在F#中的Deedle时间序列中使用缺失值(1) [英] Working with missing values in Deedle Time Series in F# (1)
问题描述
这是一个小例子,我想处理系列自定义函数中的缺失值.
here is a small example where I want to deal with missing values on custom functions on series.
假设我已经获得了一个系列
suppose that i have obtained a series
series4;;
val it : Series<int,int opt> =
1 -> 1
2 -> 2
3 -> 3
4 -> <missing>
例如,这种方式:
let series1 = Series.ofObservations [(1,1);(2,2);(3,3)]
let series2 = Series.ofObservations [(1,2);(2,2);(3,1);(4,4)]
let series3 = series1.Zip(series2,JoinKind.Outer);;
let series4 = series3 |> Series.mapValues fst
然后,如果我这样做,
Series.mapAll (fun v -> match v with
| Some a -> (a>1)
| _-> false) series4
失败
System.Exception:由于操作较早而无法完成 错误``int选项''类型与``int opt''类型不匹配.看 还要输入.fsx(4,42)-(4,49).在4,42
System.Exception: Operation could not be completed due to earlier error The type 'int option' does not match the type 'int opt'. See also input.fsx(4,42)-(4,49). at 4,42
我希望结果是
val it : Series<int,bool opt> =
1 -> false
2 -> true
3 -> true
4 -> false
甚至更好的是能够得到类似的结果
even better would be to able to get a result like
val it : Series<int,int opt> =
1 -> false
2 -> true
3 -> true
4 -> <missing>
那里正确的语法是什么?理想情况下,如果有一个<missing>
值,我想为新系列中的同一键提供一个<missing>
值
what would be the right syntax there ? ideally if there is a <missing>
value, i would like a <missing>
value for the same key in the new series
基本上我需要对int opt
类型
奖金问题:对于像>"这样的常用运算符,Deedle中是否有矢量化运算符? (系列1>系列2),如果两个系列具有相同的键类型,则会返回一个新的布尔(选项?)类型系列
Bonus question: is there are vectorized operator in Deedle for some usual operators such like ">" ? (series1 > series2) when both series have the same key types would return a new series of boolean (option ?)type
谢谢
推荐答案
您可以这样做:
let series5 =
series4
|> Series.mapValues(OptionalValue.map(fun x -> x > 1))
您可以在文档中了解有关模块OptionalValue
的信息.
You can read about module OptionalValue
in the documentation
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