在python中使用旋转进行高斯消除-Java 学习之路

-4

我正在尝试编写一个函数来解决线性系统使用高斯消除与旋转 . 我不允许使用任何模块 . 有人可以帮帮我吗？我不知道我做错了什么 . 提前致谢：）：）：）

首先我设置了增强矩阵M，然后我进行了旋转和行操作，最后我进行了后置替换

def linearsolver(A,b):   
  n = len(A) 
  M = A 

  i = 0 
  for x in M:
   x.append(b[i]) 
   i += 1 

  for k in range(n):   
   for i in range(k,n): 
     if abs(M[i][k]) > abs(M[k][k]):  
        M[k], M[i] = M[i],M[k]  
     else: 
        pass  

   for j in range(k+1,n):
       q = M[j][k] / M[k][k] 
       for m in range(k, n+1):         
          M[j][m] +=  q * M[k][m]

  x = [0 for i in range(n)]

  x[n] =float(M[n][n+1])/M[n][n]
  for i in range (n-1,-1,-1): 
    z = 0 
    for j in range(i+1,n):
        z = z  + float(M[i][j])*x[j] 
    x[i] = float(M[i][n+1] - z)/M[i][i]
  print x

1 回答

让它变得有教育意义，因为它显然是你的功课，并解释代码中的错误，不是吗？也许你可以在这个过程中学到一些东西 .

假设有以下矩阵要解决：

A = [[3, 2, -4], [2, 3, 3], [5, -3, 1]]
b = [3, 15, 14]

你的初始代码：

def linearsolver(A,b):   
  n = len(A) 
  M = A 

  i = 0 
  for x in M:
   x.append(b[i]) 
   i += 1 

  for k in range(n):   
   for i in range(k,n): 
     if abs(M[i][k]) > abs(M[k][k]):  
        M[k], M[i] = M[i],M[k]  
     else: 
        pass  

   for j in range(k+1,n):
       q = M[j][k] / M[k][k] 
       for m in range(k, n+1):         
          M[j][m] +=  q * M[k][m]

  x = [0 for i in range(n)]

  x[n] =float(M[n][n+1])/M[n][n]
  for i in range (n-1,-1,-1): 
    z = 0 
    for j in range(i+1,n):
        z = z  + float(M[i][j])*x[j] 
    x[i] = float(M[i][n+1] - z)/M[i][i]
  print x

首先你得到一个错误，关于索引超出范围或某事

Traceback (most recent call last):
  File "solver.py", line 32, in <module>
    print linearsolver([[3, 2, -4], [2, 3, 3], [5, -3, 1]], [3, 15, 14])
  File "solver.py", line 24, in linearsolver
    x[n] =float(M[n][n+1])/M[n][n]
IndexError: list index out of range

你怎么调试这个？打印当前的计算状态：

x = [0 for i in range(n)]

  print "n = ", n
  print "x = ", x
  for row in M:
    print row

  x[n] =float(M[n][n+1])/M[n][n]

你会得到输出：

n =  3
x =  [0, 0, 0]
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]
Traceback (most recent call last):
  File "solver.py", line 37, in <module>
    print linearsolver([[3, 2, -4], [2, 3, 3], [5, -3, 1]], [3, 15, 14])
  File "solver.py", line 29, in linearsolver
    x[n] =float(M[n][n+1])/M[n][n]
IndexError: list index out of range

如果算正确，则尝试写入 x[3] 并访问 M[3][4] 和 M[3][3] . 但是，python从0开始计数，这意味着最后一个元素比预期的小-1 . 最后一个元素是 x[2] 和 M[2][3] . 这可以通过对所有索引应用-1来解决：

x[n-1] =float(M[n-1][n])/M[n-1][n-1]

跑吧......

n =  3
x =  [0, 0, 0]
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]
Traceback (most recent call last):
  File "solver.py", line 37, in <module>
    print linearsolver([[3, 2, -4], [2, 3, 3], [5, -3, 1]], [3, 15, 14])
  File "solver.py", line 34, in linearsolver
    x[i] = float(M[i][n+1] - z)/M[i][i]
IndexError: list index out of range

该死的，另一个追溯！但是，我们正在取得进展，现在它在更高的线上！当你仔细观察它时，再次使用 n+1 . 最初 A 有 n 行和 n 列 . 因为您将 b 附加到矩阵，所以有 n+1 列 . 由于从0开始索引，最后一个元素是总列数的-1： n+1-1 = n 我们修复它并且有代码：

def linearsolver(A,b):
  n = len(A)
  M = A

  i = 0
  for x in M:
   x.append(b[i])
   i += 1

  for k in range(n):
   for i in range(k,n):
     if abs(M[i][k]) > abs(M[k][k]):
        M[k], M[i] = M[i],M[k]
     else:
        pass

   for j in range(k+1,n):
       q = M[j][k] / M[k][k]·
       for m in range(k, n+1):
          M[j][m] +=  q * M[k][m]

  x = [0 for i in range(n)]

  print "n = ", n
  print "x = ", x
  for row in M:
    print row

  x[n-1] =float(M[n-1][n])/M[n-1][n-1]
  for i in range (n-1,-1,-1):·
    z = 0·
    for j in range(i+1,n):
        z = z  + float(M[i][j])*x[j]·
    x[i] = float(M[i][n] - z)/M[i][i]
  print x

当我们运行它时，我们得到结果!!! 1！

n =  3
x =  [0, 0, 0]
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]
[6.4, 5.75, -0.75]

只需检查它是否有效：

3 * 6.4 + 2 * 5.75 - 4 * -0.75 = 33.7 (not 3)
2 * 6.4 + 3 * 5.75 + 3 * -0.75 = 27.8 (not 15)
5 * 6.4 - 3 * 5.75 + 1 * -0.75 = 14.0 (correct)

