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问如何压缩2 Deedle框架并处理丢失的值？
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Stack Overflow用户

提问于 2020-04-16 13:16:36

回答 3查看 123关注 0票数 1

给定一个价格框架，priceFrame作为

                       28881              29021              29399
2010-01-01 00:00:00 -> 123.535878499576   195.28635425580265 189.92210186152082
2010-01-04 00:00:00 -> 124.19087548338847 198.10448102247753 190.1571733631235
2010-01-05 00:00:00 -> 123.82028508465247 197.8259452373992  190.31388769752525
2010-01-06 00:00:00 -> 124.17363872065654 197.80956077945342 189.98478759528152
2010-01-07 00:00:00 -> 123.4583130672824  197.58017836821244 190.31388769752527
2010-01-08 00:00:00 -> 124.23396739021821 198.10448102247756 190.25120196376457
2010-01-11 00:00:00 -> 125.12166067091142 197.87509861123658 190.73701640041008
2010-01-12 00:00:00 -> 124.9234378994945  195.0569718445617  191.41088803833776
2010-01-13 00:00:00 -> 125.06133200134975 195.64681233060992 191.50491663897884
2010-01-14 00:00:00 -> 124.97514818769021 196.28580619049552 191.56760237273951
2010-01-15 00:00:00 -> 123.71686450826103 192.5829186947483  192.08475967626538
2010-01-18 00:00:00 -> 123.71686450826103 194.10667328370621 192.31983117786805
2010-01-19 00:00:00 -> 123.15666971947407 195.87619474185092 191.94371677530378
2010-01-20 00:00:00 -> 121.5622691667727  191.79646471335064 192.82131704795376
2010-01-21 00:00:00 -> 121.5450324040408  188.38849746062752 192.9937028157957
2010-01-22 00:00:00 -> 121.81220222638535 186.8647428716696  192.9937028157957
2010-01-25 00:00:00 -> 121.94147794687466 184.83307008639233 192.9937028157957
2010-01-26 00:00:00 -> 121.38990153945363 185.9799821425972  193.19743145051802
2010-01-27 00:00:00 -> 120.94174570842405 184.91499237612123 193.3541457849198
2010-01-28 00:00:00 -> 120.44187958919875 182.5392459739825  193.22877431739838
2010-01-29 00:00:00 -> 119.4938576389439  183.75169586197052 193.35414578491978

和红利框架divFrame作为

                       28881     29021     29399
2010-01-04 00:00:00 -> 1.3       <missing> <missing>
2010-01-13 00:00:00 -> <missing> 1.3       <missing>
2010-01-22 00:00:00 -> <missing> <missing> 1.3

我想把它们结合起来，在有红利的情况下，我有价格+红利，否则就保持原来的价格。

以下两次尝试

let dfZipped1 = priceFrame.Zip(divFrame, JoinKind.Left, JoinKind.Left, Lookup.Exact, false, fun (p:float) d -> p + d)
        dfZipped1.Print()
        let dfZipped2 = priceFrame.Zip(divFrame, JoinKind.Left, JoinKind.Left, Lookup.Exact, true, fun (p:float) d -> p + d)
        dfZipped2.Print()

结果相同的输出

                       28881              29021              29399
2010-01-01 00:00:00 -> <missing>          <missing>          <missing>
2010-01-04 00:00:00 -> 125.49087548338846 <missing>          <missing>
2010-01-05 00:00:00 -> <missing>          <missing>          <missing>
2010-01-06 00:00:00 -> <missing>          <missing>          <missing>
2010-01-07 00:00:00 -> <missing>          <missing>          <missing>
2010-01-08 00:00:00 -> <missing>          <missing>          <missing>
2010-01-11 00:00:00 -> <missing>          <missing>          <missing>
2010-01-12 00:00:00 -> <missing>          <missing>          <missing>
2010-01-13 00:00:00 -> <missing>          196.94681233060993 <missing>
2010-01-14 00:00:00 -> <missing>          <missing>          <missing>
2010-01-15 00:00:00 -> <missing>          <missing>          <missing>
2010-01-18 00:00:00 -> <missing>          <missing>          <missing>
2010-01-19 00:00:00 -> <missing>          <missing>          <missing>
2010-01-20 00:00:00 -> <missing>          <missing>          <missing>
2010-01-21 00:00:00 -> <missing>          <missing>          <missing>
2010-01-22 00:00:00 -> <missing>          <missing>          194.2937028157957
2010-01-25 00:00:00 -> <missing>          <missing>          <missing>
2010-01-26 00:00:00 -> <missing>          <missing>          <missing>
2010-01-27 00:00:00 -> <missing>          <missing>          <missing>
2010-01-28 00:00:00 -> <missing>          <missing>          <missing>
2010-01-29 00:00:00 -> <missing>          <missing>          <missing>

没有丢失的数字是正确的。但我想保留那些没有红利的价格。

let dfZipped3 = priceFrame.Zip(divFrame, JoinKind.Left, JoinKind.Left, Lookup.Exact, false, fun (p:float) d -> p + (d |> Option.defaultValue 0.0))
        dfZipped3.Print()

