可能的重复项：
python中的矩阵乘法？
我已经写了一个程序可以乘以两个矩阵。但是如果两个矩阵有不同的列（错误答案），这些代码不起作用。
C = [[4,1,9], [6,2,8], [7,3,5]]
D = [[2,9], [5,2], [1,0]]
M=[]
for i in range(len(C)):
Z.append([])
for j in range(len(D[0])):
Z[i].append(0)
for k in range(len(D[0])):
M[i][j]+=C[i][k]*D[k][j]
print (M)
在最后一个循环中，您希望k从最多 len(D) 或 len(C[0]) 开始（必须相等，否则乘法没有任何意义）。下面是修复它的代码的最小修改：
C = [[4,1,9], [6,2,8], [7,3,5]]
D = [[2,9], [5,2], [1,0]]
assert(len(C[0]) == len(D)) # sanity check!
M=[]
for i in range(len(C)):
M.append([])
for j in range(len(D[0])):
M[i].append(0)
for k in range(len(D)): # fixed value
M[i][j]+=C[i][k]*D[k][j]
print (M)
这是使用 numpy 的其他版本
好吧，我只是放了一些代码，但这里是整个代码：
# matrices.py
# This optional case study solves each of the following problems. Each
# solution uses the solutions to the preceding problems.
# 1) Multiply two matrices
# 2) Invert a matrix
# 3) Solve a system of linear equations
# 4) Fit an exact polynomial to a list of points
# 5) Find the coefficients of the polynomial for a power sum
# This last problem needs a little explaining. Consider this power sum:
# 1 + 2 + ... + n. We know this equals n(n+1)/2. We can represent this
# quadratic function by its coefficients alone, as [1/2, 1/2, 0] (ignoring
# integer division in this explanation).
# Now, the sum of the squares is: 1**2 + 2**2 + ... + n**2. This is a cubic
# equation. It happens to be [1/3, 1/2, 1/6, 0]. That is:
# 1**2 + ... + n**2 = (1/3)n**3 + (1/2)n**2 + (1/6)n
# In fact, for positive integer k, 1**k + ... + n**k is a (k+1)-degree
# polynomial. The problem we are solving is to write a function which takes
# this number k, and returns the coefficients of that polynomial.
# Also, our approach for matrix multiplication needs some slight explaining.
# We basically use Gauss-Jordan Elimination to get reduced row echelon form
# while applying the same elementary row operations to an identity matrix -- in
# so doing, the identity matrix is transformed into the inverse of the original
# matrix. Amazing!
import copy
from fractions import Fraction
def copyMatrix(m):
return copy.deepcopy(m)
def makeIdentity(n):
result = make2dList(n,n)
for i in xrange(n):
result[i][i] = Fraction(1, 1)
return result
def testMakeIdentity():
print "Testing makeIdentity...",
i3 = [[1,0,0],[0,1,0],[0,0,1]]
assert(i3 == makeIdentity(3))
print "Passed!"
def multiplyMatrices(a, b):
# confirm dimensions
aRows = len(a)
aCols = len(a[0])
bRows = len(b)
bCols = len(b[0])
assert(aCols == bRows) # belongs in a contract
rows = aRows
cols = bCols
# create the result matrix c = a*b
c = make2dList(rows, cols)
# now find each value in turn in the result matrix
for row in xrange(rows):
for col in xrange(cols):
dotProduct = Fraction(0, 1)
for i in xrange(aCols):
dotProduct += a[row][i]*b[i][col]
c[row][col] = dotProduct
return c
def testMultiplyMatrices():
print "Testing multiplyMatrices...",
a = [ [ 1, 2, 3],
[ 4, 5, 6 ] ]
b = [ [ 0, 3],
[ 1, 4],
[ 2, 5] ]
c = [ [ 8, 26],
[17, 62 ] ]
observedC = multiplyMatrices(a, b)
assert(observedC == c)
print "Passed!"
def multiplyRowOfSquareMatrix(m, row, k):
n = len(m)
rowOperator = makeIdentity(n)
rowOperator[row][row] = k
return multiplyMatrices(rowOperator, m)
def testMultiplyRowOfSquareMatrix():
print "Testing multiplyRowOfSquareMatrix...",
a = [ [ 1, 2 ],
[ 4, 5 ] ]
assert(multiplyRowOfSquareMatrix(a, 0, 5) == [[5, 10], [4, 5]])
assert(multiplyRowOfSquareMatrix(a, 1, 6) == [[1, 2], [24, 30]])
print "Passed!"
def addMultipleOfRowOfSquareMatrix(m, sourceRow, k, targetRow):
# add k * sourceRow to targetRow of matrix m
n = len(m)
rowOperator = makeIdentity(n)
rowOperator[targetRow][sourceRow] = k
return multiplyMatrices(rowOperator, m)
def testAddMultipleOfRowOfSquareMatrix():
print "Testing addMultipleOfRowOfSquareMatrix...",
a = [ [ 1, 2 ],
[ 4, 5 ] ]
assert(addMultipleOfRowOfSquareMatrix(a, 0, 5, 1) == [[1, 2], [9, 15]])
assert(addMultipleOfRowOfSquareMatrix(a, 1, 6, 0) == [[25, 32], [4, 5]])
print "Passed!"
