Returns a vector whose observations are moved ahead by
the specified number of observations.
Namespace:
Extreme.DataAnalysis
Assembly:
Extreme.Numerics (in Extreme.Numerics.dll) Version: 8.1.1
public static Vector<T> Lag<T>(
this Vector<T> vector,
int lag,
T paddedValue
)
<ExtensionAttribute>
Public Shared Function Lag(Of T) (
vector As Vector(Of T),
lag As Integer,
paddedValue As T
) As Vector(Of T)
public:
[ExtensionAttribute]
generic<typename T>
static Vector<T>^ Lag(
Vector<T>^ vector,
int lag,
T paddedValue
)
[<ExtensionAttribute>]
static member Lag :
vector : Vector<'T> *
lag : int *
paddedValue : 'T -> Vector<'T>
Parameters
- vector
- Type: Extreme.MathematicsVectorT
The vector to transform. - lag
- Type: SystemInt32
The number of observations to shift the series by. - paddedValue
- Type: T
The valuie to be used for values in the new series that don't exist
in the original series.
Type Parameters
- T
Return Value
Type:
VectorTA vector.
Usage Note
In Visual Basic and C#, you can call this method as an instance method on any object of type
VectorT. When you use instance method syntax to call this method, omit the first parameter. For more information, see
Extension Methods (Visual Basic) or
Extension Methods (C# Programming Guide).
A positive value of lag indicates that the observations are shifted forward.
If lag equals 1, then each new observation is
the observation before the current observation.
The first lag observations are set to paddedValue.
If lag equals -1, then each new observation is
the observation after the current observation.
The last abs(lag) observations are
set to paddedValue.
The name of the new vector depends on whether lag is positive or negative.
If lag is positive, the new vector gets the name lag<n>(<name>), where
<n> is the lag and <name> is the name of the original vector.
If lag is positive, the new vector gets the name lead<abs(n)>(<name>).
Reference