What does this result mean?

General Usage
1 
julia> A = [2, 2, 2]
3-element Vector{Int64}:
 2
 2
 2

julia> all(A, dims = 1)
1-element Vector{Bool}:
 0

Why does this code return 0, not 1?

2

In Julia unlike python for example values are not considered truthy. See a discussion about this concept here: On the arbitrariness of truth(iness).
If you would like to test whether all values are nonzero for example you could provide a function to the all function:

julia> all(!iszero, [2,2,2], dims=1)
1-element Vector{Bool}:
 1
2 Likes
3

Thanks, but why does this code return 1?

julia> all([3, 3, 3], dims = 1)
1-element Vector{Bool}:
 1
2 Likes
4 
julia> for i in 0:10
           A = i * ones(Int, 3)
          @show  all(A, dims = 1)
       end
all(A, dims = 1) = Bool[0]
all(A, dims = 1) = Bool[1]
all(A, dims = 1) = Bool[0]
all(A, dims = 1) = Bool[1]
all(A, dims = 1) = Bool[0]
all(A, dims = 1) = Bool[1]
all(A, dims = 1) = Bool[0]
all(A, dims = 1) = Bool[1]
all(A, dims = 1) = Bool[0]
all(A, dims = 1) = Bool[1]
all(A, dims = 1) = Bool[0]
1 Like
5

Wow, that’s wild. Looks like a bug to me. This should probably just error, like it does without the dims keyword argument:

all([1, 1, 1])
ERROR: TypeError: non-boolean (Int64) used in boolean context
Stacktrace:
 [1] _all
   @ .\reduce.jl:1161 [inlined]
 [2] _all
   @ .\reducedim.jl:903 [inlined]
 [3] #all#750
   @ .\reducedim.jl:901 [inlined]
 [4] all(a::Vector{Int64})
   @ Base .\reducedim.jl:901
 [5] top-level scope
   @ REPL[21]:1
2 Likes
6

This boils down to

julia> mapreduce(identity, &, [3,3,3]; dims=1)
1-element Vector{Bool}:
 1

julia> mapreduce(identity, &, [2,2,2]; dims=1)
1-element Vector{Bool}:
 0

and definitely is a bug.

7 Likes
7

Related:

julia> any([1, 2], dims=1)
ERROR: InexactError: Bool(3)

Like, I guess it’s good that it gives an error, but it’s not really the right error.

3 Likes
8

Could you provide some tips how to get there? Thanks.

9

@edit all([3, 3, 3], dims = 1) and scroll down to

$(_fname)(f, A, dims; kw...) = mapreduce(f, $(op), A; dims=dims, kw...)

You could also @enter all([3, 3, 3], dims = 1) (in VS Code or with Debugger.jl loaded) and then step around a bit until you end up in that function (which may be annoying because the kwarg wrapper heuristics are kinda borked).

1 Like
10

Thanks Sebastian. And sorry, my brain is overheating: I cannot see the connection between “all vector elements being true” and that unfamiliar mapreduce expression. (if all vector elements are booleans, it is a bit more easy to see)

11

Well,

any(x) == mapreduce(is_truthy, |, x) == is_truthy(x[1]) | is_truthy(x[2]) | ... | is_truthy(x[end])
all(x) == mapreduce(is_truthy, &, x) == is_truthy(x[1]) & is_truthy(x[2]) & ... & is_truthy(x[end])

For booleans, is_truthy == identity. This of course breaks down when the eltype of x is e.g. an integer like in the example above, because in that case |/& means bitwise-or/and.

4 Likes
12

Thanks a lot for your explanation, it is very clear now.

13 
julia> mapreduce(is_truthy, &, [3, 3, 3])
ERROR: UndefVarError: is_truthy not defined
Stacktrace:
 [1] top-level scope
   @ REPL[91]:1

I get an error. is_truthy not defined.

14

This is a julia bug. @Raymond, do you want to have the honor of filing an issue since you found it? :slight_smile:

15

Oh, Yes.
So it is a bug. I will file an issue in github.

1 Like