# WorksheetFunction.Rank_Eq method (Excel)

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.

## Syntax

*expression*.**Rank_Eq** (*Arg1*, *Arg2*, *Arg3*)

*expression* A variable that represents a **WorksheetFunction** object.

## Parameters

Name | Required/Optional | Data type | Description |
---|---|---|---|

Arg1 |
Required | Double |
Number - The number whose rank you want to find. |

Arg2 |
Required | Range |
Ref - An array of, or a reference to, a list of numbers. Non-numeric values in reference are ignored. |

Arg3 |
Optional | Variant |
Order - A number that specifies how to rank the number. |

## Return value

**Double**

## Remarks

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

If the order is any non-zero value, Excel ranks the number as if the reference were 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, 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 because

**Rank_Eq**would not have to be adjusted for a tie.

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