or rather,
julia> cumfoldl(f, x0, itr) = foldl((a, b) -> push!(a, f(a[end], b)), [x0],
itr)
julia> cumfoldl(*, "0", map(string, 1:10))
11-element Array{ASCIIString,1}:
"0"
"01"
"012"
"0123"
"01234"
"012345"
"0123456"
"01234567"
"012345678"
"0123456789"
"012345678910"
On Tue, Sep 9, 2014 at 9:15 PM, Shashi Gowda <[email protected]>
wrote:
> In this particular case, cumsum does exactly this,
>
> julia> cumsum([1:10])
> 10-element Array{Int64,1}:
> 1
> 3
> 6
> 10
> 15
> 21
> 28
> 36
> 45
> 55
>
> I guess that would be idiomatic Julia ;)
>
> An equivalent foldl would be
> foldl((a, b) -> push!(a, a[end]+b), Int[1], [2:10])
> 10-element Array{Int64,1}:
> 1
> 3
> 6
> 10
> 15
> 21
> 28
> 36
> 45
> 55
>
> A generic cumfoldl would be
>
> julia> cumfoldl(f, x0, itr) = foldl((a, b) -> push!(a, f(a[end], b)),
> [itr[1]], itr[2:end])
> cumfoldl (generic function with 1 method)
>
> julia> cumfoldl(*, 0, map(string, 1:10))
> 10-element Array{ASCIIString,1}:
> "1"
> "12"
> "123"
> "1234"
> "12345"
> "123456"
> "1234567"
> "12345678"
> "123456789"
> "12345678910"
>
>
> I often find it useful to remind myself that foldl(push!, eltype(list)[],
> list) constructs the same list and take it from there.
>
