本文整理汇总了Python中matutil.mat2rowdict函数的典型用法代码示例。如果您正苦于以下问题：Python mat2rowdict函数的具体用法？Python mat2rowdict怎么用？Python mat2rowdict使用的例子？那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了mat2rowdict函数的15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: QR_solve
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
'''
from triangular import triangular_solve
from QR import factor
from matutil import mat2rowdict
Q,R = factor(A)
rowlist = [mat2rowdict(R)[v] for v in mat2rowdict(R)]
label_list = sorted(A.D[1], key = repr)
return triangular_solve(rowlist, label_list, b*Q)示例2: find_triangular_matrix_inverse
def find_triangular_matrix_inverse(A):
'''
input: An upper triangular Mat, A, with nonzero diagonal elements
output: Inverse of A
>>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
>>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
True
'''
(v,k)=(list(mat2rowdict(A).values()),list(mat2rowdict(A).keys()))
return coldict2mat([triangular_solve(v,k,Vec(A.D[1], {e:1})) for e in A.D[1]])示例3: dot_prod_mat_mat_mult
def dot_prod_mat_mat_mult(A, B):
assert A.D[1] == B.D[0]
return Mat(
(A.D[0], B.D[1]),
{
(row, col): matutil.mat2rowdict(A)[row] * matutil.mat2coldict(B)[col]
for row in matutil.mat2rowdict(A)
for col in matutil.mat2coldict(B)
},
)示例4: find_triangular_matrix_inverse
def find_triangular_matrix_inverse(A):
'''
input: An upper triangular Mat, A, with nonzero diagonal elements
output: Inverse of A
>>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
>>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
True
'''
cols = []
i = identity(A.D[0], 1)
i_cols = mat2coldict(i)
for k in A.D[0]:
cols.append( triangular_solve([ mat2rowdict(A)[k] for k in mat2rowdict(A) ], list(range(len(A.D[0]))), i_cols[k]) )
return coldict2mat(cols)示例5: QR_solve
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 = factor(A)
return triangular_solve(mat2rowdict(R),list(A.D[1]), Q.transpose()*b)
# test
#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 = factor(A)
#b = Vec(domain[0], {'a': 1, 'b': -1})
#x = QR_solve(A, b)
#result = A.transpose()*(b-A*x)
#print(result * result < 1E-10)示例6: dot_product_mat_vec_mult
def dot_product_mat_vec_mult(M, v):
assert M.D[1] == v.D
result = Vec(M.D[0], {c: 0 for c in M.D[0]})
rows = mat2rowdict(M)
for k in M.D[0]:
result.f[k] = rows[k] * v
return result示例7: lin_comb_vec_mat_mult
def lin_comb_vec_mat_mult(v, M):
assert v.D == M.D[0]
result = Vec(M.D[1], {r: 0 for r in M.D[1]})
rows = mat2rowdict(M)
for k, value in v.f.items():
result = result + value * rows[k]
return result示例8: dot_product_mat_vec_mult
def dot_product_mat_vec_mult(M, v):
assert(M.D[1] == v.D)
res = Vec(M.D[0], {})
dct = matutil.mat2rowdict(M)
for k,n in dct.items():
res[k] = n * v
return res示例9: lin_comb_vec_mat_mult
def lin_comb_vec_mat_mult(v, M):
assert(v.D == M.D[0])
res = Vec(M.D[1], {})
dct = matutil.mat2rowdict(M)
for k in v.D:
res = res + v[k]*dct[k]
return res示例10: vM_mat_mat_mult
def vM_mat_mat_mult(A, B):
assert A.D[1] == B.D[0]
from matutil import mat2rowdict, rowdict2mat
a = mat2rowdict(A)
m = rowdict2mat({k: v * B for (k, v) in a.items()})
return m示例11: find_triangular_matrix_inverse
def find_triangular_matrix_inverse(A):
"""
input: An upper triangular Mat, A, with nonzero diagonal elements
output: Inverse of A
>>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
>>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
True
"""
cd = list()
label_list = list(mat2rowdict(A).keys())
rowlist = list(mat2rowdict(A).values())
for j in label_list:
c = triangular_solve(rowlist, label_list, list2vec([1 if i == j else 0 for i in label_list]))
cd.append(c)
return coldict2mat(cd)示例12: find_triangular_matrix_inverse
def find_triangular_matrix_inverse(A):
"""
input: An upper triangular Mat, A, with nonzero diagonal elements
output: Inverse of A
>>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
>>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
True
"""
I_col_dict = {i: Vec(A.D[0], {i: 1}) for i in A.D[1]}
# def triangular_solve(rowlist, label_list, b):
rowlist = list()
label_list = list()
from matutil import mat2rowdict
rowdict = mat2rowdict(A)
for key in rowdict.keys():
label_list.append(key)
rowlist.append(rowdict[key])
B_coldict = dict()
from triangular import triangular_solve
for key in label_list:
B_coldict[key] = triangular_solve(rowlist, label_list, I_col_dict[key])
return coldict2mat(B_coldict)示例13: QR_solve
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
'''
from QR import factor
from triangular import triangular_solve
from mat import transpose
from matutil import mat2rowdict
Q, R = factor(A)
qT = Q.transpose()
c = qT*b
rows = [rowVec for index,rowVec in mat2rowdict(R).items()]
colLabels = sorted(A.D[1], key=repr)
x_hat = triangular_solve(rows,colLabels,c)
return x_hat示例14: dot_prod_mat_mat_mult
def dot_prod_mat_mat_mult(A, B):
assert A.D[1] == B.D[0]
import vec, matutil
A_dict = matutil.mat2rowdict(A)
B_dict = matutil.mat2coldict(B)
dp = {(i,j): A_dict[i]*B_dict[j] for j in B_dict.keys() for i in A_dict.keys()}
return Mat((set(A_dict.keys()), set(B_dict.keys())), dp)示例15: find_triangular_matrix_inverse
def find_triangular_matrix_inverse(A):
'''
Supporting GF2 is not required.
Input:
- A: an upper triangular Mat with nonzero diagonal elements
Output:
- Mat that is the inverse of A
Example:
>>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
>>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
True
'''
I = {i:Vec(A.D[0],{i:1}) for i in A.D[1]}
rl = list()
ll = list()
cd = dict()
rd = mat2rowdict(A)
for k,i in rd.items():
ll.append(k)
rl.append(rd[k])
for k in ll:
cd[k]= triangular_solve(rl,ll,I[k])
return coldict2mat(cd)