我一直在与F#斗争,特别是与Deedle.Frame和Deedle.Series合作。
我当前的解决方案非常难看,基本上根据每个if语句过滤Frame,并返回一个仅包含新列的新Frame。然后我把它合并到原始的Frame中。这种情况起作用或相互排斥。我需要插入一个新行的情况是这样的,我将重新索引新帧,以便在合并期间它只是附加在原始帧的底部(在最后一行之后)。
但是,由于不同的if-cases /模式很多,我必须处理大量的框架合并,并确保处理IF语句的补充(否则会丢失值)。 / p>
// ****** PSEUDO CODE ******
// Create new columns as copies.
Add column ["NEWTYPE"] = ["TYPE"]
Add column ["NEWVALUE"] = ["VALUE"]
// Iterate over rows.
foreach row in table
IF row["GROUP"] == A
typeParts = row["TYPE"].Split('/')
// Swap order X/Y -> Y/X.
IF typeParts[0] == "X" && typeParts[1] != "X"
row["NEWTYPE"] = typeParts[1] + "/" + "X"
row["NEWVALUE"] = row["VALUE"] / 10.0
// Split row into two rows Z/Y -> {Z/X, Y/X}.
ElseIF typeParts[0] != "X" && typeParts[1] != "X"
Insert extraRow = row
extraRow["NEWTYPE"] = typeParts[0] + "/" + "X"
extraRow.["NEWVALUE"] = row["VALUE"] * 2.0
row["NEWTYPE"] = typeParts[1] + "/" + "X"
row["NEWVALUE"] = row["VALUE"] * 3.0
ELSE
// Do nothing, new columns already copied.
ELSE
// Do nothing, new columns already copied.
任何能够提出良好解决方案的人的金星。我想这可以通过返回一个框架列表来解决(因为可能会创建一个新行)然后展平并合并所有?
***** Here is my current ugly F# code: *****
let subA =
inputFrame
|> Frame.filterRowValues(fun row -> row.GetAs<string>("GROUP") = "A")
let grpSwap =
subA
|> Frame.filterRowValues(fun row ->
let typeParts = row.GetAs<string>("TYPE").Split('/')
typeParts.[0] = "X" && typeParts.[1] <> "X")
|> Frame.mapRowValues(fun r ->
let typeParts = r.GetAs<string>("TYPE").Split('/')
series ["NEWTYPE" => box (typeParts.[1] + "/" + typeParts.[0]); "NEWVALUE" => box (r.GetAs<float>("VALUE") / 10.0)])
|> Frame.ofRows
let grpCopy =
subA
|> Frame.filterRowValues(fun row ->
let typeParts = row.GetAs<string>("TYPE").Split('/')
typeParts.[0] <> "X" && typeParts.[1] = "X")
|> Frame.mapRowValues(fun r ->
series ["NEWTYPE" => box (r.GetAs<string>("TYPE")); "NEWVALUE" => box (r.GetAs<float>("VALUE"))])
|> Frame.ofRows
let rowsToSplit =
subA
|> Frame.filterRowValues(fun row ->
let typeParts = row.GetAs<string>("TYPE").Split('/')
typeParts.[0] <> "X" && typeParts.[1] <> "X")
let grpSplit1 =
rowsToSplit
|> Frame.mapRowValues(fun r ->
let typeParts = r.GetAs<string>("TYPE").Split('/')
series ["NEWTYPE" => box (typeParts.[0] + "/" + "X"); "NEWVALUE" => box (r.GetAs<float>("VALUE") * 2.0)])
|> Frame.ofRows
let grpSplit2 =
rowsToSplit
|> Frame.mapRowValues(fun r ->
let typeParts = r.GetAs<string>("TYPE").Split('/')
series ["NEWTYPE" => box (typeParts.[1] + "/" + "X"); "NEWVALUE" => box (r.GetAs<float>("VALUE") * 3.0)])
|> Frame.ofRows
let grpAComplement =
inputFrame
|> Frame.filterRowValues(fun row ->
row.GetAs<string>("GROUP") <> "A")
|> Frame.mapRowValues(fun r ->
series ["NEWTYPE" => box (r.GetAs<string>("TYPE")); "NEWVALUE" => box (r.GetAs<float>("VALUE"))])
|> Frame.ofRows
let outputFrame =
let final0 = Frame.mergeAll([inputFrame; grpSwap; grpCopy; grpSplit1; grpAComplement])
let appendFromIndex = (final0.RowCount)
let appendToIndex = appendFromIndex + (grpSplit2.RowCount-1)
let newRow = grpSplit2 |> Frame.merge rowsToSplit
newRow |> Frame.indexRowsWith [appendFromIndex..appendToIndex] |> Frame.merge final0
答案 0 :(得分:4)
我会尝试这样的事情
首先定义一些类型以使事情更容易处理
type Group =
| A
| B
| C
type Source = {
Group: Group
Typ: string
Value: float
}
type Target = {
Group: Group
Typ: string
Value: float
NType: string
NValue: float
}
创建初始化您的初始列表
let xs : List<Source> = createFromWhereEver()
定义转换函数
诀窍是此函数返回Target
个对象的列表。可以是一个项目,也可以是两个项目。
let transform (x:Source) : List<Target> =
if x.Group = A then
let init x ntype nvalue =
{
Group = x.Group
Typ = x.Typ
Value = x.Value
NType = ntype
NValue = nvalue
}
let tp0 :: tp1 :: _ = x.Typ.split('/')
// Swap order X/Y -> Y/X.
if tp0 = "X" && tp1 <> "X" then
[init x (tp1 + "/X") (x.value / 10)]
// Split row into two rows Z/Y -> {Z/X, Y/X}.
elif tp0 <> "X" && tp1 <> "X"
[
init x (tp0 + "/X") (x.value * 2)
init x (tp1 + "/X") (x.value * 3)
]
else
[x]
else
[x]
最后通过地图抽取你的清单和资源,最后连续这些清单
xs
|> List.map transform
//will give you a List<List<Target>>
|> List.concat