三对角矩阵算法(TDMA)，又名托马斯算法，使用Python和NumPy数组

Question

我在MATLAB中找到了thomas算法或TDMA的实现。

function x = TDMAsolver(a,b,c,d)
    %a, b, c are the column vectors for the compressed tridiagonal matrix, d is the right vector
    n = length(b); % n is the number of rows
    % Modify the first-row coefficients
    c(1) = c(1) / b(1);    % Division by zero risk.
    d(1) = d(1) / b(1);    % Division by zero would imply a singular matrix.
    for i = 2:n-1
        temp = b(i) - a(i) * c(i-1);
        c(i) = c(i) / temp;
        d(i) = (d(i) - a(i) * d(i-1))/temp;
    end
    d(n) = (d(n) - a(n) * d(n-1))/( b(n) - a(n) * c(n-1));
    % Now back substitute.
    x(n) = d(n);
    for i = n-1:-1:1
        x(i) = d(i) - c(i) * x(i + 1);
    end
end

我需要在python中使用numpy数组，这是我在python中首次尝试该算法。

import numpy
aa = (0.,8.,9.,3.,4.)
bb = (4.,5.,9.,4.,7.)
cc = (9.,4.,5.,7.,0.)
dd = (8.,4.,5.,9.,6.)
ary = numpy.array
a = ary(aa)
b = ary(bb)
c = ary(cc)
d = ary(dd)
n = len(b)## n is the number of rows
## Modify the first-row coefficients
c[0] = c[0]/ b[0]    ## risk of Division by zero.
d[0] = d[0]/ b[0]
for i in range(1,n,1):
    temp = b[i] - a[i] * c[i-1]
    c[i] = c[i]/temp
    d[i] = (d[i] - a[i] * d[i-1])/temp
d[-1] = (d[-1] - a[-1] * d[-2])/( b[-1] - a[-1] * c[-2])
## Now back substitute.
x = numpy.zeros(5)
x[-1] = d[-1]
for i in range(-2, -n-1, -1):
    x[i] = d[i] - c[i] * x[i + 1]

他们给出了不同的结果，那么我做错了什么呢？

Answer 1

我之所以这么做，是因为没有一个python的在线实现能真正工作。我已经针对内置的矩阵反转对其进行了测试，结果是匹配的。

此处为a = Lower Diag、b = Main Diag、c = Upper Diag、d = solution vector

import numpy as np
def TDMA(a,b,c,d):
    n = len(d)
    w= np.zeros(n-1,float)
    g= np.zeros(n, float)
    p = np.zeros(n,float)
    
    w[0] = c[0]/b[0]
    g[0] = d[0]/b[0]
    for i in range(1,n-1):
        w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
    for i in range(1,n):
        g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
    p[n-1] = g[n-1]
    for i in range(n-1,0,-1):
        p[i-1] = g[i-1] - w[i-1]*p[i]
    return p

要轻松提高大型矩阵的性能，请使用numba！此代码在我的测试中优于np.linalg.inv()：

import numpy as np
from numba import jit    
@jit
def TDMA(a,b,c,d):
    n = len(d)
    w= np.zeros(n-1,float)
    g= np.zeros(n, float)
    p = np.zeros(n,float)
    
    w[0] = c[0]/b[0]
    g[0] = d[0]/b[0]
    for i in range(1,n-1):
        w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
    for i in range(1,n):
        g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
    p[n-1] = g[n-1]
    for i in range(n-1,0,-1):
        p[i-1] = g[i-1] - w[i-1]*p[i]
    return p

Answer 2

两者之间至少有一个区别：

for i in range(1,n,1):

在Python中，从索引1迭代到最后一个索引n-1，而

for i = 2:n-1

从索引1（基于零）迭代到last-1索引，因为Matlab具有基于一的索引。

Answer 3

在python中写这样的东西会非常慢。你最好使用LAPACK来完成数字繁重的工作，并使用python来处理它周围的一切。LAPACK经过编译，因此它的运行速度将比python快得多，它的优化程度也比我们大多数人所能匹配的要高得多。

SciPY为LAPACK提供了低级别的包装器，因此您可以非常简单地从python中调用它，您正在寻找的包装器可以在这里找到：

https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.lapack.dgtsv.html#scipy.linalg.lapack.dgtsv

Answer 4

在循环中，Matlab版本对倒数第二个元素进行迭代。要在Python中执行同样的操作，您需要：

for i in range(1,n-1):

（正如voithos的评论中所指出的，这是因为range函数排除了最后一个索引，所以除了更改为0索引之外，还需要对此进行更正）。