def invertMatrix(m):
n = len(m)
assert(len(m) == len(m[0]))
inverse = makeIdentity(n) # this will BECOME the inverse eventually
for col in xrange(n):
# 1. make the diagonal contain a 1
diagonalRow = col
assert(m[diagonalRow][col] != 0) # @TODO: actually, we could swap rows
# here, or if no other row has a 0 in
# this column, then we have a singular
# (non-invertible) matrix. Let's not
# worry about that for now. :-)
k = Fraction(1,m[diagonalRow][col])
m = multiplyRowOfSquareMatrix(m, diagonalRow, k)
inverse = multiplyRowOfSquareMatrix(inverse, diagonalRow, k)
# 2. use the 1 on the diagonal to make everything else
# in this column a 0
sourceRow = diagonalRow
for targetRow in xrange(n):
if (sourceRow != targetRow):
k = -m[targetRow][col]
m = addMultipleOfRowOfSquareMatrix(m, sourceRow, k, targetRow)
inverse = addMultipleOfRowOfSquareMatrix(inverse, sourceRow,
k, targetRow)
# that's it!
return inverse
def testInvertMatrix():
print "Testing invertMatrix...",
a = [ [ 1, 2 ], [ 4, 5 ] ]
aInverse = invertMatrix(a)
identity = makeIdentity(len(a))
assert (almostEqualMatrices(identity, multiplyMatrices(a, aInverse)))
a = [ [ 1, 2, 3], [ 2, 5, 7 ], [3, 4, 8 ] ]
aInverse = invertMatrix(a)
identity = makeIdentity(len(a))
assert (almostEqualMatrices(identity, multiplyMatrices(a, aInverse)))
print "Passed!"
def solveSystemOfEquations(A, b):
return multiplyMatrices(invertMatrix(A), b)
def testSolveSystemOfEquations():
print "Testing solveSystemOfEquations...",
# 3x + 2y - 2z = 10
# 2x - 4y + 8z = 0
# 4x + 4y - 7z = 13
# x = 2, y = 3, z = 1
A = [ [3, 2, -2],
[2, -4, 8],
[4, 4, -7] ]
b = [ [ 10 ],
[ 0 ],
[ 13 ] ]
observedX = solveSystemOfEquations(A,b)
expectedX = [ [ 2 ],
[ 3 ],
[ 1 ] ]
assert(almostEqualMatrices(observedX, expectedX))
print "Passed!"
def fitExactPolynomial(pointList):
n = len(pointList)
degree = n - 1
# 1. make A
A = make2dList(n,n)
for row in xrange(n):
for col in xrange(n):
x = pointList[row][0]
exponent = degree - col
A[row][col] = x**exponent
# 2. make b
b = make2dList(n,1)
for row in xrange(n):
y = pointList[row][1]
b[row][0] = y
# use system solver to find solution
return solveSystemOfEquations(A, b)
def testFitExactPolynomial():
print "Testing fitPolynomialExactly...",
def f(x): return 3*x**3 + 2*x**2 + 4*x + 1
expected = [[3], [2], [4], [1]]
pointList = [(1,f(1)), (2,f(2)), (5,f(5)), (-3,f(-3))]
observed = fitExactPolynomial(pointList)
assert(almostEqualMatrices(observed, expected))
print "Passed!"
def findCoefficientsOfPowerSum(k):
# Assume f(n) = 1**k + ... + n**k
# We argued by handwaving-ish-calculusy-stuff that f(n) is
# a polynomial of degree (k+1)
# We need (k+2) points to fit just such a polynomial
pointList = []
y = 0
for n in xrange(1,k+3):
x = Fraction(n, 1)
# y = 1**k + ... + n**k
y += x**k
pointList += [(x,y)]
return fitExactPolynomial(pointList)
def testFindCoefficientsOfPowerSum():
print "Testing findCoefficientsOfPowerSum..."
# Not a formal test here, just printing the answers.
# Check here for expected values:
# http://mathworld.wolfram.com/PowerSum.html
for k in xrange(10):
print "k = %d:" % k,
printMatrix(findCoefficientsOfPowerSum(k))
print "Passed!"
def almostEqualMatrices(m1, m2):
# verifies each element in the two matrices are almostEqual to each other
# (and that the two matrices have the same dimensions).
if (len(m1) != len(m2)): return False
if (len(m1[0]) != len(m2[0])): return False
for row in xrange(len(m1)):
for col in xrange(len(m1[0])):
if not almostEqual(m1[row][col], m2[row][col]):
return False
return True
def almostEqual(d1, d2):
epsilon = 0.00001
return abs(d1 - d2) < epsilon
def make2dList(rows, cols):
a=[]
for row in xrange(rows): a += [[0]*cols]
return a
def printMatrix(a):
def valueStr(value):
if (isinstance(value, Fraction)):
(num, den) = (value.numerator, value.denominator)
if ((num == 0) or (den == 1)): return str(num)
else: return str(num) + "/" + str(den)
else:
return str(value)
def maxItemLength(a):
maxLen = 0
rows = len(a)
cols = len(a[0])
for row in xrange(rows):
for col in xrange(cols):
maxLen = max(maxLen, len(valueStr(a[row][col])))
return maxLen
if (a == []):
# So we don't crash accessing a[0]
print []
return
rows = len(a)
cols = len(a[0])
fieldWidth = maxItemLength(a)
print "[",
for row in xrange(rows):
if (row > 0) and (len(a[row-1]) > 1): print "n ",
print "[",
for col in xrange(cols):
if (col > 0): print ",",
# The next 2 lines print a[row][col] with the given fieldWidth
format = "%" + str(fieldWidth) + "s"
print format % valueStr(a[row][col]),
print "]",
print "]"
def main():
testMakeIdentity()
testMultiplyMatrices()
testMultiplyRowOfSquareMatrix()
testAddMultipleOfRowOfSquareMatrix()
testInvertMatrix()
testSolveSystemOfEquations()
testFitExactPolynomial()
testFindCoefficientsOfPowerSum()
main()
天哪，只需将for k in range(len(D[0])):中的D更改为C即可。该程序将起作用。谢谢你们！