结果：

                       28881              29021              29399
2010-01-01 00:00:00 -> 123.535878499576   195.28635425580265 189.92210186152082
2010-01-04 00:00:00 -> 124.19087548338847 198.10448102247753 190.1571733631235
2010-01-05 00:00:00 -> 123.82028508465247 197.8259452373992  190.31388769752525
2010-01-06 00:00:00 -> 124.17363872065654 197.80956077945342 189.98478759528152
2010-01-07 00:00:00 -> 123.4583130672824  197.58017836821244 190.31388769752527
2010-01-08 00:00:00 -> 124.23396739021821 198.10448102247756 190.25120196376457
2010-01-11 00:00:00 -> 125.12166067091142 197.87509861123658 190.73701640041008
2010-01-12 00:00:00 -> 124.9234378994945  195.0569718445617  191.41088803833776
2010-01-13 00:00:00 -> 125.06133200134975 195.64681233060992 191.50491663897884
2010-01-14 00:00:00 -> 124.97514818769021 196.28580619049552 191.56760237273951
2010-01-15 00:00:00 -> 123.71686450826103 192.5829186947483  192.08475967626538
2010-01-18 00:00:00 -> 123.71686450826103 194.10667328370621 192.31983117786805
2010-01-19 00:00:00 -> 123.15666971947407 195.87619474185092 191.94371677530378
2010-01-20 00:00:00 -> 121.5622691667727  191.79646471335064 192.82131704795376
2010-01-21 00:00:00 -> 121.5450324040408  188.38849746062752 192.9937028157957
2010-01-22 00:00:00 -> 121.81220222638535 186.8647428716696  192.9937028157957
2010-01-25 00:00:00 -> 121.94147794687466 184.83307008639233 192.9937028157957
2010-01-26 00:00:00 -> 121.38990153945363 185.9799821425972  193.19743145051802
2010-01-27 00:00:00 -> 120.94174570842405 184.91499237612123 193.3541457849198
2010-01-28 00:00:00 -> 120.44187958919875 182.5392459739825  193.22877431739838
2010-01-29 00:00:00 -> 119.4938576389439  183.75169586197052 193.35414578491978

所有的价格都在那里，但红利都没有增加。

let dfZipped4 = priceFrame.Zip(divFrame, JoinKind.Left, JoinKind.Left, Lookup.Exact, true, fun (p:float) d -> p + (d |> Option.defaultValue 0.0))
        dfZipped4.Print()

只会导致价值的缺失。

当股息一致时，我该如何将价格加到股息中，但如果不是这样，价格就会保持不变吗？

更新

我已经安排了执行Frocha和zuzhu的每一个答案的时间。合述的第二个答案并没有产生正确的结果。

对于每种技术的1000次连续运行，我得到的是典型的

frocha1: 572.974400
frocha2: 562.867600
zyzhu1: 1099.057100

frocha2始终比frocha1稍快。zyzhu1总是比其他的慢。所以现在我接受弗罗查的回答。

然而，如果zyzhu2可以工作，它可能是最快的，因为它是最简单的。在这种情况下，我将更改接受的答案。

dataframe

deedle

回答 3

Stack Overflow用户

回答已采纳

发布于 2020-04-16 16:42:20

我的方法不考虑速度限制，如下所示: 1)重命名列以便能够在没有错误的情况下执行联接，2)加入框架。3)用零替换缺失值。4)相应列之和。5)取消股利栏。6)可选:如果不需要"string“转换，则将priceFrame的col名称更改为原始类型。

module Frame =
    //I usually add this handy function to the Frame module
    let mapReplaceCol col f frame =
        frame
        |> Frame.replaceCol col (Frame.mapRowValues f frame)

let priceFrame' = priceFrame |> Frame.mapColKeys string

//appends a "D" in the col key to eliminate col with same name
let dividends' =
    dividends
    |> Frame.mapColKeys (string >> (+) "D") 

let joinedFrame =
    priceFrame'
    |> Frame.join JoinKind.Right dividends'
    |> Frame.fillMissingWith 0.

(joinedFrame,priceFrame'.ColumnKeys |> List.ofSeq)
||> List.fold (fun acc elem ->
    acc|> Frame.mapReplaceCol elem (fun row ->
        row.GetAs<float>("D" + elem) + row.GetAs<float>(elem))
    |> Frame.dropCol ("D" + elem))

编辑

另一种使用Zip的方法。

//generate a dividends frame with the same rows of priceFrame
let dividends2 =
    (priceFrame,priceFrame.ColumnKeys |> List.ofSeq)
    ||> List.fold (fun acc elem -> acc|> Frame.dropCol elem) //empty frame
    |> Frame.join JoinKind.Outer dividends
    |> Frame.fillMissingWith 0.

(priceFrame,dividends2) ||> Frame.zip (fun (p : float) (d : float) -> p + d)

票数 0

Stack Overflow用户

发布于 2020-04-17 17:22:10

若要使用zip，两个框架将逐列匹配，并将两个系列相加在一起。

在您的例子中，divFrame的观察量比priceFrame少。当两个系列没有相同的观测量并加在一起时，无与伦比的结果就会丢失。

这是我的解决方案，创建一个虚拟框架，以便divFrame首先与priceFrame对齐。

let divFrame2 =
  let dummy =
    priceFrame.RowKeys
    |> Seq.collect(fun row -> divFrame.ColumnKeys |> Seq.map(fun col -> row, col, 0) )
    |> Frame.ofValues
  (dummy + divFrame).FillMissing(0.)
priceFrame + divFrame2

票数 0

Stack Overflow用户

发布于 2020-04-23 01:57:59

我有类似的问题，我的解决方案与zyzhu相似，但可以处理多个帧

let zipAll (dfs:Frame<_,_>[]) = 
    let outerKeys = dfs |> Array.collect (fun df -> df.RowKeys |> Array.ofSeq) |> Array.distinct |> Array.sort
    let dfsNew = 
        dfs
        |> Array.map ( Frame.indexRowsWith outerKeys >> Frame.mapRowValues (Series.fillMissingWith 0.) >> Frame.ofRows)
    Array.fold (Frame.zip (+)) (Array.head dfsNew) (Array.tail dfsNew)

[|priceFrame;dividends2|] |> zipAll

它可能很慢，但它可以处理多个帧。

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/61251194

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