Repeat with y as k

New to Julia
1 
d = .05
n = .01
g = .015
s = .3
k=6
yn( k, s, d, g,) = k-(d+n+g)*k + s*k^(1/3)

I have this function, and it calculates yn= 6.09514 . Now I would like to replace k with yn, and find a new yn, and repeat that 50 times.

2

Use a for-loop?

julia> function do_stuff(N)
           d = 0.05
           n = 0.01
           g = 0.015
           s = 0.3
           k = 6.0
           yn(k, s, d, g) = k - (d + n + g) * k + s * k^(1/3)
           for _ in 1:N
               k = yn(k, s, d, g)
           end
           return k
       end
do_stuff (generic function with 1 method)

julia> do_stuff(1)
6.095136177849642

julia> do_stuff(50)
7.839026641775517
4 Likes
3

Thanks, that was easy, I’ll be sure to remember than in the future. Now I need to put each value of do_stuff in a data frame, that shows each value of do_stuff.

4 
function do_stuffdf(N)
    d = 0.05
    n = 0.01
    g = 0.015
    s = 0.3
    k = 6.0
    yn(k, s, d, g) = k - (d + n + g) * k + s * k^(1/3)

    df = DataFrame([Float64], [:k], 0)
    for _ in 1:N
        k = yn(k, s, d, g)
        push!(df, (k,))
    end
    return df
end

e.g.

1 Like
5

it’s saying k not defined, not sure why, since I started with k=6 and keep repeating k.

6

Ok, yeah I put in the whole think, but it still only gives me one point.

7

Oh, to get the DataFrame just define the function as posted and do:

mydataframe = do_stuffdf(50)
1 Like
8

[There was a warning that you should not push Float64 into a DataFrame? I changed push!(df, k)to push!(df, (k,)) and the warning is gone.]
One point?

julia> do_stuffdf(50)
50×1 DataFrame
│ Row │ k       │
│     │ Float64 │
├─────┼─────────┤
│ 1   │ 6.09514 │
│ 2   │ 6.186   │
│ 3   │ 6.27277 │
│ 4   │ 6.35558 │
│ 5   │ 6.43461 │
│ 6   │ 6.51001 │
│ 7   │ 6.58192 │
│ 8   │ 6.6505  │
⋮
│ 42  │ 7.7578  │
│ 43  │ 7.76985 │
│ 44  │ 7.7813  │
│ 45  │ 7.79218 │
│ 46  │ 7.80253 │
│ 47  │ 7.81236 │
│ 48  │ 7.8217  │
│ 49  │ 7.83059 │
│ 50  │ 7.83903 │
1 Like
10

Cool, work perfect, except that it starts with k =0, instead of k=6

11

The first k was missing. So this is an alternative, and k starts with 6.0

julia> function do_stuffdf(N)
           d = 0.05
           n = 0.01
           g = 0.015
           s = 0.3
           k = 6.0
           yn(k, s, d, g) = k - (d + n + g) * k + s * k^(1/3)

           df = DataFrame(k = k)
           for _ in 1:N
                   k = yn(k, s, d, g)
               push!(df, (k,))
           end
           return df
       end
do_stuffdf (generic function with 1 method)

julia> do_stuffdf(50)
51×1 DataFrame
│ Row │ k       │
│     │ Float64 │
├─────┼─────────┤
│ 1   │ 6.0     │
│ 2   │ 6.09514 │
│ 3   │ 6.186   │
│ 4   │ 6.27277 │
│ 5   │ 6.35558 │
│ 6   │ 6.43461 │
│ 7   │ 6.51001 │
│ 8   │ 6.58192 │
⋮
│ 43  │ 7.7578  │
│ 44  │ 7.76985 │
│ 45  │ 7.7813  │
│ 46  │ 7.79218 │
│ 47  │ 7.80253 │
│ 48  │ 7.81236 │
│ 49  │ 7.8217  │
│ 50  │ 7.83059 │
│ 51  │ 7.83903 │
12

AWESOME!!! That’s it.

13

If you want to create the DataFrame at top-level, you can put k in the context of the definition of yn:

julia> let k = 6.0
           global yn, setk
           function yn()
               d = 0.05
               n = 0.01
               g = 0.015
               s = 0.3
               r = k
               k = k - (d + n + g) * k + s * k^(1/3)
               return r
           end

           setk(kinit) = (k = kinit)
       end
setk (generic function with 1 method)

julia> setk(6.0)
6.0

julia> dfyn = DataFrame(yn = [yn() for i in 1:50])
50×1 DataFrame
│ Row │ yn      │
│     │ Float64 │
├─────┼─────────┤
│ 1   │ 6.0     │
│ 2   │ 6.09514 │
│ 3   │ 6.186   │
│ 4   │ 6.27277 │
│ 5   │ 6.35558 │
│ 6   │ 6.43461 │
│ 7   │ 6.51001 │
│ 8   │ 6.58192 │
⋮
│ 42  │ 7.74512 │
│ 43  │ 7.7578  │
│ 44  │ 7.76985 │
│ 45  │ 7.7813  │
│ 46  │ 7.79218 │
│ 47  │ 7.80253 │
│ 48  │ 7.81236 │
│ 49  │ 7.8217  │
│ 50  │ 7.83059 │