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#1
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How can I find the closest match larger( or smaller) than a reference value
in a row of unsorted data |
#2
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Hi!
Try this: Entered as an array using the key combo of CTRL,SHIFT,ENTER: A5 = lookup_value =INDEX(A1:J1,MATCH(MIN(ABS(A1:J1-A5)),ABS(A1:J1-A5),0)) Biff "Nick Krill" <Nick wrote in message ... How can I find the closest match larger( or smaller) than a reference value in a row of unsorted data |
#3
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Biff:
This is very nice. Can you explain or interpret how the formula is treating or bringing back the both the match(row) and the match(column) arguments of the index function? Confused here... TIA,, "Biff" wrote in message ... Hi! Try this: Entered as an array using the key combo of CTRL,SHIFT,ENTER: A5 = lookup_value =INDEX(A1:J1,MATCH(MIN(ABS(A1:J1-A5)),ABS(A1:J1-A5),0)) Biff "Nick Krill" <Nick wrote in message ... How can I find the closest match larger( or smaller) than a reference value in a row of unsorted data |
#4
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Hi!
It's only referencing a single (row) array so there is no column argument. Simply determine the minimum deviation from the lookup_value. Since the OP stated that it could be either above or below the lookup_value we have to use the ABS function so that negative deviations are made equal to positive deviations. Biff "JMay" wrote in message news:hnSpf.61035$WH.17922@dukeread01... Biff: This is very nice. Can you explain or interpret how the formula is treating or bringing back the both the match(row) and the match(column) arguments of the index function? Confused here... TIA,, "Biff" wrote in message ... Hi! Try this: Entered as an array using the key combo of CTRL,SHIFT,ENTER: A5 = lookup_value =INDEX(A1:J1,MATCH(MIN(ABS(A1:J1-A5)),ABS(A1:J1-A5),0)) Biff "Nick Krill" <Nick wrote in message ... How can I find the closest match larger( or smaller) than a reference value in a row of unsorted data |
#5
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Thanks for the clarification Biff;
Jim "Biff" wrote: Hi! It's only referencing a single (row) array so there is no column argument. Simply determine the minimum deviation from the lookup_value. Since the OP stated that it could be either above or below the lookup_value we have to use the ABS function so that negative deviations are made equal to positive deviations. Biff "JMay" wrote in message news:hnSpf.61035$WH.17922@dukeread01... Biff: This is very nice. Can you explain or interpret how the formula is treating or bringing back the both the match(row) and the match(column) arguments of the index function? Confused here... TIA,, "Biff" wrote in message ... Hi! Try this: Entered as an array using the key combo of CTRL,SHIFT,ENTER: A5 = lookup_value =INDEX(A1:J1,MATCH(MIN(ABS(A1:J1-A5)),ABS(A1:J1-A5),0)) Biff "Nick Krill" <Nick wrote in message ... How can I find the closest match larger( or smaller) than a reference value in a row of unsorted data |
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