Excel INDEX MATCH with multiple criteria - formula examples

The tutorial shows how to lookup with multiple criteria in Excel using INDEX and MATCH and a few other ways.

Although Microsoft Excel provides special functions for vertical and horizontal lookup, expert users normally replace them with INDEX MATCH, which is superior to VLOOKUP and HLOOKUP in many ways. Among other things, it can look up two or more criteria in columns and rows. This tutorial explains the syntax and inner mechanics in full detail so that you can easily adjust the formula for your particular needs. To make the examples easier to follow, you are welcome to download our sample workbook.

Excel INDEX MATCH with multiple criteria

When working with large databases, you may sometimes find yourself in a situation when you need to find something but don't have a unique identifier for the search. In this case, lookup with several conditions is the only solution.

To look up a value based on multiple criteria in separate columns, use this generic formula:

{=INDEX(return_range, MATCH(1, (criteria1=range1) * (criteria2=range2) * (…), 0))}

Where:

  • Return_range is the range from which to return a value.
  • Criteria1, criteria2, … are the conditions to be met.
  • Range1, range2, … are the ranges on which the corresponding criteria should be tested.

Important note! This is an array formula and it must be completed with Ctrl + Shift + Enter. This will enclose your formula in {curly brackets}, which is a visual sign of an array formula in Excel. Do not try typing the braces manually, that won't work!

The formula is an advanced version of the iconic INDEX MATCH that returns a match based on a single criterion. To evaluate multiple criteria, we use the multiplication operation that works as the AND operator in array formulas. Below, you will find a real-life example and the detailed explanation of the logic.

Tip. In Excel 365 and 2021, you can use XLOOKUP formula with multiple criteria.

INDEX MATCH with several criteria - formula example

For this example, we will be using a table in the so-called "flat-file" format with each separate criteria combination (region-month-item in our case) on its own row. Our goal is to retrieve the sales figure for a certain item in a specific region and month.

With the source data and criteria in the following cells:

  • Return_range (sales) - D2:D13
  • Criteria1 (target region) - G1
  • Criteria2 (target month) - G2
  • Criteria3 (target item) - G3
  • Range1 (regions) - A2:A13
  • Range2 (months) - B2:B13
  • Range3 (items) - C2:C13

The formula takes the following shape:

=INDEX(D2:D13, MATCH(1, (G1=A2:A13) * (G2=B2:B13) * (G3=C2:C13), 0))

Enter the formula, say in G4, complete it by pressing Ctrl + Shift + Enter and you will get the following result: Excel INDEX MATCH formula with multiple criteria

How this formula works

The trickiest part is the MATCH function, so let's figure it out first:

MATCH(1, (G1=A2:A13) * (G2=B2:B13) * (G3=C2:C13), 0))

As you may remember, MATCH(lookup_value, lookup_array, [match_type]) searches for the lookup value in the lookup array and returns the relative position of that value in the array.

In our formula, the arguments are as follows:

  • Lookup_value: 1
  • Lookup_array: (G1=A2:A13) * (G2=B2:B13) * (G3=C2:C13)
  • Match_type: 0

The 1st argument is crystal clear - the function searches for the number 1. The 3rd argument set to 0 means an "exact match", i.e. the formula returns the first found value that is exactly equal to the lookup value.

The question is - why do we search for "1"? To get the answer, let's have a closer look at the lookup array where we compare each criterion against the corresponding range: the target region in G1 against all regions (A2:A13), the target month in G2 against all months (B2:B13) and the target item in G3 against all items (C2:C13). An intermediate result is 3 arrays of TRUE and FALSE where TRUE represents values that meet the tested condition. To visualize this, you can select the individual expressions in the formula and press the F9 key to see what each expression evaluates to: How INDEX MATCH with multiple criteria works

The multiplication operation transforms the TRUE and FALSE values into 1's and 0's, respectively:

{1;1;1;1;1;1;0;0;0;0;0;0} * {0;0;1;1;0;0;0;0;1;1;0;0} * {1;0;1;0;1;0;1;0;1;0;1;0}

And because multiplying by 0 always gives 0, the resulting array has 1's only in the rows that meet all the criteria:

{0;0;1;0;0;0;0;0;0;0;0;0}

The above array goes to the lookup_array argument of MATCH. With lookup_value of 1, the function returns the relative position of the row for which all the criteria are TRUE (row 3 in our case). If there are several 1's in the array, the position of the first one is returned.

