为什么我没有匹配Complex(:: Sym)的方法?

时间:2019-10-26 14:01:34

标签: julia sympy pycall

我有一个要执行的函数,它已经被编译,但是执行时显示了一个MethodError,这里是函数

为此功能,我使用的是SymPy

function op_mat(op)
           op = op.as_poly(domain="C")
           op_a = op.x.gens
           nab = op.length()
           op_ab = ones(SymPy.Sym, nab)
           coef = zeros(Complex, nab)
           mat = zeros(Int64, length(op_a), nab)
           for (i, (ps, c)) in enumerate(op.as_dict())
               for (j, p) in enumerate(ps)
                   mat[j, i] = p
                   op_ab[i] = op_a[j]^p * op_ab[i]
               end
               coef[i] = c
           end

           return op_a, op_ab, mat, coef
       end

我收到的错误消息是这样的:

  Complex(::T<:Number) where T<:Number at boot.jl:718
  Complex(::Real) at complex.jl:16
  Complex(::T<:Real, ::T<:Real) where T<:Real at complex.jl:12
  ...
Stacktrace:
 [1] convert(::Type{Complex}, ::Sym) at ./number.jl:7
 [2] setindex!(::Array{Complex,1}, ::Sym, ::Int64) at ./array.jl:766
 [3] op_mat(::Sym) at ./REPL[3]:13
 [4] top-level scope at REPL[7]:1

1 个答案:

答案 0 :(得分:1)

由于c的类型为type room struct { L int W int A int } func newRoom(length, width, int) room { return room{ L: length, W: width, A: length * width, } } ,我只应将其类型更改为复杂的Symcoef[i]=complex(c)