I have a row of values - 1,2,3,8,35,7,8,3,5,7,x
X is where I want the formula to be
I'd like to somehow get the value of the row which 8 is the closest to X (so not row 4 in this case, but row 7)
If I use match("8",A:A,0) I get the first match it finds.
How do I find the closest match to the cell where the calculation occurs?
You may use:
{=MATCH(2,1/(A1:A10=8))}
just remember it is an array function, so CTRL+SHIFT+ENTER must be used
The answer is based on a trick and behaviour of the MATCH function. I will allow myself to copy the explanation from ozgrid:
The magic here is actually in the MATCH function more than the array. Two interesting properties of the match function are being used here:
1) With MATCH, if no match is found, then the function will return the position of the last value in the array, so if you did =MATCH(8,{1,2,3,4,5,6,7,6,5,4,3,2,1}), your result is = 13, since there was no 8 to find in the array
2) MATCH will return the position of the last value, but won't return the position of an error (or of a blank value), so if you have =MATCH(8,{1,2,3,4,#DIV/0!,#DIV/0!,7,6,5,4,3,#DIV/0!,#DIV/0!}), your result is = 11, as the 3 in the 11th position is the last value in the array
So daddylonglegs' formula checks each cell against the target value with (A1:A13=B1) in the array formula, giving an array with TRUE (or 1) for the positions where the cells match, and with FALSE (or 0) for the rest. Dividing 1 by this results in 1s wherever that array was TRUE, and #DIV/0! wherever that array was false. So his formula is evaluated as
=MATCH(2,{#DIV/0!,1,#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!,1,#DIV/0!,1,#DIV/0!,#DIV/0!,#DIV/0!,#DIV/0!})
Since no '2' is found in the array, and the last value (1) is found in the 9th position, the MATCH returns 9
If you want the last match you can use this formula in A11
=MATCH(2,INDEX(1/(A1:A10=8),0))
[Note: if the values are numeric then 8 is OK - if they are text formatted numbers then you need "8"]
....or do you have values either side of the calculation cell (so nearest might be up or down)?
Related
A
B
C
D
4
1
6
5649
3
8
10
9853
5
2
7
1354
I have two worksheets, for example column A in sheet 1 and columns B-D in sheet 2.
What I want to do is to take one value in Column A, and scan both columns B and C and it is between those two values, then display the corresponding value from column D in a new worksheet.
There could be multiple matches for each of the cell in column A and if there is no match, to skip it and not have anything displayed. Is there a way to code this and somehow create a loop to do all of column A? I tried using this formula, but I think it only matches for each row and not how I want it to.
=IF(AND([PQ.xlsx]Sheet1!$A2>=[PQ.xlsx]Sheet2!$B2,[PQ.xlsx]Sheet1!$A2<[PQ.xlsx]Sheet2!$C2),[PQ.xlsx]Sheet2!$D$2,"")
How do I do this?
Thank you.
I'm not positive if I understood exactly what you intended. In this sheet, I have taken each value in A:A and checked to see if it was between any pair of values in B:C, and then returned each value from D:D where that is true. I did keep this all on a single tab for ease of demonstration, but you can easily change the references to match your own layout. I tested in Excel and then transferred to this Google Sheet, but the functions should work the same.
https://docs.google.com/spreadsheets/d/1-RR1UZC8-AVnRoj1h8JLbnXewmzyDQKuKU49Ef-1F1Y/edit#gid=0
=IFERROR(TRANSPOSE(FILTER($D$2:$D$15, ($A2>=$B$2:$B$15)*($A2<=$C$2:$C$15))), "")
So what I have done is FILTEREDed column D on the two conditions that Ax is >= B:B and <= C:C, then TRANSPOSED the result so that it lays out horizontally instead of vertically, and finally wrapped it in an error trap to avoid #CALC where there are no results returned.
I added some random data to test with. Let me know if this is what you were looking at, or if I misunderstood your intent.