The number returned by MATCH goes directly to the row_num argument of the INDEX(array, row_num, [column_num]) function:

=INDEX(D2:D13, 3)

And it yields a result of $115, which is the 3rd value in the D2:D13 array.

Non-array INDEX MATCH formula with multiple criteria

The array formula discussed in the previous example works nice for experienced users. But if you are building a formula for someone else and that someone does not know array functions, they may inadvertently break it. For example, a user may click your formula to examine it, and then press Enter instead of Ctrl + Shift + Enter. In such cases, it would be wise to avoid arrays and use a regular formula that is more bulletproof:

INDEX(return_range, MATCH(1, INDEX((criteria1=range1) * (criteria2=range2) * (..), 0, 1), 0))

For our sample dataset, the formula goes as follows:

=INDEX(D2:D13, MATCH(1, INDEX((G1=A2:A13) * (G2=B2:B13) * (G3=C2:C13), 0, 1), 0)) Non-array INDEX MATCH formula with multiple criteria

How this formula works

As the INDEX function can process arrays natively, we add another INDEX to handle the array of 1's and 0's that is created by multiplying two or more TRUE/FALSE arrays. The second INDEX is configured with 0 row_num argument for the formula to return the entire column array rather than a single value. Since it's a one-column array anyway, we can safely supply 1 for column_num:

INDEX({0;0;1;0;0;0;0;0;0;0;0;0}, 0, 1) returns {0;0;1;0;0;0;0;0;0;0;0;0}

This array is passed to the MATCH function:

MATCH(1, {0;0;1;0;0;0;0;0;0;0;0;0}, 0)

MATCH finds the row number for which all the criteria are TRUE (more precisely, the the relative position of that row in the specified array) and passes that number to the row_num argument of the first INDEX:

=INDEX(D2:D13, 3)

INDEX MATCH with multiple criteria in rows and columns

This example shows how to perform lookup by testing two or more criteria in rows and columns. In fact, it's a more complex case of the so-called "matrix lookup" or "two-way lookup" with more than one header row.

Here's the generic INDEX MATCH formula with multiple criteria in rows and columns:

{=INDEX(table_array, MATCH(vlookup_value, lookup_column, 0), MATCH(hlookup_value1 & hlookup_value2, lookup_row1 & lookup_row2, 0))}

Where:

Table_array - the map or area to search within, i.e. all data values excluding column and rows headers.

Vlookup_value - the value you are looking for vertically in a column.

Lookup_column - the column range to search in, usually the row headers.

Hlookup_value1, hlookup_value2, … - the values you are looking for horizontally in rows.

Lookup_row1, lookup_row2, … - the row ranges to search in, usually the column headers.

Important note! For the formula to work correctly, it must be entered as an array formula with Ctrl + Shift + Enter.

It is a variation of the classic two-way lookup formula that searches for a value at the intersection of a certain row and column. The difference is that you concatenate several hlookup values and ranges to evaluate multiple column headers. To better understand the logic, please consider the following example.

Matrix lookup with multiple criteria - formula example

In the sample table below, we'll be searching for a value based on the row headers (Items) and 2 column headers (Regions and Vendors). To make the formula easier to build, let's first define all the criteria and ranges:

  • Table_array - B3:E4
  • Vlookup_value (target item) - H1
  • Lookup_column (Row headers: items) - A3:A4
  • Hlookup_value1 (target region) - H2
  • Hlookup_value2 (target vendor) - H3
  • Lookup_row1 (Column headers 1: regions) - B1:E1
  • Lookup_row2 (Column headers 2: vendors) - B2:E2

And now, supply the arguments into the generic formula explained above, and you will get this result:

=INDEX(B3:E5, MATCH(H1,A3:A5,0), MATCH(H2&H3,B1:E1&B2:E2,0))

Remember to complete the formula by pressing the Ctrl + Shift + Enter shortcut, and your matrix lookup with multiple criteria will be done successfully: INDEX MATCH with multiple criteria in rows and columns

How this formula works

As we are searching vertically and horizontally, we need to supply both the row and column numbers for the INDEX(array, row_num, column_num) function.

