VectorExtensions.Lag<T>(Vector<T>, Int32) Method

Returns a vector whose observations are moved ahead by the specified number of observations.

Definition

Namespace: Extreme.DataAnalysis
Assembly: Extreme.Numerics (in Extreme.Numerics.dll) Version: 8.1.23
C#
public static Vector<T> Lag<T>(
	this Vector<T> vector,
	int lag
)

Parameters

vector  Vector<T>
The vector to transform.
lag  Int32
The number of observations to shift the series by.

Type Parameters

T

Return Value

Vector<T>
A vector.

Usage Note

In Visual Basic and C#, you can call this method as an instance method on any object of type Vector<T>. 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).

Remarks

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 NaN. If lag equals -1, then each new observation is the observation after the current observation. The last abs(lag) observations are set to NaN.

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>).

See Also