Marc has a worksheet in which he keeps produce prices for a community group. Each row represents a produce item (apples, carrots, lettuce, etc.) and each column represents a purchase date. Thus, the intersection of each row and column represents the price at which a produce item was purchased on a particular date. Not all items are purchased on any given date. Marc needs a formula that will, for a particular produce item, return the rightmost value (price) for that item.

An easy way to handle this is to use the LOOKUP function. Let's say that your dates columns are B:N. If you want to figure out the latest price (rightmost value) in row 2, you could use the following:

=LOOKUP(999,B2:N2)

This works because LOOKUP will work with only two parameters. The first is the value you want to look up and the second is the array (range of cells) in which to look for the value. If the value in the first parameter is found, then the value is returned. In other words, if you had the value 999 in one of the cells B2:N2, then 999 is returned.

However, the trick here is to specify a value in the first parameter that is much larger than the largest anticipated purchase price. If you anticipated that you might purchase something for $999 or more, then you could simply increase the value in the first parameter:

=LOOKUP(9^9,B2:N2)

When LOOKUP cannot find the desired value, it returns the last value in the array—the last price shown in the range of cells.

The biggest drawback to this approach is that you'll need to remember to update the range of cells when you add more columns of data. You could get around this by simply making the range much larger than what you need, as shown here:

=LOOKUP(9^9,B2:AZ2)

Of course, if you ever go past column AZ with your data, you still run into the same need to update the range specification. If you don't want to worry about needing to do this, you could combine use of the INDEX and MATCH functions, in this manner:

=INDEX(2:2,MATCH(9^9,2:2))

In this formula, the MATCH function tries to match, within row 2, a value that equals 9^9. If that cannot be located, then MATCH returns the column number of the last value in the row. This is then used as an offset for the INDEX function, which returns the value at the cell offset in row 2. This is, of course, the very rightmost value in the row.

For additional possible formula ideas, see this tip:

https://tips.net/T11250

The biggest difference between the ideas in that tip and this one is that it needs to account for the possibility of zero-value formula results being in the rightmost cells of a row, whereas Marc's situation doesn't. (This means that Marc can use simpler formulas than are included in the other tip.)

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2021-07-17 18:59:52

Roy

MAX(Range)+1

Admittedly, that will fail if the largest value Excel can use is already in the "Range" but then no other number would work either so...

The one way in which it seriously fails is if there could be errors in the "Range" and switching it to LARGE() doesn't help. So if a range is not well-behaved enough to be sure there are no errors, this won't be very helpful. The traditional method of selecting some humungous value WILL work though being sure it's high enough is probably usually no more certain than that the range is well-behaved enough to have no errors. Probably. I imagine.

(How did I not see that there could be no 0's as cell values when I read the email Help section? Ah well, "reading for comprehension" is still alive and well in the repertoires of successful people I guess...)

2021-07-17 10:19:04

Tomek

2021-07-17 07:05:44

Does not matter how many rows or columns in the worksheet.

====================================================

Sub LastValueInColumn()

Dim iRowCount As Long

Dim i As Long

Dim lcol As Long

Dim ColumnLetter As String

' get the last row in column "A" that is not blank

iRowCount = Cells(Rows.Count, 1).End(xlUp).Row

' get the row number to search

i = InputBox("Enter row number between 2 and " & iRowCount, "Get row number.")

' get the last column in row "i" that is not blank

lcol = Cells(i, Columns.Count).End(xlToLeft).Column

' convert column number to a letter

ColumnLetter = Split(Cells(1, lcol).Address, "$")(1)

MsgBox Cells(i, 1) & " - " & Format(Cells(i, lcol), "$.00") & " in cell " & """" & ColumnLetter & i & """" & " on " & Cells(1, lcol)

End Sub

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