data.table rleid() 函数 Julia 等效

Question

>有人知道 data.table：：rleid(( 函数可用于 Julia 吗？

https://www.rdocumentation.org/packages/data.table/versions/1.10.4-2/topics/rleid

Answer 1

我不知道任何库函数。但在这里你有两个选择。

将missing视为有效条目：

function rleid(x::AbstractVector)
    isempty(x) && return Int[]
    rle = similar(x, Int)
    idx = 1
    rle[1] = idx
    prev = x[1]
    for i in 2:length(x)
        this = x[i]
        if ismissing(this)
            if !ismissing(prev)
                prev = this
                idx += 1    
            end
        else
            if ismissing(prev) || this != prev
                prev = this
                idx += 1
            end
        end
        rle[i] = idx
    end
    rle
end

跳过missing将missing放入输出向量：

function rleid_missing(x::AbstractVector)
    isempty(x) && return Union{Int,Missing}[]
    rle = similar(x, Union{Int, Missing})
    start_i = 1
    while start_i <= length(x) && ismissing(x[start_i])
        rle[start_i] = missing
        start_i += 1
    end
    if start_i <= length(x)
        idx = 1
        rle[start_i] = idx
        prev = x[start_i]
        start_i += 1
        for i in start_i:length(x)
            this = x[i]
            if ismissing(this)
                rle[i] = missing
            else
                if this != prev
                    prev = this
                    idx += 1
                end
                rle[i] = idx
            end
        end
    end
    rle
end

这是一个测试：

Main> rleid([missing,3,4,4,missing,1,1,missing,missing,6])
10-element Array{Int64,1}:
 1
 2
 3
 3
 4
 5
 5
 6
 6
 7
Main> rleid_missing([missing,3,4,4,missing,1,1,missing,missing,6])
10-element Array{Union{Int64, Missings.Missing},1}:
  missing
 1
 2
 2
  missing
 3
 3
  missing
  missing
 4
Main> rleid_missing([missing,3,4,4,missing,1,1,missing,missing,1,6])
11-element Array{Union{Int64, Missings.Missing},1}:
  missing
 1
 2
 2
  missing
 3
 3
  missing
  missing
 3
 4

(在最后一种情况下，请注意missing被视为不存在 - 如果您想要不同的东西，很容易调整行为(。

Julia 的美妙之处在于这些功能将很快 - 无需在用 C++ 编写的外部库中实现它们。