Row_num is delivered by MATCH(H1, A3:A5, 0) that compares the target item (Apples) in H1 against the row headers in A3:A5. This gives a result of 1 because "Apples" is the 1st item in the specified range.

Column_num is worked out by concatenating 2 lookup values and 2 lookup arrays: MATCH(H2&H3, B1:E1&B2:E2, 0))

The key factor for success is that the lookup values should match the column headers exactly and be concatenated in the same order. To visualize this, select the first two arguments in the MATCH formula, press F9, and you will see what each argument evaluates to:

MATCH("NorthVendor 2", {"NorthVendor 1", "NorthVendor 2", "SouthVendor 1", "SouthVendor 2"}, 0)

As "NorthVendor 2" is the second element in the array, the function returns 2.

At this point, our lengthy two-dimensional INDEX MATCH formula transforms into this simple one:

=INDEX(B3:E5, 1, 2)

And returns a value at the intersection of the 1st row and 2nd column in the range B3:E5, which is the value in the cell C3.

That's how to look up multiple criteria in Excel. I thank you for reading and hope to see you on our blog next week!

Practice workbook for download

Excel INDEX MATCH multiple criteria (.xlsx file)

How to look up several conditions in Excel

Table of contents

283 comments

  1. Hi! Thank you very much, this has been very useful. Can I please have some advice? Please excuse my lack of technical terms, I'm new to Excel. I would like to input a figure (commodity code) into cell D2 and have the formula search for all exact matches in column B (commodity codes for each product on purchase order) once the formula has located each cell containing the commodity code in column B it will add up all matching data in column A and output the total figure into C2?

    FOR EXAMPLE:

    A B C D
    Net Weight Commodity Code Total Net Weight for Specific Commodity Code Specific Commodity Code
    0.36 4202990090 12.6 4202990090
    7.60 4202990090
    0.10 9004909000
    0.08 9004909000
    4.20 4202990090

    Apologies if I haven't explained my question very well.
    If you can please help me with my query I will be very grateful !

    Thanks
    Connor

  2. I'm creating a team schedule from logged future appointments across multiple people and I want to return the appointment name next to the relevant time slots that it spans over.

    =IFERROR(INDEX('CRM Schedule'!B:B,MATCH(1,INDEX((Schedule!D7>='CRM Schedule'!L:L)*(Schedule!D7<='CRM Schedule'!M:M)*(Schedule!$E$2='CRM Schedule'!J:J),0,1),0)),"Free")

    with the first two criteria above being a schedule sheet and listed appointment times from a CRM system.

    The formula works where there is only one appointment attendee listed in the CRM Schedule Sheet, Column J:J - this part of the formula, with Schedule!E2 being a Name: (Schedule!$E$2='CRM Schedule'!J:J)

    It does not work in the case where there are multiple attendees listed within that cell. Is there a way to add this part of the criteria as a wildcard?

    • Hi! To determine a partial matching of a name in E2 and a record in the CRM Schedule Sheet, use these guidelines: How to find substring in Excel.
      Replace formula (Schedule!$E$2='CRM Schedule'!J:J) with ISNUMBER(SEARCH(Schedule!$E$2,'CRM Schedule'!J:J)). This will identify a partial match between the values in E2 and column J.
      I can't check your entire formula as I don't have your data.

  3. Just wanted to say thanks for this excellent article. Explained a fairly complex idea in an easy to understand way and solved a problem I had trying to extract some data out of an Excel data table that had been causing me some problems . Cracked it thanks to this article.

  4. How to search, compare and match a value from a cell in one template with an instance of a value (not a full cell match) in a row from a second template and return a value from different columns from the second template

      • It partially helps, but still not what I was looking for.

