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WorksheetFunction.Rank_Eq(Double, Range, Object) Method

Definition

Returns the rank of a number in a list of numbers. The rank of a number is its size relative to other values in a list. (If you were to sort the list, the rank of the number would be its position.)

public double Rank_Eq (double Arg1, Microsoft.Office.Interop.Excel.Range Arg2, object Arg3);
Public Function Rank_Eq (Arg1 As Double, Arg2 As Range, Optional Arg3 As Object) As Double

Parameters

Arg1
Double

Number - The number whose rank you want to find.

Arg2
Range

Ref - An array of, or a reference to, a list of numbers. Non-numeric values in reference are ignored.

Arg3
Object

Order - A number that specifies how to rank the number.

Returns

Remarks

If the order is 0 (zero) or omitted, Microsoft Excel ranks the number as if the reference was a list sorted in descending order.

If the order is any non-zero value, Microsoft Excel ranks the number as if the reference was a list sorted in ascending order.

Rank_Eq gives duplicate numbers the same rank. However, the presence of duplicate numbers affects the ranks of subsequent numbers. For example, in a list of integers sorted in ascending order, if the number 10 appears twice and has a rank of 5, then 11 would have a rank of 7 (no number would have a rank of 6).

For some purposes you might want to use a definition of rank that takes ties into account. In the previous example, you would want a revised rank of 5.5 for the number 10. To do this, add the following correction factor to the value returned by Rank_Eq. This correction factor is appropriate both for the case where rank is computed in descending order (order = 0 or omitted) or ascending order (order = nonzero value).

Correction factor for tied ranks=[COUNT(ref) + 1 – RANK_EQ(number, ref, 0) – RANK_EQ(number, ref, 1)]/2.In the following example, RANK_EQ(A2,A1:A5,1) equals 3. The correction factor is (5 + 1 – 2 – 3)/2 = 0.5 and the revised rank that takes ties into account is 3 + 0.5 = 3.5. If number occurs only once in ref, the correction factor will be 0, since Rank_Eq would not have to be adjusted for a tie.

Applies to