本文整理汇总了Python中sympy.matrices.Matrix.zeros方法的典型用法代码示例。如果您正苦于以下问题：Python Matrix.zeros方法的具体用法？Python Matrix.zeros怎么用？Python Matrix.zeros使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.matrices.Matrix的用法示例。
在下文中一共展示了Matrix.zeros方法的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: test_hstack
# 需要导入模块: from sympy.matrices import Matrix [as 别名]
# 或者: from sympy.matrices.Matrix import zeros [as 别名]
def test_hstack():
m = ShapingOnlyMatrix(4, 3, lambda i, j: i*3 + j)
m2 = ShapingOnlyMatrix(3, 4, lambda i, j: i*3 + j)
assert m == m.hstack(m)
assert m.hstack(m, m, m) == ShapingOnlyMatrix.hstack(m, m, m) == Matrix([
[0, 1, 2, 0, 1, 2, 0, 1, 2],
[3, 4, 5, 3, 4, 5, 3, 4, 5],
[6, 7, 8, 6, 7, 8, 6, 7, 8],
[9, 10, 11, 9, 10, 11, 9, 10, 11]])
raises(ShapeError, lambda: m.hstack(m, m2))
assert Matrix.hstack() == Matrix()
# test regression #12938
M1 = Matrix.zeros(0, 0)
M2 = Matrix.zeros(0, 1)
M3 = Matrix.zeros(0, 2)
M4 = Matrix.zeros(0, 3)
m = ShapingOnlyMatrix.hstack(M1, M2, M3, M4)
assert m.rows == 0 and m.cols == 6示例2: zeros
# 需要导入模块: from sympy.matrices import Matrix [as 别名]
# 或者: from sympy.matrices.Matrix import zeros [as 别名]
def zeros(r, c=None, cls=None):
"""Returns a matrix of zeros with ``r`` rows and ``c`` columns;
if ``c`` is omitted a square matrix will be returned.
See Also
========
ones
eye
diag
"""
if cls is None:
from .dense import Matrix as cls
return cls.zeros(r, c)示例3: test_sparse_matrix
# 需要导入模块: from sympy.matrices import Matrix [as 别名]
# 或者: from sympy.matrices.Matrix import zeros [as 别名]
def test_sparse_matrix():
def sparse_eye(n):
return SparseMatrix.eye(n)
def sparse_zeros(n):
return SparseMatrix.zeros(n)
# creation args
raises(TypeError, lambda: SparseMatrix(1, 2))
a = SparseMatrix((
(1, 0),
(0, 1)
))
assert SparseMatrix(a) == a
from sympy.matrices import MutableSparseMatrix, MutableDenseMatrix
a = MutableSparseMatrix([])
b = MutableDenseMatrix([1, 2])
assert a.row_join(b) == b
assert a.col_join(b) == b
assert type(a.row_join(b)) == type(a)
assert type(a.col_join(b)) == type(a)
# make sure 0 x n matrices get stacked correctly
sparse_matrices = [SparseMatrix.zeros(0, n) for n in range(4)]
assert SparseMatrix.hstack(*sparse_matrices) == Matrix(0, 6, [])
sparse_matrices = [SparseMatrix.zeros(n, 0) for n in range(4)]
assert SparseMatrix.vstack(*sparse_matrices) == Matrix(6, 0, [])
# test element assignment
a = SparseMatrix((
(1, 0),
(0, 1)
))
a[3] = 4
assert a[1, 1] == 4
a[3] = 1
a[0, 0] = 2
assert a == SparseMatrix((
(2, 0),
(0, 1)
))
a[1, 0] = 5
assert a == SparseMatrix((
(2, 0),
(5, 1)
))
a[1, 1] = 0
assert a == SparseMatrix((
(2, 0),
(5, 0)
))
assert a._smat == {(0, 0): 2, (1, 0): 5}
# test_multiplication
a = SparseMatrix((
(1, 2),
(3, 1),
(0, 6),
))
b = SparseMatrix((
(1, 2),
(3, 0),
))
c = a*b
assert c[0, 0] == 7
assert c[0, 1] == 2
assert c[1, 0] == 6
assert c[1, 1] == 6
assert c[2, 0] == 18
assert c[2, 1] == 0
try:
eval('c = a @ b')
except SyntaxError:
pass
else:
assert c[0, 0] == 7
assert c[0, 1] == 2
assert c[1, 0] == 6
assert c[1, 1] == 6
assert c[2, 0] == 18
assert c[2, 1] == 0
x = Symbol("x")
c = b * Symbol("x")
assert isinstance(c, SparseMatrix)
assert c[0, 0] == x
assert c[0, 1] == 2*x
assert c[1, 0] == 3*x
assert c[1, 1] == 0
c = 5 * b
assert isinstance(c, SparseMatrix)
#.........这里部分代码省略.........
示例4: exp
# 需要导入模块: from sympy.matrices import Matrix [as 别名]
# 或者: from sympy.matrices.Matrix import zeros [as 别名]
#
# Solution given in Cartesian space, but expressed in spectral space
# Remove 2*pi for "2pi-sized" domain
#
U_fromSpec = exp(I*2*pi*(k[0]*x[0]/s[0]+k[1]*x[1]/s[1])) * U_spec
print
print
print("L(U):")
M = getL(U_fromSpec)
pprint(M)
# Remove spectral basis components (This should be possible in general)
Lik = Matrix.zeros(3,3)
Lik[:,0] = simplify(M[:,0]/U_fromSpec[0])
Lik[:,1] = simplify(M[:,1]/U_fromSpec[1])
Lik[:,2] = simplify(M[:,2]/U_fromSpec[2])
if True:
print
print
print("Lik(U):")
pprint(Lik)
print
print
print("Using omega(k)^2:")