在 SymPy 中打印四舍五入到 3 位小数的输出

Question

我有一个 SymPy 矩阵 M

In [1]: from sympy import *
In [2]: M = 1/10**6 * Matrix(([1, 10, 100], [1000, 10000, 100000]))
In [3]: M
Out[3]: 
Matrix([
[1.0e-6, 1.0e-5, 0.0001],
[ 0.001,   0.01,    0.1]])

我想打印四舍五入到 3 位小数的输出，如下所示：

In [3]: M
Out[3]: 
Matrix([
[ 0.000, 0.000, 0.000],
[ 0.001, 0.010, 0.100]])

在普通的 Python 中，我会这样：

In [5]: '{:.3f}'.format(1/10**6)
Out[5]: '0.000'

但是如何在 SymPy 矩阵中做呢？

此外，更一般的情况是一个还包含符号的表达式

x = symbols('x')
M = 1/10**6 * Matrix(([1, 10*x, 100], [1000, 10000*x, 100000]))

首选输出是

In [3]: M
Out[3]: 
Matrix([
[ 0.000, 0.000*x, 0.000],
[ 0.001, 0.010*x, 0.100]])

Answer 1

要舍入表达式中的每个数字，请使用以下函数

def round_expr(expr, num_digits):
    return expr.xreplace({n : round(n, num_digits) for n in expr.atoms(Number)})

It can be applied to any SymPy expression, including matrices.

x = symbols('x')
M = 1/10**6 * Matrix(([1, 10*x, 100], [1000, 10000*x, 100000]))
round_expr(M, 3)

yields

Matrix([
[  0.0,      0, 0.0],
[0.001, 0.01*x, 0.1]])

For numeric matrices, the following is simpler and preserves trailing zeros:

>>> M.applyfunc(lambda x: Symbol('{:.3f}'.format(x)))
Matrix([
[0.000, 0.000, 0.000],
[0.001, 0.010, 0.100]])

Here, applyfunc applies the given function to each entry of M. A natural thing to try would be lambda x: '{:.3f}'.format(x) but SymPy matrices aren't really meant to hold strings: the strings get parsed back into numbers and the trailing zeros get dropped, resulting in

Matrix([
[  0.0,  0.0, 0.0],
[0.001, 0.01, 0.1]])

So I wrap each string in Symbol, thus making a new symbol out of it. Symbols can appear in matrices, and they printed as their names, in this case the names are "0.000" and so on.