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#1
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Problem with MATCH () function . . .
FIRST, is it possible to use the MATCH () function on an unsorted range of
data? We just need a match from the range, regardless of whether it is the first, last, or inbetween match. In turn, this match echos the column heading label associated with that amount using an INDEX () function. SECOND, is it possible to return all the matches in an unsorted series, grab their associated column headings, and then concatenate them into one string? Any help would be appreciated . . . |
#2
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Problem with MATCH () function . . .
Q1 - Yes, it is. You would do so like this:
=MATCH(A2,B1:H1,0) this would return the relative position of the first exactly matching value in B1:H1 compared with A2. Note that the zero at the end forces it to look for an exact match. If A2 is not in the range, then #N/A will be returned. Q2 - yes, this would be possible. However, you need to give some more detail about your data so that a formula can be offered that is more likely to suit your set up directly. Hope this helps. Pete On Dec 11, 1:13*am, "Blue Max" wrote: FIRST, is it possible to use the MATCH () function on an unsorted range of data? *We just need a match from the range, regardless of whether it is the first, last, or inbetween match. *In turn, this match echos the column heading label associated with that amount using an INDEX () function. SECOND, is it possible to return all the matches in an unsorted series, grab their associated column headings, and then concatenate them into one string? Any help would be appreciated . . . |
#3
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Problem with MATCH () function . . .
Hello Pete,
Thanks for sharing the information on the missing argument. Shortly after our post, I did find the help for that function option. As to the more complex issue, let me give a sample case: The formula would need to find all the matches in a row of prices from different vendors and then list the vendor names matching the lowest prices. In essence, this is a price comparison table where the rows represent line-items for specific products and the columns each represent a different vendor. As such, we use the MIN() function to extract the lowest price for each row (line-item) and then we need to find all the amounts in that row that match that row's lowest price. In turn, we need to use the location of each match to identify the name of the matching vendors. Vendor names all reside in the same row at the top of the columns. Finally, the matching vendor names, separated by commas, would be concatenated into a stringfor display in the formula cell. There will always be at least one match, but sometimes several. The SUMIF() function does something similar for numbers (selects numbers based on a criteria), but, in this case, we need to use numbers to find text that is concatenated rather than summed. In this case, however, it may be a question of a worksheet formula where a an operation could be performed repeatedly on each of the values of an array which is also calculated within the formula, a loop, so to speak? Any suggestions? *************** "Pete_UK" wrote in message ... Q1 - Yes, it is. You would do so like this: =MATCH(A2,B1:H1,0) this would return the relative position of the first exactly matching value in B1:H1 compared with A2. Note that the zero at the end forces it to look for an exact match. If A2 is not in the range, then #N/A will be returned. Q2 - yes, this would be possible. However, you need to give some more detail about your data so that a formula can be offered that is more likely to suit your set up directly. Hope this helps. Pete On Dec 11, 1:13 am, "Blue Max" wrote: FIRST, is it possible to use the MATCH () function on an unsorted range of data? We just need a match from the range, regardless of whether it is the first, last, or inbetween match. In turn, this match echos the column heading label associated with that amount using an INDEX () function. SECOND, is it possible to return all the matches in an unsorted series, grab their associated column headings, and then concatenate them into one string? Any help would be appreciated . . . |
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