group-by - 在 Julia 中按组创建滞后/提前时间序列?
问题描述
我想知道是否有一种简单的方法可以根据分组或条件在 Julia 中创建时间序列变量的滞后(或超前)?例如:我有一个如下形式的数据集
julia> df1 = DataFrame(var1=["a","a","a","a","b","b","b","b"],
var2=[0,1,2,3,0,1,2,3])
8×2 DataFrame
│ Row │ var1 │ var2 │
│ │ String │ Int64 │
├─────┼────────┼───────┤
│ 1 │ a │ 0 │
│ 2 │ a │ 1 │
│ 3 │ a │ 2 │
│ 4 │ a │ 3 │
│ 5 │ b │ 0 │
│ 6 │ b │ 1 │
│ 7 │ b │ 2 │
│ 8 │ b │ 3 │
我想创建一个lag2
包含var2
滞后 2 的值的变量。但是,这应该按 var1 分组完成,以便“b”组中的前两个观察值不会得到“a”组的最后两个值. 相反,它们应该设置为缺失值或零或某个默认值。
我尝试了以下代码,它会产生以下错误。
julia> df2 = df1 |> @groupby(_.var1) |> @mutate(lag2 = lag(_.var2,2)) |> DataFrame
ERROR: MethodError: no method matching merge(::Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}}, ::NamedTuple{(:lag2,),Tuple{ShiftedArray{Int64,Missing,1,QueryOperators.GroupColumnArrayView{Int64,Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},:var2}}}})
Closest candidates are:
merge(::NamedTuple{,T} where T<:Tuple, ::NamedTuple) at namedtuple.jl:245
merge(::NamedTuple{an,T} where T<:Tuple, ::NamedTuple{bn,T} where T<:Tuple) where {an, bn} at namedtuple.jl:233
merge(::NamedTuple, ::NamedTuple, ::NamedTuple...) at namedtuple.jl:249
...
Stacktrace:
[1] (::var"#437#442")(::Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}}) at /Users/kayvon/.julia/packages/Query/AwBtd/src/query_translation.jl:58
[2] iterate at /Users/kayvon/.julia/packages/QueryOperators/g4G21/src/enumerable/enumerable_map.jl:25 [inlined]
[3] iterate at /Users/kayvon/.julia/packages/Tables/TjjiP/src/tofromdatavalues.jl:45 [inlined]
[4] buildcolumns at /Users/kayvon/.julia/packages/Tables/TjjiP/src/fallbacks.jl:185 [inlined]
[5] columns at /Users/kayvon/.julia/packages/Tables/TjjiP/src/fallbacks.jl:237 [inlined]
[6] #DataFrame#453(::Bool, ::Type{DataFrame}, ::QueryOperators.EnumerableMap{Union{},QueryOperators.EnumerableIterable{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},QueryOperators.EnumerableGroupBy{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},String,NamedTuple{(:var1, :var2),Tuple{String,Int64}},QueryOperators.EnumerableIterable{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.DataValueRowIterator{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.Schema{(:var1, :var2),Tuple{String,Int64}},Tables.RowIterator{NamedTuple{(:var1, :var2),Tuple{Array{String,1},Array{Int64,1}}}}}},var"#434#439",var"#435#440"}},var"#437#442"}) at /Users/kayvon/.julia/packages/DataFrames/S3ZFo/src/other/tables.jl:40
[7] DataFrame(::QueryOperators.EnumerableMap{Union{},QueryOperators.EnumerableIterable{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},QueryOperators.EnumerableGroupBy{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},String,NamedTuple{(:var1, :var2),Tuple{String,Int64}},QueryOperators.EnumerableIterable{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.DataValueRowIterator{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.Schema{(:var1, :var2),Tuple{String,Int64}},Tables.RowIterator{NamedTuple{(:var1, :var2),Tuple{Array{String,1},Array{Int64,1}}}}}},var"#434#439",var"#435#440"}},var"#437#442"}) at /Users/kayvon/.julia/packages/DataFrames/S3ZFo/src/other/tables.jl:31
[8] |>(::QueryOperators.EnumerableMap{Union{},QueryOperators.EnumerableIterable{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},QueryOperators.EnumerableGroupBy{Grouping{String,NamedTuple{(:var1, :var2),Tuple{String,Int64}}},String,NamedTuple{(:var1, :var2),Tuple{String,Int64}},QueryOperators.EnumerableIterable{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.DataValueRowIterator{NamedTuple{(:var1, :var2),Tuple{String,Int64}},Tables.Schema{(:var1, :var2),Tuple{String,Int64}},Tables.RowIterator{NamedTuple{(:var1, :var2),Tuple{Array{String,1},Array{Int64,1}}}}}},var"#434#439",var"#435#440"}},var"#437#442"}, ::Type) at ./operators.jl:854
[9] top-level scope at none:0
感谢这种方法或替代方法的任何帮助。谢谢。
解决方案
你肯定有正确的想法 - 我不使用 Query.jl 但这可以通过基本的 DataFrames 语法轻松完成:
julia> using DataFrames, ShiftedArrays
julia> df1 = DataFrame(var1=["a","a","a","a","b","b","b","b"],
var2=[0,1,2,3,0,1,2,3]);
julia> by(df1, :var1, var2_l2 = :var2 => Base.Fix2(lag, 2)))
8×2 DataFrame
│ Row │ var1 │ var2_l2 │
│ │ String │ Int64⍰ │
├─────┼────────┼─────────┤
│ 1 │ a │ missing │
│ 2 │ a │ missing │
│ 3 │ a │ 0 │
│ 4 │ a │ 1 │
│ 5 │ b │ missing │
│ 6 │ b │ missing │
│ 7 │ b │ 0 │
│ 8 │ b │ 1 │
请注意,我Base.Fix2
在这里用来获取lag
. 这本质上与定义您自己的l2(x) = lag(x, 2)
然后l2
在by
调用中使用相同。如果您确实定义了自己的l2
函数,您还可以设置默认值,例如l2(x) = lag(x, 2, default = -1000)
如果您想避免缺失值:
julia> l2(x) = lag(x, 2, default = -1000)
l2 (generic function with 1 method)
julia> by(df1, :var1, var2_l2 = :var2 => l2)
8×2 DataFrame
│ Row │ var1 │ var2_l2 │
│ │ String │ Int64 │
├─────┼────────┼─────────┤
│ 1 │ a │ -1000 │
│ 2 │ a │ -1000 │
│ 3 │ a │ 0 │
│ 4 │ a │ 1 │
│ 5 │ b │ -1000 │
│ 6 │ b │ -1000 │
│ 7 │ b │ 0 │
│ 8 │ b │ 1 │
编辑
在 DataFrames.jl 0.22.2 下,正确的语法是:
julia> combine(groupby(df1, :var1), :var2 => Base.Fix2(lag, 2) => :var2_l2)
8×2 DataFrame
Row │ var1 var2_l2
│ String Int64?
─────┼─────────────────
1 │ a missing
2 │ a missing
3 │ a 0
4 │ a 1
5 │ b missing
6 │ b missing
7 │ b 0
8 │ b 1
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