Python 矩阵乘法的麻烦
我正在尝试对可选作业进行QR分解,但我没有使用numpy:Python 矩阵乘法的麻烦,python,list,linear-algebra,Python,List,Linear Algebra,我正在尝试对可选作业进行QR分解,但我没有使用numpy: def multiply(matrix1, matrix2): rows_A = len(matrix1) cols_A = len(matrix1[0]) rows_B = len(matrix2) cols_B = len(matrix2[0]) new_matrix = [[0 for row in range(cols_B)] for col in range(rows_A)]
def multiply(matrix1, matrix2):
rows_A = len(matrix1)
cols_A = len(matrix1[0])
rows_B = len(matrix2)
cols_B = len(matrix2[0])
new_matrix = [[0 for row in range(cols_B)] for col in range(rows_A)]
for i in range(len(matrix1)):
for j in range(len(matrix2[0])):
for k in range(len(matrix2)):
new_matrix[i][j] += matrix1[i][k]*matrix2[k][j]
return new_matrix
def transpose(matrix):
newmatrix = []
for i in range(len(matrix)):
newline = []
for j in range(len(matrix[i])):
newline.append(matrix[j][i])
newmatrix.append(newline)
for k in range(len(newmatrix)):
print(newmatrix[k])
return matrix
# Returns the Gramm-Schmidt orthogonalization of matrix X
def gramm_schmidt(X, inplace = False):
if not inplace:
V = [row[:] for row in X] # make a copy.
else:
V = X
k = len(X[0]) # number of columns.
n = len(X) # number of rows.
for j in range(k):
for i in range(j):
# D = < Vi, Vj>
D = sum([V[p][i]*V[p][j] for p in range(n)])
for p in range(n):
# Note that the Vi's already have length one!
# Vj = Vj - <Vi,Vj> Vi/< Vi,Vi >
V[p][j] -= (D * V[p][i])
# Normalize column V[j]
invnorm = 1.0 / sqrt(sum([(V[p][j])**2 for p in range(n)]))
for p in range(n):
V[p][j] *= invnorm
return V
def QR(matrix):
Q = gramm_schmidt(matrix)
Q_transpose = transpose(Q)
R = multiply(Q_transpose, matrix)
QR = multiply(Q, R)
print ("Q:\n")
for row in Q:
print (row)
print ("\n")
print ("R:\n")
for row in R:
print (row)
print ("\n")
print ("QR:\n")
for row in QR:
print (row)
print ("\n")
我得到这个错误:
rows_A = len(matrix1)
TypeError: object of type 'NoneType' has no len()
我不知道为什么,因为当我做标准乘法时,乘法函数本身就可以工作……如果有人能帮忙,请提前感谢
编辑:我在转置中添加了一个return语句,是的,它确实给出了一个转置矩阵。看来是乘法运算造成的。以下是它应该做的:
A:
[[12, -51, 4],
[6, 167, -68],
[-4, 24, -41]]
Q:
[[0.8571428571428571, 0.39428571428571435, -0.33142857142857135],
[0.4285714285714286, -0.9028571428571429, 0.034285714285714114],
[-0.28571428571428575, -0.17142857142857126, -0.942857142857143]]
R:
[[13.999999999999998, 21.00000000000001, -14.000000000000004],
[-5.506706202140776e-16, -175.00000000000003, 70.0],
[3.0198066269804245e-16, -3.552713678800501e-14, 35.000000000000014]]
但是我只能让Q正常工作。在
transpose()
中没有return
,所以返回None
。因此,您正在执行Q_transpose=None
,因为当前情况下,如果您仍然收到相同的错误,那么transpose()函数仍然返回None类型,并且在转置中返回转置,但从未定义它。也许这是一个输入错误?可能是这样,但转置确实有效。我已经更新了我的问题,请参考。@Ratman2050我还看到你键入了返回矩阵
-我想你的意思是新矩阵
。哦,对了,我把它改成了返回矩阵,但你说它仍然没有返回是什么意思?当我打印Q_转置时,它会为我打印转置
A:
[[12, -51, 4],
[6, 167, -68],
[-4, 24, -41]]
Q:
[[0.8571428571428571, 0.39428571428571435, -0.33142857142857135],
[0.4285714285714286, -0.9028571428571429, 0.034285714285714114],
[-0.28571428571428575, -0.17142857142857126, -0.942857142857143]]
R:
[[13.999999999999998, 21.00000000000001, -14.000000000000004],
[-5.506706202140776e-16, -175.00000000000003, 70.0],
[3.0198066269804245e-16, -3.552713678800501e-14, 35.000000000000014]]