SUPPORT FOR EXCEL VERSIONS WITHOUT DYNAMIC ARRAY FUNCTIONS
You can duplicate this effect with array functions in pre-dynamic array versions of Excel. This is an array function, so it has be finished with SHFT+ENTER. Put it in F2, SHFT+ENTER, and then drag it to fill F2:O15:
=IFERROR(INDEX($D$2:$D$15, SMALL(IF(($A2>=$B$2:$B$15)*($A2<=$C$2:$C$15), ROW($A$2:$A$15)-MIN(ROW($A$2:$A$15))+1), COLUMNS($F$2:F2))),"")
reformatted for easier explanation:
=IFERROR(
INDEX(
$D$2:$D$15,
SMALL(
IF(
($A2>=$B$2:$B$15)*($A2<=$C$2:$C$15),
ROW($A$2:$A$15) - MIN(ROW($A$2:$A$15))+1
),
COLUMNS($F$2:F2)
)
),
"")
From the inside out: ROW($A$2:$A$15) creates an array from 2 to 15, and MIN(ROW($A$2:$A$15))+1 scales it so that no matter which row the range starts in it will return the numbers starting from 1, so ROW($A$2:$A$15) - MIN(ROW($A$2:$A$15))+1 returns an array from 1 to 14.
We use this as the second argument in the IF clause, what to return if TRUE. For the first argument, the logical conditions, we take the same two conditions from the original formula: ($A2>=$B$2:$B$15)*($A2<=$C$2:$C$15). As before, this returns an array of true/false values. So the output of the entire IF clause is an array that consists of the row numbers where the conditions are true or FALSE where the conditions aren't met.
Take that array and pass it to SMALL. SMALL takes an array and returns the kth smallest value from the array. You'll use COLUMNS($F$2:F2) to determine k. COLUMNS returns the number of columns in the range, and since the first cell in the range reference is fixed and the second cell is dynamic, the range will expand when you drag the formula. What this will do is give you the 1st, 2nd, ... kth row numbers that contain matches, since FALSE values aren't returned by SMALL (as a matter of fact they generate an error, which is why the whole formula is wrapped in IFERROR).
Finally, we pass the range with the numbers we want to return (D2:D15 in this case) to INDEX along with the row number we got from SMALL, and INDEX will return the value from that row.
So FILTER is a lot simpler to look at, but you can get it done in an older version. This will also work in Google Sheets, and I added a second tab there with this formula, but array formulas work a little different there. Instead of using SHFT+ENTER to indicate an array formula, Sheets just wraps the formula in ARRAY_FORMULA(). Other than that, the two formulas are the same.
Since FALSE values aren't considered, it will skip those.
I know how to use index and match formulas to get the value or location of a matching cell. But what I don't know how to do is get that information when the cell I'm looking for isn't going to be the first match.
Take the image below for example. I want to get the location of the cell that says "Successful Deliveries". In this example there's a cell that matches that in rows 11 and 30. These locations can vary in the future so I need a formula that's smart enough to handle that.
How would I get the location of the second instance of "Successful Deliveries"? I figured I could use the "Combination 2 Stats" value from row 24 as a starting point.
I tried using this formula:
=MATCH("Successful Deliveries:",A24:A1000,0)
But it returns a row number of 7 which is just relative to the A24 cell I started my match at.
My end goal here is to get the value from the cell directly to the right of the second match of "Successful Deliveries".
In your formula, with no further intelligence, you can simply add 23 to adjust 7 to the result:
=MATCH("Successful Deliveries:",A24:A1000,0) + 23
You know that 23 is the number to add because you started your search on row 24.
The full answer is here:
https://exceljet.net/formula/get-nth-match-with-index-match
You use this formula:
=INDEX(B1:B100,SMALL(if(A1:A100 = "Successful Deliveries:",ROW(A1:A100) - ROW(INDEX(A1:A100,1,1))+1),2))
...where 2 is the instance you want.
Make sure to finish typing the formula by hitting ctrl-shift-enter. (You know you did this right because the formula gets curly brackets {})
HOW IT WORKS
Normally, we use INDEX / MATCH to find a value. The Index function gives you the nth value in a range, and the Match function determines which "n" is a match for our criteria.
Here we use INDEX the same way, but we need more intelligence to find that "n", since it's the second one that matches the criteria. That's where SMALL comes in. The Small function "gets the nth smallest value in an array". So we give Small the number of the desired instance (2 in this case) and we give it an array of blanks and the rows numbers of the rows we like.
We obtained the array of blanks and row numbers using the If function, asking it to check for our criterion (="Successful...") and making it return the row number where the criterion passes (=Row(A1:A100)). By using the If function as an array function (by giving it arrays and using ctrl-shift-enter) it can deliver a whole list of values.
Our final value is just one number because the Small function used the array from the IF to return just one thing: the second-smallest row we gave to it.
I'm trying to understand some legacy Excel file (it works, but I would really like to understand how/why it's working).
