therefore necessary to specify \(a-1\) and \(b-1\)
>>> from sympy.tensor.array import tensorcontraction
>>> C = Array([[x, y], [z, t]])
The matrix trace is equivalent to the contraction of a rank-2 array:
\(A_{m,n} \implies \sum_k A_{k,k}\)
>>> tensorcontraction(C, (0, 1))
t + x
Matrix product is equivalent to a tensor product of two rank-2 arrays, followed by a contraction of the 2nd and 3rd axes (in Python indexing axes number 1, 2).
\(A_{m,n}\cdot B_{i,j} \implies \sum_k A_{m, k}\cdot B_{k, j}\)
>>> D = Array([[2, 1], [0, -1]])
>>> tensorcontraction(tensorproduct(C, D), (1, 2))
[[2*x, x - y], [2*z, -t + z]]
One may verify that the matrix product is equivalent:
>>> from sympy import Matrix
>>> Matrix([[x, y], [z, t]])*Matrix([[2, 1], [0, -1]])
Matrix([
[2*x, x - y],
[2*z, -t + z]])
or equivalently
>>> C.tomatrix()*D.tomatrix()
Matrix([
[2*x, x - y],
[2*z, -t + z]])
Derivatives by array¶
The usual derivative operation may be extended to support derivation with respect to arrays, provided that all elements in the that array are symbols or expressions suitable for derivations.
The definition of a derivative by an array is as follows: given the array \(A_{i_1, \ldots, i_N}\) and the array \(X_{j_1, \ldots, j_M}\) the derivative of arrays will return a new array \(B\) defined by
\(B_{j_1,\ldots,j_M,i_1,\ldots,i_N} := \frac{\partial A_{i_1,\ldots,i_N}}{\partial X_{j_1,\ldots,j_M}}\)
The function derive_by_array performs such an operation:
>>> from sympy.tensor.array import Array, tensorcontraction, derive_by_array
>>> from sympy.abc import x, y, z, t
>>> from sympy import sin, exp, symbols, Function
With scalars, it behaves exactly as the ordinary derivative:
>>> derive_by_array(sin(x*y), x)
y*cos(x*y)
Scalar derived by an array basis:
>>> derive_by_array(sin(x*y), [x, y, z])
[y*cos(x*y), x*cos(x*y), 0]
Deriving array by an array basis: \(B^{nm} := \frac{\partial A^m}{\partial x^n}\)
>>> basis = [x, y, z]
>>> ax = derive_by_array([exp(x), sin(y*z), t], basis)
>>> ax
[[exp(x), 0, 0], [0, z*cos(y*z), 0], [0, y*cos(y*z), 0]]
Contraction of the resulting array: \(\sum_m \frac{\partial A^m}{\partial x^m}\)
>>> tensorcontraction(ax, (0, 1))
z*cos(y*z) + exp(x)
