NthMostFrequentbeforereadingbasicsyntaxcrystalsyntax

Computes the n^th most frequent element of a field or array. If multiple values with the same frequency occur, the smallest value is used as value with the greatest frequency.
It is possible to specify a condition which records to include.

Usage:

NthMostFrequent( n, fld )
NthMostFrequent( n, fld, condFld )
NthMostFrequent( n, fld, condFld, cond )

Parameters:

n	index of which largest element to take.
fld	The name of a field or array variable whose elements should be used to calculate the n^th most frequent element.
condFld	The name of the field that groups the elements of fld (if fld isn't an array).
cond	A string declaring the type of grouping to be used on condFld. This only makes sense for groups based on Date, DateTime, Time, and Boolean fields. See Summary Field Conditions for a list of strings you can use here.

Returns:

A number.

Examples:

// returns 7 because it is the most frequently occurring smallest value
nthMostFrequent(1, [3,12,7,6,5,12,7])

// returns 12 because it is the most frequently occurring second smallest value
nthMostFrequent(2, [3,12,7,6,5,12,7])

// returns the first most frequent value of the array
nthMostFrequent(1, [2,5,3,2,9,2,4,3,2,5,7,8,5,1,2,6,4])  // returns 2

// returns the first most frequent value of the database field {Failures.Reason}
nthMostFrequent(1, {Failures.Reason})

// returns the second most frequent Failures.Reason grouped by Vendor
nthMostFrequent(2, {Failures.Reason}, {Vendors.Vendor})

// a formula that depends on the 10th most frequent value of Failures.Reason
numberVar nth = nthMostFrequent(10, {Failures.Reason});
if {Failures.Reason} > nth then
	red
else
	black

Notes:

If the values in fld should be grouped, a group based on condFld is needed. If the group does not exist in the report, you can add it using i-net Designer or using API method: Engine.addGroup.
If there are several values with the same frequency of occurrence, these group will be sorted in ascending order. So, the 3rd most frequent of [5,4,3,2,1] would be 3.