Ok, no exeption any more if I use j.call instead of j.run.
But the result is wrong:
import julia
j=julia.Julia()
In [3]: j.call("2+2")
Out[3]: 49655584
Any idea?
Uwe
On Tuesday, September 9, 2014 1:49:55 PM UTC+2, Isaiah wrote:
>
> I think you need `julia.call` instead of `julia.run`.
>
> On Tue, Sep 9, 2014 at 7:16 AM, Uwe Fechner <uwe.fec...@gmail.com
> <javascript:>> wrote:
>
>> Hello,
>>
>> I have a function, that is running much faster in Julia then in Python.
>>
>> Now I want to call it from my (large) Python program.
>>
>> I tried to do this, using pyjulia from
>> https://github.com/JuliaLang/pyjulia .
>>
>> I am using Ubuntu 12.04, 64 bits and Julia 0.3 and tried to install
>> pyjulia in the following way:
>>
>> cd ~
>> mkdir 00Software
>> cd 00Software
>> git clone https://github.com/JuliaLang/pyjulia.git
>> cd pyjulia
>> sudo python setup.py install
>>
>> Then I started ipython and tried the following code:
>>
>> import julia
>> j=julia.Julia()
>> j.run("2+2")
>>
>> I get the following error message:
>>
>> In [3]: j.run("2+2")
>>
>> ---------------------------------------------------------------------------
>> RuntimeError Traceback (most recent call
>> last)
>> <ipython-input-3-4f45159364ef> in <module>()
>> ----> 1 j.run("2+2")
>>
>> RuntimeError: Julia exception: MethodError(run,("2+2",))
>>
>> In [4]:
>>
>> Any idea what is going wrong?
>>
>> Best regards:
>>
>> Uwe Fechner
>>
>
>
