Question

我正在玩Julia并且我正在使用Sympy我认为使用PyCall来调用Python。

当我运行下面的脚本时，我收到一个很长的错误。在这里发布所有内容太长了，但这是它的开始：

LoadError: PyError (ccall(@pysym(:PyObject_Call), PyPtr, (PyPtr, PyPtr, 

PyPtr), o, arg, C_NULL)) <type 'exceptions.RuntimeError'>
RuntimeError('maximum recursion depth exceeded while calling a Python object',)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\cache.py", line 93, in wrapper
    retval = cfunc(*args, **kwargs)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\compatibility.py", line 809, in wrapper
    result = user_function(*args, **kwds)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\function.py", line 427, in __new__
    result = super(Function, cls).__new__(cls, *args, **options)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\cache.py", line 93, in wrapper
    retval = cfunc(*args, **kwargs)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\compatibility.py", line 809, in wrapper
    result = user_function(*args, **kwds)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\function.py", line 250, in __new__
    evaluated = cls.eval(*args)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\functions\elementary\integers.py", line 25, in eval
    if arg.is_imaginary or (S.ImaginaryUnit*arg).is_real:
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\decorators.py", line 91, in __sympifyit_wrapper
    return func(a, b)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\decorators.py", line 132, in binary_op_wrapper
    return func(self, other)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\expr.py", line 140, in __mul__
    return Mul(self, other)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\cache.py", line 93, in wrapper
    retval = cfunc(*args, **kwargs)
  File "d:\Users\OEM\AppData\Local\JuliaPro-0.6.0.1\pkgs-0.6.0.1\v0.6\Conda\deps\usr\lib\site-packages\sympy\core\compatibility.py", line 809, in wrapper
    result = user_function(*args, **kwds)

正如你可能看到的那样，最后它会重复：看到最后的第93行，然后是第140行，然后第93行......

这是我的代码：

function oddPeriodSquareRoots()
#=
  Get the length of the continued fraction for square root of for the number i.
  E.g. √7=[2;(1,1,1,4)]
=#


irrationalNumber, intPart, fractionalPart = symbols(string("irrationalNumber intPart fractionalPart"))

for i in [6451]

    # For perfect squares, the period is 0
    irrationalNumber = BigFloat(sqrt(BigFloat(i)))
    if irrationalNumber == floor(irrationalNumber)
        continue
    end

    # Get the continued fraction using symbolic programming
    irrationalNumber = sqrt(Sym(i))

    continuedFractionLength = 0
    while true

        intPart = Sym(BigInt(floor(irrationalNumber)))
        if continuedFractionLength == 0
            firstContinuedFractionTimes2 = intPart*2
        end

        continuedFractionLength += 1
        if intPart == firstContinuedFractionTimes2
            break
        end

        fractionalPart = irrationalNumber - intPart
        irrationalNumber = 1 / fractionalPart

    end

    continuedFractionLength -= 1 # We ignore the first term.


end


return continuedFractionLength
end

此例程计算某个数字的平方根的连续分数的长度。对于数字6451，它给出了错误。

所以我的问题是，这可以解决吗？

Answer 1

我很高兴找到了递归限制解决方案。这在以前没有见过。这个评论是关于如何简化你的SymPy代码，因为你似乎对此感到困惑。基本上，你只需要使你的初始值符号化，然后Julia的方法应该（在大多数情况下）处理其余的。这是一个轻微的重写：

using SymPy

使用PyCall @pyimport sys sys.setrecursionlimit（10000）

“”” 获得数字i的平方根的连续分数的长度。例如。 √7= [2;（1,1,1,4-）] “”” function oddPeriodSquareRoots（n）

i = Sym(n)
# For perfect squares, the period is 0
continuedFractionLength = 0

irrationalNumber = sqrt(i)
if is_integer(irrationalNumber)
    return continuedFractionLength
end

# Get the continued fraction using symbolic programming

while true

    intPart = floor(irrationalNumber)
    if continuedFractionLength == 0
        firstContinuedFractionTimes2 = intPart*2
    end

        continuedFractionLength += 1
    if intPart == firstContinuedFractionTimes2
        break
    end

    fractionalPart = irrationalNumber - intPart
    irrationalNumber = 1 / fractionalPart

end

continuedFractionLength -= 1 # We ignore the first term.


return continuedFractionLength

端

Answer 2

非常感谢大家的投入。我设法通过将这些行放在文件的顶部来解决这个问题（除了我一直使用的“使用Sympy”行之外）：

using SymPy

using PyCall
@pyimport sys
sys.setrecursionlimit(10000)

这在Python中设置了递归限制。不知道为什么它必须如此大才能发挥作用。

我还删除了一些类型转换等。我认为这可能有助于错误和/或速度。但事实并非如此。

此外，删除我用符号声明变量的行不会阻止代码工作。

irrationalNumber, intPart, fractionalPart = symbols(string("irrationalNumber intPart fractionalPart"))

Python相同。所以不确定它的重点是什么。

但是在朱莉娅，无论哪种方式，我必须围绕这两行包含Sym（）包装：

irrationalNumber = sqrt(Sym(i))
...
intPart = Sym(floor(irrationalNumber))

通过检查这些类型，使用typeof，我可以看到它们是象征性的，而不是浮动的。没有它们，一切都会变成花车，所以我不会象征性地做它。

从Julia调用Python时出现PyCall错误

2 个答案: