<p>您可以尝试使用库<a href="http://mpmath.org/" rel="nofollow noreferrer">mpmath</a>,它可以对任意精度的数字进行简单的矩阵代数和其他类似的问题。在</p>
<p><strong>几个注意事项:</strong>它几乎肯定比使用numpy慢,而且,正如Lutzl在<a href="https://stackoverflow.com/a/29875756/4657412">his answer to this question</a>中指出的那样,这个问题很可能在数学上没有很好的定义。另外,你需要在开始之前决定你想要的精度。在</p>
<p>一些简单的示例代码</p>
<pre><code>from mpmath import mp, matrix
# set the precision - see http://mpmath.org/doc/current/basics.html#setting-the-precision
mp.prec = 5000 # set it to something big at the cost of speed.
# Ideally you'd precalculate what you need.
# a quick trial with 100*100 showed that 5000 works and 500 fails
# see the documentation at http://mpmath.org/doc/current/matrices.html
# where xtotal is the output from arrayit
my_matrix = matrix(xtotal) # I think this should work. If not you'll have to create it and copy
# do the inverse
xinverted = my_matrix**-1
coeff = xinverted*matrix(ylist)
# note that as lutlz pointed out you really want to use solve instead of calculating the inverse.
# I think this is something like
from mpmath import lu_solve
coeff = lu_solve(my_matrix,matrix(ylist))
</code></pre>
<p>我怀疑你真正的问题是数学而不是软件,所以我怀疑这对你是否有效,但它总是有可能的!在</p>