        My example:

        Template 1

        COLUMN A COLUMN B
        ROW 1: Last Name First Name
        ROW 2: Doe John

        Template 2

        COLUMN A COLUMN B
        ROW 1: Last Name, First Name Date Of Birth
        ROW 2: Doe, John 11/19/1954

        If I wanted to compare values/data from COLUMNS A (Row 2) & B (Row 2) in TEMPLATE 1 with the values/data from ROW 2 in Template 2, and if the match is true for both sets of data (COLUMNS A (Row 2) & B (Row 2) from TEMPLATE 1), to return the data on COLUMN B (Row 2) in Template 2.

        I have used the following formulas, but it will not give me the desired result:

        =IFERROR(INDEX('Template 2'!$B$2, MATCH(TRUE, ISNUMBER(SEARCH(A2, Sheet2!$2:$2)) * ISNUMBER(SEARCH(B2, Sheet2!$2:$2)), 0)), "Not Found")

        =IFERROR(INDEX('Template 2'!$B$2, MATCH(1, (COUNTIF('Template 2'!$2:$2, "*"&A2&"*") > 0) * (COUNTIF('Template 2'!$2:$2, "*"&B2&"*") > 0), 0)), "Not Found")

        Appreciate the help.

        • Hi! When you multiply two logical values, you get 1 or 0. Try changing the formula

          =IFERROR(INDEX('Template 2'!$B$2, MATCH(1, ISNUMBER(SEARCH(A2, Sheet2!$2:$2)) * ISNUMBER(SEARCH(B2, Sheet2!$2:$2)), 0)), "Not Found")

  5. Hi,
    I am designing a beam on excel, and have data sheets I need to refer to. I need to INDEX match where I am specifying criteria in both rows and columns, here are my criteria:
    "1.00"='Buckling Resistance - UB'!D8:D756
    'Beam Calculations'!AL21=E7:U7
    'Beam Calculations'!J21='Buckling Resistance - UB'!A8:A756
    'Beam Calculations'!K21='Buckling Resistance - UB'!B8:B756
    Thank you in advanced,
    Millie

  6. Hello,

    Can someone please help me with a formula that averages based on multiple criteria but also identifies the offset values.

    x y x
    April 2 4 5
    May 3 6 4
    April 2 5 2

    I would want to average April but only figures in x columns to get 5.5. Can anyone assist me with this please!!

      • 2.75 sorry

      • Hi,

        My version of Excel does not have the filter formula. Is there a substitute formula I can use?

  7. =INDEX(C3:C46,MATCH(MAX(E3:E46),E3:E46,FALSE),)&" Scored "&MAX(E3:E46)& " in "& "ENGLISH"

    im using this function to find out name of the student who scored maximum marks. if two children got same top score its not showing both names what should i do?

  8. Hi, looking for help to see if this is possible without using =if(and formula. I made it work with that, but exceeded the 8100 characters in a single cell.
    In a table A1:F1 there are 6 random numbers between 1 and 48 that do not repeat
    I would like to be able to match 3 of the 6 numbers in each row and then populate the 3 remaining numbers that go with that set of 6.
    example
    4,9,10,19,25,43
    look up 4,19,25 that would then populate 9,10,43 in another column
    Thanks

      • Thank you so much for your help and the quick response. This pointed me in the right direction and was able to make this work!! If I wanted to learn this stuff where would be a good place to start? Take care

  9. Hi! Is there a way to execute the matrix function where I am matching two criteria based on the column and a third based on a row? I am trying to use the below to return dollars based on 3 matched criteria, but one of those criteria uses the header of the dollar columns

    =INDEX($P$72:$R$148, MATCH(1,($I$64=$N$72:$N$148)*($I68=$O$72:$O$148)*($J$64=$P$71:$R$71),0))

    I am looking to make a dynamic graph using manufacturer names, driven by a specific name and implant chosen. It's healthcare data so I cannot share an example, but I hope I've explained it well enough.

    I have the manufacturer names on the left, and I want to return the spend by physician, for each manuf. using the dropdowns that contain surname and joint type as the variable. I want the dollars from either joint or the "Both" column.

