列主元高斯的python實現
相對于順序高斯只是每次循環的時候增加了一個選擇列主元的過程。
選擇列主元也就是找到余下的列中最大的一行,并以此行為主元
代碼如下:
# coding: utf8
import numpy as np
def getInput1():
matrix_a = np.mat([[0.726, 0.81, 0.9],
[1, 1, 1],
[1.331, 1.21, 1.1]
], dtype=float)
matrix_b = np.mat([0.6867, 0.8338, 1])
# 答案:-2 0 1 1
return matrix_a, matrix_b
# 設置矩陣
def getInput():
matrix_a = np.mat([[2, 3, 11, 5],
[1, 1, 5, 2],
[2, 1, 3, 2],
[1, 1, 3, 4]], dtype=float)
matrix_b = np.mat([2, 1, -3, -3])
# 答案:-2 0 1 1
return matrix_a, matrix_b
# 交換
def swap(mat, num):
print(num)
print("調換前")
print(mat)
maxid = num
for j in range(num, mat.shape[0]):
if mat[j, num] > mat[num, num]:
maxid = j
if maxid is not num:
mat[[maxid,num],:] = mat[[num,maxid],:]
else:pass
print("調換后")
print(maxid)
print(mat)
return mat
def SequentialGauss(mat_a):
for i in range(0, (mat_a.shape[0])-1):
swap(mat_a, i)
if mat_a[i, i] == 0:
print("終斷運算:")
print(mat_a)
break
else:
for j in range(i+1, mat_a.shape[0]):
mat_a[j:j+1 , :] = mat_a[j:j+1,:] - \
(mat_a[j,i]/mat_a[i,i])*mat_a[i, :]
return mat_a
# 回帶過程
def revert(new_mat):
# 創建矩陣存放答案 初始化為0
x = np.mat(np.zeros(new_mat.shape[0], dtype=float))
number = x.shape[1]-1
# print(number)
b = number+1
x[0,number] = new_mat[number,b]/new_mat[number, number]
for i in range(number-1,-1,-1):
try:
x[0, i] = (new_mat[i,b]-np.sum(np.multiply(new_mat[i,i+1:b],x[0,i+1:b])))/(new_mat[i,i])
except:
print("錯誤")
print(x)
if __name__ == "__main__":
mat_a, mat_b = getInput()
# 合并兩個矩陣
print("原矩陣")
print(np.hstack((mat_a, mat_b.T)))
new_mat = SequentialGauss(np.hstack((mat_a, mat_b.T)))
print("三角矩陣")
print(new_mat)
print("方程的解")
revert(new_mat)

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