本文实例讲述了Python实现的矩阵类。分享给大家供大家参考，具体如下：
科学计算离不开矩阵的运算。当然，python已经有非常好的现成的库：numpy(numpy的简单安装与使用可参考http://www.zzvips.com/article/80457.html)。
我写这个矩阵类，并不是打算重新造一个*，只是作为一个练习，记录在此。
注：这个类的函数还没全部实现，慢慢在完善吧。
全部代码：
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import copy
class Matrix:
'''矩阵类'''
def __init__(self, row, column, fill=0.0):
self.shape = (row, column)
self.row = row
self.column = column
self._matrix = [[fill]*column for i in range(row)]
# 返回元素m(i, j)的值: m[i, j]
def __getitem__(self, index):
if isinstance(index, int):
return self._matrix[index-1]
elif isinstance(index, tuple):
return self._matrix[index[0]-1][index[1]-1]
# 设置元素m(i,j)的值为s: m[i, j] = s
def __setitem__(self, index, value):
if isinstance(index, int):
self._matrix[index-1] = copy.deepcopy(value)
elif isinstance(index, tuple):
self._matrix[index[0]-1][index[1]-1] = value
def __eq__(self, N):
'''相等'''
# A == B
assert isinstance(N, Matrix), "类型不匹配，不能比较"
return N.shape == self.shape # 比较维度，可以修改为别的
def __add__(self, N):
'''加法'''
# A + B
assert N.shape == self.shape, "维度不匹配，不能相加"
M = Matrix(self.row, self.column)
for r in range(self.row):
for c in range(self.column):
M[r, c] = self[r, c] + N[r, c]
return M
def __sub__(self, N):
'''减法'''
# A - B
assert N.shape == self.shape, "维度不匹配，不能相减"
M = Matrix(self.row, self.column)
for r in range(self.row):
for c in range(self.column):
M[r, c] = self[r, c] - N[r, c]
return M
def __mul__(self, N):
'''乘法'''
# A * B (或：A * 2.0)
if isinstance(N, int) or isinstance(N,float):
M = Matrix(self.row, self.column)
for r in range(self.row):
for c in range(self.column):
M[r, c] = self[r, c]*N
else:
assert N.row == self.column, "维度不匹配，不能相乘"
M = Matrix(self.row, N.column)
for r in range(self.row):
for c in range(N.column):
sum = 0
for k in range(self.column):
sum += self[r, k] * N[k, r]
M[r, c] = sum
return M
def __div__(self, N):
'''除法'''
# A / B
pass
def __pow__(self, k):
'''乘方'''
# A**k
assert self.row == self.column, "不是方阵，不能乘方"
M = copy.deepcopy(self)
for i in range(k):
M = M * self
return M
def rank(self):
'''矩阵的秩'''
pass
def trace(self):
'''矩阵的迹'''
pass
def adjoint(self):
'''伴随矩阵'''
pass
def invert(self):
'''逆矩阵'''
assert self.row == self.column, "不是方阵"
M = Matrix(self.row, self.column*2)
I = self.identity() # 单位矩阵
I.show()#############################
# 拼接
for r in range(1,M.row+1):
temp = self[r]
temp.extend(I[r])
M[r] = copy.deepcopy(temp)
M.show()#############################
# 初等行变换
for r in range(1, M.row+1):
# 本行首元素(M[r, r])若为 0，则向下交换最近的当前列元素非零的行
if M[r, r] == 0:
for rr in range(r+1, M.row+1):
if M[rr, r] != 0:
M[r],M[rr] = M[rr],M[r] # 交换两行
break
assert M[r, r] != 0, '矩阵不可逆'
# 本行首元素(M[r, r])化为 1
temp = M[r,r] # 缓存
for c in range(r, M.column+1):
M[r, c] /= temp
print("M[{0}, {1}] /= {2}".format(r,c,temp))
M.show()
# 本列上、下方的所有元素化为 0
for rr in range(1, M.row+1):
temp = M[rr, r] # 缓存
for c in range(r, M.column+1):
if rr == r:
continue
M[rr, c] -= temp * M[r, c]
print("M[{0}, {1}] -= {2} * M[{3}, {1}]".format(rr, c, temp,r))
M.show()
# 截取逆矩阵
N = Matrix(self.row,self.column)
for r in range(1,self.row+1):
N[r] = M[r][self.row:]
return N
def jieti(self):
'''行简化阶梯矩阵'''
pass
def transpose(self):
'''转置'''
M = Matrix(self.column, self.row)
for r in range(self.column):
for c in range(self.row):
M[r, c] = self[c, r]
return M
def cofactor(self, row, column):
'''代数余子式（用于行列式展开）'''
assert self.row == self.column, "不是方阵，无法计算代数余子式"
assert self.row >= 3, "至少是3*3阶方阵"
assert row <= self.row and column <= self.column, "下标超出范围"
M = Matrix(self.column-1, self.row-1)
for r in range(self.row):
if r == row:
continue
for c in range(self.column):
if c == column:
continue
rr = r-1 if r > row else r
cc = c-1 if c > column else c
M[rr, cc] = self[r, c]
return M
def det(self):
'''计算行列式(determinant)'''
assert self.row == self.column,"非行列式，不能计算"
if self.shape == (2,2):
return self[1,1]*self[2,2]-self[1,2]*self[2,1]
else:
sum = 0.0
for c in range(self.column+1):
sum += (-1)**(c+1)*self[1,c]*self.cofactor(1,c).det()
return sum
def zeros(self):
'''全零矩阵'''
M = Matrix(self.column, self.row, fill=0.0)
return M
def ones(self):
'''全1矩阵'''
M = Matrix(self.column, self.row, fill=1.0)
return M
def identity(self):
'''单位矩阵'''
assert self.row == self.column, "非n*n矩阵，无单位矩阵"
M = Matrix(self.column, self.row)
for r in range(self.row):
for c in range(self.column):
M[r, c] = 1.0 if r == c else 0.0
return M
def show(self):
'''打印矩阵'''
for r in range(self.row):
for c in range(self.column):
print(self[r+1, c+1],end=' ')
print()
if __name__ == '__main__':
m = Matrix(3,3,fill=2.0)
n = Matrix(3,3,fill=3.5)
m[1] = [1.,1.,2.]
m[2] = [1.,2.,1.]
m[3] = [2.,1.,1.]
p = m * n
q = m*2.1
r = m**3
#r.show()
#q.show()
#print(p[1,1])
#r = m.invert()
#s = r*m
print()
m.show()
print()
#r.show()
print()
#s.show()
print()
print(m.det())
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希望本文所述对大家Python程序设计有所帮助。
原文链接：http://www.cnblogs.com/hhh5460/p/4314231.html