There is a sheet for data input (input sheet)and some code that is called to process data in the input sheet. I found out that number of rows in the input sheet is determined using a Lookup formula like this:
=LOOKUP(2;1/('Input sheet'!E1:E52863<>"");ROW(A:A))
"E" column contains names for import items and column is NOT sorted
"A" column does not contain anything special - I can replace it with B, C or whatever column and it does not affect the formula's outcome
According to what I have found about Lookup behaviour: •If the LOOKUP function can not find an exact match, it chooses the largest value in the lookup_range that is less than or equal to the value.
What does this ^-1 operation to the specified range? If E(x) is not empty -> it should turn into 1, but if it is empty - then it would be 1/0 -> that should produce #DIV/0! error...
1/('Input sheet'!E1:E52863<>"")
The outcome is the same, if I replace 2 with any positive number (ok, tried only some, but it looks like this is the case). If I change lookup value to 0, then I get #N/A error -> •If the value is smaller than all of the values in the lookup_range, then the LOOKUP function will return #N/A
I am stuck... can anyone shed some light?
LOOKUP has the rare ability to ignore errors. Conducting the 1/n operation will produce an error every time n is zero. False is the same as zero. So, for your formula, every empty cell produces an error in this calculation. All of those results are put in a vector array in the 2nd argument.
Searching for any positive value (the 1st argument) larger than 1 will result in LOOKUP finding the last non-error value in the above vector.
It also has the nice optional 3rd argument where you can specify the vector of results from which to return the lookup value. This is similar to the INDEX component of the the INDEX/MATCH combo.
In the case of your formula, the 3rd argument is an array that looks like this: {1;2;3;4;5;6;7;8;9;...n} where n is the last row number of the worksheet, which in modern versions of Excel is 1048576.
So LOOKUP returns the value from the vector in the 3rd argument that corresponds to the last non-error (non-blank cell) in the 2nd argument.
Note that this method of determining the last row will ignore cells that have formulas that result in a zero-length string. Such cells look blank but of course they are not. Depending on the situation, this may be precisely what you want. If, on the other hand you want to find the last row in column E that has a formula in it even if it results in a zero-length string, then this will do that:
=MATCH("";'Input sheet'!E:E;)
You might get some idea what the formula is doing (or any other formula) if you apply Evaluate Formula. Though since the principle is the same whether 3 rows or 52863 I'd suggest limiting the range, to speed things up if choosing Evaluate Formula. As usual with trying to explain formulae, it is best to start from the inside and work outwards. This:
'Input Sheet'!E1:E52863<>""
returns an array with a result for every entry in ColumnE from Row1 to Row52863. Since it is a comparison (<> does not equal) the result is Boolean - ie TRUE (not empty) or FALSE (is empty). So if only the first half of E1 to E52863 is populated, the result is {TRUE;TRUE;TRUE; ... and a LOT more TRUE; ... and FALSE ... and a LOT more ;FALSE and finally }.
Working outwards, the next step is to divide this array into 1. In arithmetic operations Boolean TRUE is treated as 1 and FALSE as 0, so the resultant array is {1;1;1; ... and a LOT more 1; ... and #DIV/0!... and a LOT more ;#DIV/0! and finally }.
This then becomes the lookup_vector within which LOOKUP seeks the lookup_value. The lookup_value you show is 2. But the array comprises either 1 or #DIV/0! - so 2 will never be found in it. As you have noticed, that 2 could just as well be 3, or 45 or 123 - anything as long as not a value present in the array.
That (not present) is necessary because LOOKUP stops searching when it finds a match. The fact that there is no match forces it to the end of the (valid) possibilities - ie the last 1. At this point, in my opinion, it would be logical to return "not found" but - I suspect merely a quirk, though very convenient - it returns that 1 - by its index number in the list, ie 52863 if all cells in E1:E52863 are populated.
Although the result_vector (Row(A:A)) is optional for LOOKUP it is required in this usage in effect to fix the start point for the index (effectively Row1, since an entire column). You might change that to say A3:A.. and the result would be the number of the highest populated row number in ColumnE plus 2 (3 -1).
Here is the screenshot of my excel workbook
I do not understand why the value in cell j7 is 44 ?
j7 formula is =LOOKUP(1,(TRIM($D$2:$D$9)=TRIM(H7))/(TRIM($E$2:$E$9)=TRIM(I7)),$F$2:$F$9)
The result of the two arrays division is the following
{TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE}/
{TRUE;FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE} =
{1;#DIV/0!;0;#DIV/0!;#DIV/0!;0;#DIV/0!;#DIV/0!}
Right ?
