Question

我尝试使用此http://www.irma-international.org/viewtitle/41011/算法来反转nxn矩阵。

我在这个矩阵上运行了这个函数

[[1.0, -0.5],

 [-0.4444444444444444, 1.0]]

并获得输出

[[ 1.36734694,  0.64285714]

 [ 0.57142857,  1.28571429]]

正确的输出是

[[ 1.28571429,  0.64285714]

 [ 0.57142857,  1.28571429]]

我的功能：

def inverse(m):
    n = len(m)
    P = -1
    D =  1
    mI = m
    while True:
        P += 1
        if m[P][P] == 0:
           raise Exception("Not Invertible")
        else:
            D = D * m[P][P]
            for j in range(n):
                if j != P:
                    mI[P][j] =  m[P][j] / m[P][P]
            for i in range(n):
                if i != P:
                    mI[i][P] = -m[i][P] / m[P][P]
            for i in range(n):
                for j in range(n):
                    if i != P and j != P:
                        mI[i][j] = m[i][j] + (m[P][j] * mI[i][P])
            mI[P][P] = 1 / m[P][P]
            if P == n - 1: # All elements have been looped through
                break

    return mI

我在哪里弄错了？

Answer 1

https://repl.it/repls/PowerfulOriginalOpensoundsystem

输出

逆：[[十进制（'1.285714285714285693893862813'），十进制（ '0.6428571428571428469469314065'）]， [十进制（ '0.5714285714285713877877256260'），十进制（ '1.285714285714285693893862813'）]] numpy：[[1.28571429 0.64285714] [0.57142857 1.28571429]]

from decimal import Decimal
import numpy as np

def inverse(m):
    m = [[Decimal(n) for n in a] for a in m]
    n = len(m)
    P = -1
    D =  Decimal(1)
    mI = [[Decimal(0) for n in a] for a in m]
    while True:
        P += 1
        if m[P][P] == 0:
           raise Exception("Not Invertible")
        else:
            D = D * m[P][P]
            for j in range(n):
                if j != P:
                    mI[P][j] =  m[P][j] / m[P][P]
            for i in range(n):
                if i != P:
                    mI[i][P] = -m[i][P] / m[P][P]
            for i in range(n):
                for j in range(n):
                    if i != P and j != P:
                      mI[i][j] = m[i][j] + (m[P][j] * mI[i][P])
            mI[P][P] = 1 / m[P][P]
            m = [[Decimal(n) for n in a] for a in mI]
            mI = [[Decimal(0) for n in a] for a in m]
            if P == n - 1: # All elements have been looped through
                break

    return m

m = [[1.0, -0.5],

 [-0.4444444444444444, 1.0]]

print(inverse(m))
print(np.linalg.inv(np.array(m)))

我的思考过程：

起初，我认为你可能潜伏着浮点出现错误。事实证明这不是真的。这就是Decimal爵士乐的用途。

你的错误就在这里

mI = m # this just creates a pointer that points to the SAME list as m

在这里

for i in range(n):
                for j in range(n):
                    if i != P and j != P:
                        mI[i][j] = m[i][j] + (m[P][j] * mI[i][P])
            mI[P][P] = 1 / m[P][P] 
            # you are not copying mI to m for the next iteration
            # you are also not zeroing mI 
            if P == n - 1: # All elements have been looped through
                break

    return mI

遵循算法，每次迭代都会创建一个NEW a'矩阵，它不会继续修改相同的旧a。我推断这意味着在循环不变量中，a变为'。适用于您的测试用例，结果证明是真的。

稍微关闭python中的矩阵

1 个答案: