不使用numpy完成对矩阵的操作
众所周知，矩阵操作是进行数据分析的必要操作，而一般进行numpy便能对矩阵进行轻松操作。这里我尝试使用python手动编写对矩阵的操作。
矩阵输出
def print_matrix(A,lable=''):
if lable=='':
pass
else:
print(lable+"=")
print('[',end='')
sum=0
for i in A:
print('[',end='')
count=0
for j in i:
count+=1
if count == len(i):
print(" %2.2f " %j,end='')
else:
print(" %2.2f ,"%j, end='')
sum+=1
if sum==len(A ):
print(']',end='')
print(']')
else:
print(']')零矩阵生成
def zeros(x,y=''):
if y=='':
return [[0] * x for i in range(x)]
return [[0] * x for i in range(y)]对角矩阵
def identity_matrix(n):
list =zeros(n)
for i in range(0,n):
for j in range(0,n):
if(i==j):
list[i][j]=1
return list交换矩阵两行
def swap_rows(M, src, dest):
M[src],M[dest] = M[dest],M[src]
pass将矩阵某行乘某值
def mult_row_scalar(M, row, scalar):
for i in range(len(M[row])):
M[row][i]=M[row][i]*scalar
pass将矩阵某行加上某行
def add_row_into(M, src, dest):
for i in range(len(M[dest])):
M[dest][i] = M[dest][i]+M[src][i]
pass两个矩阵相加
def add_matrices(A, B):
assert len(A[0]) == len(B[0]), 'error'
assert len(A)==len(B) , 'error'
for i in range(len(A)):
for j in range(len(A[i])):
A[i][j]=A[i][j]+B[i][j]
return A两个矩阵相减
def sub_matrices(A, B):
assert len(A[0]) == len(B[0]), 'error'
assert len(A) == len(B), 'error'
for i in range(len(A)):
for j in range(len(A[i])):
A[i][j]=A[i][j]-B[i][j]
return A矩阵乘法
def mult_scalar(M, s):
for i in range(len(M)):
for j in range(len(M[i])):
M[i][j]=M[i][j]*s
return M矩阵点积
def dot_product(M, N):
c = []
for i in range(0, len(M)):
temp = []
for j in range(0, len(N[0])):
s = 0
for k in range(0, len(M[0])):
s += M[i][k] * N[k][j]
temp.append(s)
c.append(temp)
return c取矩阵的任意行列
def create_sub_matrix1(M, exclude_row,exclude_col ):
A=M.copy()
del A[exclude_row]
print_matrix(A)
for i in range(len(A)):
if(exclude_col==len(M[i])):
del A[i][exclude_col]
break
else:
for j in range(exclude_col,len(M[i])-1):
A[i][j]=A[i][j+1]
l=[]
for i in range(0, len(M)-1):
temp = []
for j in range(0, len(M[0])-1):
s = 0
s=A[i][j]
temp.append(s)
l.append(temp)
return l矩阵行列式
def determinant1(matrix, mul=1):
width = len(matrix)
if width == 1:
return mul * matrix[0][0]
else:
sign = -1
sum = 0
for i in range(width):
m = []
for j in range(1, width):
buff = []
for k in range(width):
if k != i:
buff.append(matrix[j][k])
m.append(buff)
sign *= -1
sum += mul * determinant1(m, sign * matrix[0][i])
return sum
def determinant(matrix):
return determinant1(matrix,1)矩阵余子式
def matrix_of_minors(M):
list=[[0] * len(M)] * len(M[0])
for i in range(len(M)):
for j in range(len(M[0])):
A=create_sub_matrix(M,i,j)
d=determinant(A)
list[i][j]+=d
return list单位矩阵
def identity(M):
row=len(M)
column=len(M[0])
assert row == column, "Non n by n matrix, no identity matrix"
for r in range(row):
for c in range(column):
M[r][ c] = 1.0 if r == c else 0.0
return M逆矩阵
def submatrix(A,i,j):
p=len(A)
q=len(A[0])
C=[[A[x][y] for y in range(q) if y!=j] for x in range(p) if x!=i]
return C
def det(A):
p=len(A)
q=len(A[0])
if(p==1 and q==1):
return A[0][0]
else:
value=0
for j in range(q):
value+=((-1)**(j+2))*A[0][j]*det(submatrix(A,0,j))
return value
import copy
def inverse_matrix(M):
p = len(A)
q = len(A[0])
C = copy.deepcopy(A)
d = det(A)
print(d)
for i in range(p):
for j in range(q):
C[i][j] = ((-1) ** (i + j + 2)) * det(submatrix(A, j, i))
C[i][j] = C[i][j] / d
return C总结
练习了对矩阵的操作，加深了对矩阵的理解同时也回顾了线性代数的知识。希望对大家有帮助。