So I am looking for 1, basically the formula becomes
LOOKUP(1,{1;#DIV/0!;0;#DIV/0!;#DIV/0!;0;#DIV/0!;#DIV/0!},$F$2:$F$9)
Hence the result should be 10 but not 44 . . . . . ?
EDIT
When i correct my formula to =LOOKUP(1,1/(TRIM($D$2:$D$9)=TRIM(H7))/(TRIM($E$2:$E$9)=TRIM(I7)),$F$2:$F$9)
it works fine . Why ? Thank you everybody for giving the alternative solutions with match and index. I just can't understand why my first formula did not work. Any why when i add 1/ it MAGICALLY works ? ? ?
If the values are not in ascending order, and you are looking for a value within the range of values (as opposed to a value larger than anything in the range), LOOKUP may give unexpected results.
A different way to return the desired result is with a combination of INDEX and MATCH:
=INDEX($F$2:$F$9,MATCH(1,(TRIM($D$2:$D$9)=TRIM(H7))/(TRIM($E$2:$E$9)=TRIM(I7)),0))
entered as an array formula with ctrl-shift-enter
Note that with MATCH, looking for an exact match, the range does not need to be sorted.
Another formula that will return the correct results, and can be normally entered (assuming no duplicate entries in the first table):
=SUMPRODUCT((TRIM(H7)=TRIM($D$2:$D$9))*(TRIM(I7)=TRIM($E$2:$E$9))*$F$2:$F$9)
The explanation is:
with the first formula, the first array in the Lookup function contains zeros after the 1 value, in the third and sixth value of the array:
Lookup expects data to be sorted ascending and will return the first item less than or equal to the search value, starting from the last value in the array. In this case that is the zero value in the sixth position of the array.
The edited formula results in an array that contains only one number "1". All other values are Div errors. So the position of that "1" value is what Lookup will use.
Further explanation:
In your first formula you divide two arrays that contain TRUE or FALSE and the result contains 1, 0 and Div error values.
{TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE}/
{TRUE;FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE} =
{1;#DIV/0!;0;#DIV/0!;#DIV/0!;0;#DIV/0!;#DIV/0!}
Including the 1/ in the formula will divide the first array of TRUE and FALSE values by 1, which returns an array that consists of either 1 or Div errors. Further dividing that array by the second array will only return 1 or Div errors, no zeros. The steps are
1/{TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE}/{TRUE;FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE} results in
{1;1;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!}/{TRUE;FALSE;TRUE;FALSE;FALSE;TRUE;FALSE;FALSE} results in
{1;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!;#DIV/0!}
No zeros!
As mentioned, LOOKUP expects the values in the lookup_vector to be in ascending order. To gain the first match of columns H & I to columns D & E, I would suggest mathematically excluding the non-matching rows. What is left would be the matching rows. The following example supplies the first double match.
For J2
=INDEX($F$2:$F$9,MIN(INDEX(ROW($1:$8)+(($D$2:$D$9<>H2)+($E$2:$E$9<>I2))*1E+99,,)))
Fill down as necessary. This is a standard formula and can be easily modified to supply the 2nd, 3rd, etc matching values by swapping SMALL() in place of MIN(). Your results should be close to the following.
I have a simple worksheet with 2 columns
I want to get all the results(on column "H") (I can get only the first occurrence,I want to know if I can get the others) that contains the value from cell G1, is that possible without a macro ? any way of doing it would be appreciated...Any ideas?
You can use ROW() and SMALL() to get those instead of MATCH() since this always gets the first match.
=IFERROR(INDEX($C$4:$C$7,SMALL(IF($D$4:$D$7=$G$1,ROW($D$4:$D$7)-(ROW()-1)),ROWS($D$4:D4))),"Null")
So, if the array $D$4:$D$7=$G$1 returns true (i.e. the value equals that in G1), you will get the row numbers of these values, in this case, you will get 4 and 6. All the other will return False.
After some processing with -(ROW()-1), the 4 and 6 become 1 and 3. those two values are what will be fed to the INDEX.
SMALL() then picks the smallest, starting with the 1st (you get 1 from ROWS($D$4:D4)) and when you drag the formula down, the ROWS become ROWS($D$4:D5) which gives 2, and SMALL ends up taking the 2nd smallest value, which is 3.
EDIT: Forgot to mention. You have to array enter the above formula. To do this, type the combination keys of Ctrl+Shift+Enter after typing the formula (edit the formula again if necessary) instead of Enter alone.