Function definition throwing UndefVarError

New to Julia
,
2

Wow, that’s really weird. Even worse is that even if you didn’t get an UndefVarError, you’d still get the wrong answer due to the issue discussed here: Conditional function definition

The fix for both problems is to use an anonymous function:

using Statistics: std, mean, median, var
function foo(a,b,c)
    """
    Do some stuff 
    """

    if a > 0
        if a == 1
            bar = x -> std(x, corrected=false)
        elseif a == 2
            bar = x -> var(x, corrected=false)
        elseif a == 3
            bar = x -> mean(abs.(x .- mean(x)))
        elseif a == 4
            bar = x -> median(abs.(x .- median(x)))    
        end
    else
        """
        Do some other stuff instead
        """
    end

    return c*bar(b)
end

(this will still error unless a is 1, 2, 3, or 4, in which case bar really is undefined.)

2 Likes