Julia filling an array of array

Question

So basically I wrote this code:

a = [[], [], []]   # a is an array of array

#### there is some code that modified a, then I want to clear array a ####

fill!(a, [])       # set every element of a back to an empty array
push!(a[1], 1)     # modify one of the element
println(a)

And then, the output becomes:

[[1], [1], [1]]

This means the code fill!(a, []) actually fills the same array reference to every index in a.

Is this a bug in Julia? If this method doesn't work, what alternative solution do I have?

See comments of this question to shed light on this behavior. — Andre Wildberg
– Andre Wildberg, Commented Jan 14, 2023 at 20:44
Keep in mind that [] is the same as Any[], which creates a Vector{Any}. This means your code will be slow. If you know the element type of the inner vectors, specify them. E.g. [Int[] for _ in 1:3]. — DNF
– DNF, Commented Jan 15, 2023 at 9:28

Dan Getz · Accepted Answer · 2023-01-15 00:34:21Z

2

Playing around, the following seems to work also:

v .= copy.(fill!(v,[]))

It is closer to the original use of fill!, but I've never seen it used before and so it is more of a curiousity than idiomatic Julia.

NOTE:

Appreciating the enthusiasm for this answer, I must mention that [ [] for _ in 1:3 ] is the faster way of getting this done. Another method is map(_ -> [], 1:3). Another really nice and efficient method is:

map!(_ -> [], v, v)

edited Jan 15, 2023 at 0:34

answered Jan 14, 2023 at 22:38

Dan Getz

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Simen Gaure · Accepted Answer · 2023-01-14 22:06:58Z

2

As Ahmed says, it's not a bug. In fill!(a, []) you only create a single array reference, which you store in every slot of a. To create a you can either use a= [ [], [], [] ], or slightly more readable:

a = [ [] for _ in 1:3 ].

To clean out every entry you can use broadcasting of the function empty!, i.e. call empty! on every element of a:

empty!.(a)

This avoids creating any new arrays.

answered Jan 14, 2023 at 22:06

Simen Gaure

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Comments

Ahmed Sbai · Accepted Answer · 2023-01-14 19:12:05Z

1

this is not a bug even in other programming languages the fill method works the same.

if you want to clean your array you can simply clean it just like the declaration :

a = [[], [], []]

and if the length can be changed you can use for i in 1: length(a) and clean each item

answered Jan 14, 2023 at 19:12

Ahmed Sbai

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