嗯......我们的矩阵没有解决，但至少我们有一行正确解决了 . 该算法要求最后一步具有特定格式的矩阵，其中大多数行以0开头 . 但事实并非如你所见 . 让我们在计算时添加额外的打印来显示矩阵：

def linearsolver(A,b):
  n = len(A)
  M = A

  i = 0
  for x in M:
   x.append(b[i])
   i += 1

  for k in range(n):
   print "Iteration ", k
   for i in range(k,n):
     if abs(M[i][k]) > abs(M[k][k]):
        M[k], M[i] = M[i],M[k]
     else:
        pass

   # Show the matrix after swapping rows
   for row in M:
     print row
   print

   for j in range(k+1,n):
       q = M[j][k] / M[k][k]
       for m in range(k, n+1):
          M[j][m] +=  q * M[k][m]

   # Show matrix after multiplying rows
   for row in M:
     print row
   print

  x = [0 for i in range(n)]

  print "n = ", n
  print "x = ", x
  for row in M:
    print row

  x[n-1] =float(M[n-1][n])/M[n-1][n-1]
  for i in range (n-1,-1,-1):
    z = 0
    for j in range(i+1,n):
        z = z  + float(M[i][j])*x[j]
    x[i] = float(M[i][n] - z)/M[i][i]
  print x

并运行它：

Iteration  0
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

Iteration  1
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

Iteration  2
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

n =  3
x =  [0, 0, 0]
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]
[6.4, 5.75, -0.75]

如您所见，行乘法根本没有发生！让我们检查一下状态：

for j in range(k+1,n):
       q = M[j][k] / M[k][k]·
       print "j=", j, "q=", q
       print "M[j][k]=", M[j][k], "M[k][k]=", M[k][k]
       for m in range(k, n+1):
          M[j][m] +=  q * M[k][m]

对于迭代0，我们得到：

Iteration  0
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

j= 1 q= 0
M[j][k]= 2 M[k][k]= 5
j= 2 q= 0
M[j][k]= 3 M[k][k]= 5
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

等一下， q 不应该是0！它应该是0.4和0.6！发生这种情况是因为两个数字都是整数而python提供的结果是整数而不是浮点数 . 要解决此问题，请将其更改为 float() ：

for j in range(k+1,n):
       q = float(M[j][k]) / M[k][k]·
       print "j=", j, "q=", q
       print "M[j][k]=", M[j][k], "M[k][k]=", M[k][k]
       for m in range(k, n+1):
          M[j][m] +=  q * M[k][m]

现在我们得到结果：

Iteration  0
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

j= 1 q= 0.4
M[j][k]= 2 M[k][k]= 5
j= 2 q= 0.6
M[j][k]= 3 M[k][k]= 5
[5, -3, 1, 14]
[4.0, 1.7999999999999998, 3.4, 20.6]
[6.0, 0.20000000000000018, -3.4, 11.4]

虽然它有不同的输出，但我们在第一列中缺少零 . 为什么？代替2，有4而不是0，代替3，有6而不是0.我知道！你应该减去而不是添加乘法行：

Iteration  0
[5, -3, 1, 14]
[2, 3, 3, 15]
[3, 2, -4, 3]

j= 1 q= 0.4
M[j][k]= 2 M[k][k]= 5
j= 2 q= 0.6
M[j][k]= 3 M[k][k]= 5
[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 3.8, -4.6, -5.4]

Iteration  1
[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 3.8, -4.6, -5.4]

j= 2 q= 0.904761904762
M[j][k]= 3.8 M[k][k]= 4.2
[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 0.0, -6.952380952380952, -13.904761904761903]

Iteration  2
[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 0.0, -6.952380952380952, -13.904761904761903]

[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 0.0, -6.952380952380952, -13.904761904761903]

n =  3
x =  [0, 0, 0]
[5, -3, 1, 14]
[0.0, 4.2, 2.6, 9.399999999999999]
[0.0, 0.0, -6.952380952380952, -13.904761904761903]
[2.9999999999999996, 0.9999999999999996, 2.0]

你能在大多数行的开头看到那些0.0吗？让我们现在检查证明：

3 * 2.9999999999999996 + 2 * 0.9999999999999996 - 4 * 2.0 = 2.9999999999999964 (almost 3)
2 * 2.9999999999999996 + 3 * 0.9999999999999996 + 3 * 2.0 = 14.999999999999998 (almost 15)
5 * 2.9999999999999996 - 3 * 0.9999999999999996 + 1 * 2.0 = 14.0 (correct)

计算机存在浮点数问题，但算法使我们得到了正确的解决方案[3,1,2]

删除帮助程序打印后，有代码：

def linearsolver(A,b):
  n = len(A)
  M = A

  i = 0
  for x in M:
   x.append(b[i])
   i += 1

  for k in range(n):
   for i in range(k,n):
     if abs(M[i][k]) > abs(M[k][k]):
        M[k], M[i] = M[i],M[k]
     else:
        pass

   for j in range(k+1,n):
       q = float(M[j][k]) / M[k][k]
       for m in range(k, n+1):
          M[j][m] -=  q * M[k][m]

  x = [0 for i in range(n)]

  x[n-1] =float(M[n-1][n])/M[n-1][n-1]
  for i in range (n-1,-1,-1):
    z = 0
    for j in range(i+1,n):
        z = z  + float(M[i][j])*x[j]
    x[i] = float(M[i][n] - z)/M[i][i]
  print x

要带走的东西：

Python从0开始计算索引，这意味着最后一个索引是n-1
注意整数和浮点除法之间的区别
Print语句可以帮助您调试问题 .

回复于 2024-05-16T00:22:41+08:00

在python中使用旋转进行高斯消除

1 回答

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