如何获得 Julia 中索引列表的补码？

Question

假设我有一个变量i = [1,3,5]，它是在我对数组应用过滤器时获得的。现在，假设这个数组有 10 个元素，我想获得“补”索引。我的意思是，我想获得ic = [2,4,6,7,8,9,10]。

有没有一种简洁的方法来获取这个索引的补充列表？

我的意思是，我可以通过常规循环来做到这一点，但是有没有办法通过列表理解来做到这一点？

Answer 1

您需要获取1:10 中不在i 中的所有元素。所以使用列表理解：

julia> i = [1,3,5];

julia> ic = [x for x ∈ 1:10 if x ∉ i]
7-element Array{Int64,1}:
  2
  4
  6
  7
  8
  9
 10

您可以使用\in Tab 在 REPL 中键入 ∈。 ∉ 可以通过\notin Tab 来实现。如果您在不允许这样做的环境中进行编码，则可以从此处复制或键入以下内容之一：

julia> ic = [x for x in 1:10 if !(x in i)]

julia> ic = [x for x in 1:10 if !in(x, i)] # operators are functions

---

性能

如果您关心性能，这里是基准：

julia> @benchmark ic = [x for x ∈ 1:10 if x ∉ i]
BenchmarkTools.Trial: 
  memory estimate:  288 bytes
  allocs estimate:  4
  --------------
  minimum time:     462.274 ns (0.00% GC)
  median time:      471.168 ns (0.00% GC)
  mean time:        497.947 ns (2.86% GC)
  maximum time:     13.115 μs (95.25% GC)
  --------------
  samples:          10000
  evals/sample:     197

Answer 2

你可以使用setdiff函数

julia> x = [1, 3, 5];
julia> y = collect(1:10);
julia> setdiff(y, x)
7-element Vector{Int64}:
  2
  4
  6
  7
  8
  9
 10

性能方面，基于循环的实现更好，因为我们可以考虑到新索引是原始索引的子集

function mysetdiff(y, x)
    res = Vector{eltype(y)}(undef, length(y) - length(x))
    i = 1
    @inbounds for el in y
        el ∈ x && continue
        res[i] = el
        i += 1
    end

    res
end

比较

using BenchmarkTools

@btime [z for z ∈ $y if z ∉ $x]
# 141.893 ns (4 allocations: 288 bytes)
@btime setdiff($y, $x)
# 477.056 ns (8 allocations: 688 bytes)
@btime mysetdiff($y, $x)
# 46.434 ns (1 allocation: 144 bytes)

Answer 3

如果你关心性能，你也可以考虑：

julia> @benchmark deleteat!([1:10;], $i) # indices must be unique and sorted
BenchmarkTools.Trial:
  memory estimate:  160 bytes
  allocs estimate:  1
  --------------
  minimum time:     53.798 ns (0.00% GC)
  median time:      60.790 ns (0.00% GC)
  mean time:        71.125 ns (1.76% GC)
  maximum time:     618.946 ns (77.28% GC)
  --------------
  samples:          10000
  evals/sample:     987

和

julia> @benchmark (x = trues(10); x[$i] .= false; findall(x)) # if Bool-array is enough for you you can skip the last step to save 50% of time
BenchmarkTools.Trial:
  memory estimate:  272 bytes
  allocs estimate:  3
  --------------
  minimum time:     124.863 ns (0.00% GC)
  median time:      134.033 ns (0.00% GC)
  mean time:        157.668 ns (6.84% GC)
  maximum time:     3.000 μs (95.44% GC)
  --------------
  samples:          10000
  evals/sample:     905

vs 之前提出的：

julia> @benchmark ic = [x for x ∈ 1:10 if x ∉ $i]
BenchmarkTools.Trial:
  memory estimate:  288 bytes
  allocs estimate:  4
  --------------
  minimum time:     170.714 ns (0.00% GC)
  median time:      196.571 ns (0.00% GC)
  mean time:        222.510 ns (4.43% GC)
  maximum time:     3.078 μs (91.01% GC)
  --------------
  samples:          10000
  evals/sample:     700

和

julia> @benchmark setdiff($[1:10;], $i)
BenchmarkTools.Trial:
  memory estimate:  672 bytes
  allocs estimate:  7
  --------------
  minimum time:     504.145 ns (0.00% GC)
  median time:      514.508 ns (0.00% GC)
  mean time:        589.584 ns (2.75% GC)
  maximum time:     8.954 μs (90.69% GC)
  --------------
  samples:          10000
  evals/sample:     193

和（来自另一篇文章的自定义实现）

julia> @benchmark mysetdiff($[1:10;], $i)
BenchmarkTools.Trial:
  memory estimate:  144 bytes
  allocs estimate:  1
  --------------
  minimum time:     44.748 ns (0.00% GC)
  median time:      46.869 ns (0.00% GC)
  mean time:        52.780 ns (1.75% GC)
  maximum time:     431.919 ns (88.49% GC)
  --------------
  samples:          10000
  evals/sample:     990

Answer 4

由于您已经使用过过滤，因此也可以反过来使用。或者，如果您有用于获取 i 的函数，则可以将其取反以获取索引的补集。

julia> i = [1,3,5];

julia> filter(x -> x ∉ i, 1:10)
7-element Array{Int64,1}:
  2
  4
  6
  7
  8
  9
 10

---

基准测试

@benchmark filter(x -> x ∉ i, 1:10)
BenchmarkTools.Trial:
  memory estimate:  304 bytes
  allocs estimate:  2
  --------------
  minimum time:     501.036 ns (0.00% GC)
  median time:      511.399 ns (0.00% GC)
  mean time:        546.609 ns (1.26% GC)
  maximum time:     7.516 μs (92.31% GC)
  --------------
  samples:          10000
  evals/sample:     193

参考解决方案

@benchmark ic = [x for x ∈ 1:10 if x ∉ i]
BenchmarkTools.Trial:
  memory estimate:  288 bytes
  allocs estimate:  4
  --------------
  minimum time:     626.036 ns (0.00% GC)
  median time:      646.154 ns (0.00% GC)
  mean time:        692.179 ns (2.72% GC)
  maximum time:     26.065 μs (95.43% GC)
  --------------
  samples:          10000
  evals/sample:     169

Answer 5

从InvertedIndices 包中添加另一个值得了解的语法（虽然不是太快）（通常它也与DataFrames 一起加载）：

(1:10)[Not(i)]

julia> @benchmark (1:10)[Not($i)]
BenchmarkTools.Trial:
  memory estimate:  1.42 KiB
  allocs estimate:  39
  --------------
  minimum time:     1.130 μs (0.00% GC)
  median time:      1.320 μs (0.00% GC)
  mean time:        1.713 μs (5.05% GC)
  maximum time:     324.000 μs (98.81% GC)
  --------------
  samples:          10000
  evals/sample:     10