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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5 Responses to “Count unique distinct records with a date and column criteria”
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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