## Count unique distinct records with a date and column criteria

re: Count records between two dates and a criterion

based on the example, i was looking for 1 date and 1 criterion. i slightly modify the formula to

=SUMPRODUCT(--($B$1:$B$9=$E$2), --($A$1:$A$9=$E$3)) + ENTER

[assuming E2 = 9-2-2010]

the result would be 1 (one 'JA' found on 9-2-2010 date)

but this is summation of records found on 1 date with 1 criterion. It will not work if there is *multiple* 'JA' criterion exist on the same date because SUMPRODUCT summed up the records found.

I'm curious to know...

1) What if I want to know the UNIQUE DISTINCT records found on 1 date with 1 criterion?

2) Working on >100k rows of data, this formula literally slows down Excel (heavy calculation and recalculations). Is there an alternative to speed it up? UDF? array formula?

thanks!

### Answer:

Array formula in cell D3:

### How to create an array formula

- Copy (Ctrl + c) and paste (Ctrl + v) array formula into formula bar.
- Press and hold Ctrl + Shift.
- Press Enter once.
- Release all keys.

### How the array formula in cell D3 works

**Step 1 - Count records**

The COUNTIFS function calculates the number of cells across multiple ranges that equals all given conditions.

COUNTIFS($B$6:$B$11, $B$6:$B$11, $C$6:$C$11, $C$6:$C$11, $D$6:$D$11, $D$6:$D$11)

becomes

COUNTIFS({"JA";"SH";"JA"; "JA";"JA";"SH"}, {"JA";"SH";"JA"; "JA";"JA";"SH"}, {40422;40423; 40423;40423;40423;40426}, {40422;40423; 40423;40423;40423;40426}, {"North";"South";"North"; "West";"West";"South"}, {"North";"South";"North"; "West";"West";"South"})

and returns array {1;1;1;2;2;1}

**Step 2 - Filter records using name and date criteria**

The IF function returns one value (argument2) if TRUE and another (argument3) if FALSE.

IF(($B$6:$B$11=B3)*($C$6:$C$11=C3), (1/COUNTIFS($B$6:$B$11, $B$6:$B$11, $C$6:$C$11, $C$6:$C$11, $D$6:$D$11, $D$6:$D$11)), 0)

becomes

IF(($B$6:$B$11=B3)*($C$6:$C$11=C3), (1/{1;1;1;2;2;1}), 0)

becomes

IF(({"JA";"SH";"JA";"JA";"JA";"SH"}="JA")*({40422;40423;40423;40423;40423;40426}=40423), (1/{1;1;1;2;2;1}), 0)

becomes

IF(({0;0;1;1;1;0}, (1/{1;1;1;2;2;1}), 0)

becomes

IF(({0;0;1;1;1;0}, {1;1;1;0,5;0,5;1}, 0)

and returns {0;0;1;0,5;0,5;0}

**Step 3 - Sum values**

The SUM function adds numbers an return the total.

=SUM(IF(($B$6:$B$11=B3)*($C$6:$C$11=C3), (1/COUNTIFS($B$6:$B$11, $B$6:$B$11, $C$6:$C$11, $C$6:$C$11, $D$6:$D$11, $D$6:$D$11)), 0))

becomes

=SUM({0;0;1;0,5;0,5;0}) and returns 2 in cell D3.

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thanks oscar,

have tried it, it works.

however, i have ~100K rows, and Excel is literally stalled when running the formula.

for the time being, i'm using a Pivottable and using a COUNTA function to count unique distinct value. Not automated but it's near-instantaneous to get the number :)

nonetheless, thanks for the solution above!

davidlim,

thanks!

The vba code provided here:

https://lazyvba.blogspot.com/2010/11/improve-your-pivot-table-to-count.html

seems to count unique values in a pivot table.

hi oscar,

have tried lazyvba's code. works fine, but it is not efficient (crawling for list more than >100K rows).

my pivottable is simple: dates and products. no other columns, formulas, etc.

any other suggestions?

davidlim,

Do you want to count unique distinct products between two dates?

davidlim,

read this post: Count unique distinct values in a large dataset with a date criterion