CODEWITHSUNDEEP
julia
davidj
davidj

Reputation: 58

Using result from a function to access a julia array

I'm new to julia and have a problem. I have an array, y. 

y = rand (5, 10)
y [1. 1] = 0

Running this gives me an error

for j=1:d   
    x_filt [j, 1] = y [j, findfirst (y [j, :])]
end

ERROR: syntax: missing separator in array expression

But this doesn't 

for j=1:d   # fix to 1st obs if 1st tick is missing
    temp = findfirst (y [j, :])
    x_filt [j, 1] = y [j, temp];        
end

Can someone explain how to make the first version work? Or at least explain why it doesn't?

Thanks!

Upvotes: 0

Views: 686

Answers (1)

Toivo Henningsson
Toivo Henningsson

Reputation: 2699

First, I guess you meant y[1, 1] = 0? I get an error if I use y [1. 1] = 0.

Julia has space sensitive syntax in some contexts, notable inside brackets []. Some examples:

julia> max(1, 2)
2

julia> max (1, 2)
2

julia> [max(1, 2)]
1-element Array{Int64,1}:
 2

julia> [max (1, 2)]
1x2 Array{Any,2}:
 max  (1,2)

julia> [1 + 2]
1-element Array{Int64,1}:
 3

julia> [1 +2]
1x2 Array{Int64,2}:
 1  2

In your first example, the call to findfirst in

x_filt [j, 1] = y [j, findfirst (y [j, :])]

is interpreted as two space-separated items, findfirst and (y [j, :])]. Julia then complains that they are separated by a space and not a comma.

In your second example, you were able to circumvent this since the call to findfirst in

temp = findfirst (y [j, :])

is no longer in a space sensitive context.

I would recommend that when writing Julia code, you should never put a space between the function name and parenthesis ( in a function call or the variable and bracket [ in indexing, because the code will be treated differently in space sensitive contexts. E.g., your first example without the extra spaces

for j=1:d   
    x_filt[j, 1] = y[j, findfirst(y[j, :])]
end

works fine (provided that you define d and x_filt appropriately first).

Upvotes: 7

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