Python QR.factor示例

示例#1

文件： hw7.py 项目： iryek219/LinearAlgebra_python

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True

I used the following as args to triangular_solve() and the grader was happy:
rowlist = mat2rowdict(R) (mentioned in the pdf)
label_list = sorted(A.D[1], key=repr) (mentioned in the pdf)
b = c vector b = Vec(domain[0], {'a': 1, 'b': -1})(mentioned above and on slide 1 of orthogonalization 8)        
    '''

    Q, R = QR.factor(A)
    rowlist = mat2rowdict(R)
    label_list = sorted(A.D[1], key = repr)
    return solve(A,b)

示例#2

文件： hw7.py 项目： airy-beam/online_courses

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q, R = QR.factor(A)
    return triangular_solve(mat2rowdict(R),sorted(A.D[1], key=repr),Q.transpose()*b)

示例#3

文件： hw7.py 项目： lfcunha/coding_the_matrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q, R = QR.factor(A)
    Rlist = list(mat2rowdict(R).values())
    return triangular_solve_n(Rlist, Q.transpose() * b)

示例#4

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q,R = QR.factor(A)
    C = transpose(A) * b
    dictR = mat2rowdict(R)
    return triangular_solve([dictR[k] for k in dictR],list(A.D[1]),C)

示例#5

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    import QR
    from triangular import triangular_solve

    (Q, R) = QR.factor(A)
    Rlist = [mat2rowdict(R)[i] for i in R.D[0]]
    return triangular_solve(Rlist, sorted(A.D[1], key=repr), b * Q)

示例#6

文件： hw7.py 项目： johnmerm/matrix

        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    labels = sorted(A.D[1], key=repr)
    Q, R = QR.factor(A)
    R_list = list(mat2rowdict(R).values())
    c = Q.transpose()*b
    x = triangular_solve(R_list,labels,c)
    return x

domain = ({'a','b','c'},{'A','B'})
A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
Q, R = QR.factor(A)
b = Vec(domain[0], {'a': 1, 'b': -1})
x = QR_solve(A, b)
result = A.transpose()*(b-A*x)