为什么这不符合我的想法:
benjamin@benjamin-VirtualBox:~$ julia -p 3
julia> @everywhere(function foom(bar::Vector{Any}, k::Integer) println(repeat(bar[2],bar[1])); return bar; end)
julia> foo={{1,"a"},{2,"b"},{3,"c"}}
julia> pmap(foom, foo, 5)
From worker 2: a
1-element Array{Any,1}:
{1,"a"}
这就是它的全部输出。我期待pmap迭代foo中的每个元组并在其上调用foom。
编辑:
当我没有传递其他参数时,它正常工作:
julia> @everywhere(function foom(bar::Vector{Any}) println(repeat(bar[2],bar[1])); return bar; end)
julia> pmap(foom, foo)
From worker 3: bb
From worker 2: a
From worker 4: ccc
3-element Array{Any,1}:
{1,"a"}
{2,"b"}
{3,"c"}
如何将更多参数传递给pmap?
答案 0 :(得分:10)
函数 pmap 确实会将任意数量的参数作为集合
function pmap(f, lsts...; err_retry=true, err_stop=false)
函数f将为每个集合发送一个参数
julia> @everywhere f(s,count)=(println("process id = $(myid()) s = $s count = $count");repeat(s,count))
julia> pmap((a1,a2)->f(a1,a2),{"a","b","c"},{2,1,3})
From worker 3: process id = 3 s = b count = 1
From worker 2: process id = 2 s = a count = 2
From worker 4: process id = 4 s = c count = 3
3-element Array{Any,1}:
"aa"
"b"
"ccc"
或者,可以将参数作为一个外部集合的参数集合发送,这些集合可以被喷射到目标函数的多个参数中
julia> pmap((args)->f(args...),{{"a",2},{"b",1},{"c",3}})
From worker 2: process id = 2 s = a count = 2
From worker 3: process id = 3 s = b count = 1
From worker 4: process id = 4 s = c count = 3
3-element Array{Any,1}:
"aa"
"b"
"ccc"