非交互地将函数参数传递给Julia [英] Pass function arguments into Julia non-interactively

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问题描述

我在文件中有一个Julia函数.可以说是下面的.现在,我想将参数传递给该函数.我尝试过

I have a Julia function in a file. Let's say it is the below. Now I want to pass arguments into this function. I tried doing

julia filename.jl randmatstat(5) 

但这会产生一个错误,即'('标记是意外的.不确定解决方案是什么.如果还有一个主要功能/如何使用Julia编写完整的解决方案,我也会有些不知所措.例如程序的开始/入口是什么?

but this gives an error that '(' token is unexpected. Not sure what the solution would be. I am also a little torn on if there is a main function / how to write a full solution using Julia. For example what is the starting / entry point of a Julia Program?

function randmatstat(t)
    n = 5
    v = zeros(t)
    w = zeros(t)
    for i = 1:t
        a = randn(n,n)
        b = randn(n,n)
        c = randn(n,n)
        d = randn(n,n)
        P = [a b c d]
        Q = [a b; c d]
        v[i] = trace((P.'*P)^4)
        w[i] = trace((Q.'*Q)^4)
    end
    std(v)/mean(v), std(w)/mean(w)
end

推荐答案

Julia没有这样的入口点". 从终端调用julia myscript.jl时,实际上是在要求julia执行脚本并退出.因此,它必须是一个脚本.如果脚本中只有一个函数定义,那么除非您稍后从脚本中调用该函数,否则它不会做很多事情.

Julia doesn't have an "entry point" as such. When you call julia myscript.jl from the terminal, you're essentially asking julia to execute the script and exit. As such, it needs to be a script. If all you have in your script is a function definition, then it won't do much unless you later call that function from your script.

对于参数,如果调用julia myscript.jl 1 2 3 4,则所有其余参数(在本例中为1、2、3和4)将成为具有特殊名称ARGS的字符串数组.您可以使用此特殊变量来访问输入参数.

As for arguments, if you call julia myscript.jl 1 2 3 4, all the remaining arguments (i.e. in this case, 1, 2, 3 and 4) become an array of strings with the special name ARGS. You can use this special variable to access the input arguments.

例如如果您有朱莉娅脚本,上面写着:

e.g. if you have a julia script which simply says:

# in julia mytest.jl
show(ARGS)

然后从linux终端调用它会得到以下结果:

Then calling this from the linux terminal will give this result:

<bashprompt> $ julia mytest.jl 1 two "three and four"
UTF8String["1","two","three and four"]

编辑:因此,据我从您的程序中了解到的那样,您可能想要执行以下操作(注意:在julia中,需要在调用函数之前对其进行定义).

So, from what I understand from your program, you probably want to do something like this (note: in julia, the function needs to be defined before it's called).

# in file myscript.jl
function randmatstat(t)
    n = 5
    v = zeros(t)
    w = zeros(t)
    for i = 1:t
        a = randn(n,n)
        b = randn(n,n)
        c = randn(n,n)
        d = randn(n,n)
        P = [a b c d]
        Q = [a b; c d]
        v[i] = trace((P.'*P)^4)
        w[i] = trace((Q.'*Q)^4)
    end
    std(v)/mean(v), std(w)/mean(w)
end

t = parse(Int64, ARGS[1])
(a,b) = randmatstat(t)
print("a is $a, and b is $b\n")

然后从您的linux终端调用它,如下所示:

And then call this from your linux terminal like so:

julia myscript.jl 5

这篇关于非交互地将函数参数传递给Julia的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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