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Python matrices.MatrixSymbol類代碼示例

本文整理匯總了Python中sympy.matrices.MatrixSymbol的典型用法代碼示例。如果您正苦於以下問題:Python MatrixSymbol類的具體用法?Python MatrixSymbol怎麽用?Python MatrixSymbol使用的例子?那麽, 這裏精選的類代碼示例或許可以為您提供幫助。


在下文中一共展示了MatrixSymbol類的8個代碼示例,這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚,您的評價將有助於係統推薦出更棒的Python代碼示例。

示例1: test_subs

def test_subs():
    A = MatrixSymbol('A', n, m)
    B = MatrixSymbol('B', m, l)
    C = MatrixSymbol('C', m, l)

    assert A.subs(n, m).shape == (m, m)

    assert (A*B).subs(B, C) == A*C

    assert (A*B).subs(l, n).is_square
開發者ID:ChaliZhg,項目名稱:sympy,代碼行數:10,代碼來源:test_matrix_exprs.py

示例2: test_subs

def test_subs():
    n, m, l = symbols('n m l', integer=True)
    A = MatrixSymbol('A', n, m)
    B = MatrixSymbol('B', m, l)
    C = MatrixSymbol('C', m, l)

    assert A.subs(n,m).shape == (m,m)

    assert (A*B).subs(B,C) == A*C

    assert (A*B).subs(l,n).is_square
開發者ID:101man,項目名稱:sympy,代碼行數:11,代碼來源:test_matrix_exprs.py

示例3: test_subs

def test_subs():
    n, m, l = symbols("n m l", integer=True)
    A = MatrixSymbol("A", n, m)
    B = MatrixSymbol("B", m, l)
    C = MatrixSymbol("C", m, l)

    assert A.subs(n, m).shape == (m, m)

    assert (A * B).subs(B, C) == A * C

    assert (A * B).subs(l, n).is_square
開發者ID:jenshnielsen,項目名稱:sympy,代碼行數:11,代碼來源:test_matrix_exprs.py

示例4: test_MatrixSymbol_determinant

def test_MatrixSymbol_determinant():
    A = MatrixSymbol('A', 4, 4)
    assert A.as_explicit().det() == A[0, 0]*A[1, 1]*A[2, 2]*A[3, 3] - \
        A[0, 0]*A[1, 1]*A[2, 3]*A[3, 2] - A[0, 0]*A[1, 2]*A[2, 1]*A[3, 3] + \
        A[0, 0]*A[1, 2]*A[2, 3]*A[3, 1] + A[0, 0]*A[1, 3]*A[2, 1]*A[3, 2] - \
        A[0, 0]*A[1, 3]*A[2, 2]*A[3, 1] - A[0, 1]*A[1, 0]*A[2, 2]*A[3, 3] + \
        A[0, 1]*A[1, 0]*A[2, 3]*A[3, 2] + A[0, 1]*A[1, 2]*A[2, 0]*A[3, 3] - \
        A[0, 1]*A[1, 2]*A[2, 3]*A[3, 0] - A[0, 1]*A[1, 3]*A[2, 0]*A[3, 2] + \
        A[0, 1]*A[1, 3]*A[2, 2]*A[3, 0] + A[0, 2]*A[1, 0]*A[2, 1]*A[3, 3] - \
        A[0, 2]*A[1, 0]*A[2, 3]*A[3, 1] - A[0, 2]*A[1, 1]*A[2, 0]*A[3, 3] + \
        A[0, 2]*A[1, 1]*A[2, 3]*A[3, 0] + A[0, 2]*A[1, 3]*A[2, 0]*A[3, 1] - \
        A[0, 2]*A[1, 3]*A[2, 1]*A[3, 0] - A[0, 3]*A[1, 0]*A[2, 1]*A[3, 2] + \
        A[0, 3]*A[1, 0]*A[2, 2]*A[3, 1] + A[0, 3]*A[1, 1]*A[2, 0]*A[3, 2] - \
        A[0, 3]*A[1, 1]*A[2, 2]*A[3, 0] - A[0, 3]*A[1, 2]*A[2, 0]*A[3, 1] + \
        A[0, 3]*A[1, 2]*A[2, 1]*A[3, 0]
開發者ID:latot,項目名稱:sympy,代碼行數:15,代碼來源:test_matrix_exprs.py

示例5: test_MatrixSymbol

def test_MatrixSymbol():
    n, m, t = symbols('n,m,t')
    X = MatrixSymbol('X', n, m)
    assert X.shape == (n, m)
    raises(TypeError, lambda: MatrixSymbol('X', n, m)(t))  # issue 5855
    assert X.doit() == X
開發者ID:ChaliZhg,項目名稱:sympy,代碼行數:6,代碼來源:test_matrix_exprs.py

示例6: test_as_explicit_nonsquare_symbol

def test_as_explicit_nonsquare_symbol():
    X = MatrixSymbol('X', 2, 3)
    assert MatPow(X, 1).as_explicit() == X.as_explicit()
    for r in [0, 2, S.Half, S.Pi]:
        raises(ShapeError, lambda: MatPow(X, r).as_explicit())
開發者ID:A-turing-machine,項目名稱:sympy,代碼行數:5,代碼來源:test_matpow.py

示例7: test_as_explicit_symbol

def test_as_explicit_symbol():
    X = MatrixSymbol('X', 2, 2)
    assert MatPow(X, 0).as_explicit() == ImmutableMatrix(Identity(2))
    assert MatPow(X, 1).as_explicit() == X.as_explicit()
    assert MatPow(X, 2).as_explicit() == (X.as_explicit())**2
開發者ID:A-turing-machine,項目名稱:sympy,代碼行數:5,代碼來源:test_matpow.py

示例8: test_inv

def test_inv():
    B = MatrixSymbol('B', 3, 3)
    assert B.inv() == B**-1
開發者ID:cklb,項目名稱:sympy,代碼行數:3,代碼來源:test_matexpr.py


注:本文中的sympy.matrices.MatrixSymbol類示例由純淨天空整理自Github/MSDocs等開源代碼及文檔管理平台,相關代碼片段篩選自各路編程大神貢獻的開源項目,源碼版權歸原作者所有,傳播和使用請參考對應項目的License;未經允許,請勿轉載。