    Manufacturer names are in column "I" which I am matching to column "N", Surname is in column "N" which I am matching to I64, the dollars are in columns "P", "Q", "R" with the headers in P71, Q71, R71. Trying to return the data to column "J"

    Thanks so much

    • Hi! I can't recommend a formula to you as I can't see your data. The MATCH formula can only work correctly with ranges of the same size.
      Your third range in this formula is different.
      Your description is not quite clear. How do you compare Manufacturer names to column N if you write "Surname is in column N". Are you using $J$64 in the formula and "Trying to return the data to column "J" ?
      Please clarify your specific problem or provide additional information to understand what you need. You can use fake data.

      • Hi! Thank you so much for replying. Does this data set help?

        New formula because I moved cells =INDEX($I$3:$K$79, MATCH(1,($B$3=$G$3:$G$79)*($B7=$H$3:$H$79)*($C$3=$I$2:$K$2),0))

        Data
        Columns B & C, using B3 & C3 as the variable
        PHYSICIAN JOINT TYPE
        SURNAME 4 HIP

        MANUFACT 1 #N/A
        MANUFACT 2 #N/A
        MANUFACT 3 #N/A
        MANUFACT 4 #N/A
        MANUFACT 5 #N/A
        MANUFACT 6 #N/A
        MANUFACT 7 #N/A
        MANUFACT 8 #N/A
        MANUFACT 9 #N/A
        MANUFACT 10 #N/A
        MANUFACT 11 #N/A

        The data set I'm trying to retrieve from, columns G:K
        SURNAME MANUFACTURER KNEE HIP BOTH
        SURNAME 1 MANUFACT 3 $3,390 $3,390
        SURNAME 2 MANUFACT 5 $300,216 $198,111 $498,327
        SURNAME 2 MANUFACT 3 $82,372 $70,121 $152,493
        SURNAME 2 MANUFACT 11 $32,273 $32,273
        SURNAME 2 MANUFACT 2 $18,673 $7,130 $25,803
        SURNAME 2 MANUFACT 4 $5,385 $5,385
        SURNAME 2 MANUFACT 9 $3,203 $639 $3,843
        SURNAME 2 MANUFACT 8 $675 $675
        SURNAME 3 MANUFACT 3 $3,350 $3,350
        SURNAME 4 MANUFACT 5 $191,005 $169,552 $360,557
        SURNAME 4 MANUFACT 3 $49,616 $39,761 $89,376
        SURNAME 4 MANUFACT 11 $74,272 $74,272
        SURNAME 4 MANUFACT 8 $7,966 $4,474 $12,440
        SURNAME 4 MANUFACT 9 $1,154 $5,586 $6,740
        SURNAME 4 MANUFACT 2 $4,200 $4,200
        SURNAME 5 MANUFACT 8 $675 $675
        SURNAME 6 MANUFACT 3 $3,011 $11,042 $14,053
        SURNAME 6 MANUFACT 5 $4,200 $4,200
        SURNAME 6 MANUFACT 9 $131 $131
        SURNAME 6 MANUFACT 7 $7 $7
        SURNAME 7 MANUFACT 8 $536,576 $22,887 $559,462
        SURNAME 7 MANUFACT 3 $330 $330
        SURNAME 7 MANUFACT 7 $12 $74 $86
        SURNAME 7 MANUFACT 9 $82 $82
        SURNAME 8 MANUFACT 3 $271,783 $271,783
        SURNAME 8 MANUFACT 9 $2,892 $2,892
        SURNAME 8 MANUFACT 11 $1,295 $1,295
        SURNAME 9 MANUFACT 8 $277,314 $42,031 $319,345
        SURNAME 9 MANUFACT 3 $19,257 $18,475 $37,732
        SURNAME 9 MANUFACT 5 $6,230 $3,450 $9,680
        SURNAME 9 MANUFACT 2 $3,540 $3,540
        SURNAME 9 MANUFACT 6 $487 $487
        SURNAME 9 MANUFACT 9 $54 $54
        SURNAME 10 MANUFACT 3 $1,583,232 $895,278 $2,478,509
        SURNAME 10 MANUFACT 8 $56,814 $70,019 $126,833
        SURNAME 10 MANUFACT 11 $71,930 $46,747 $118,677
        SURNAME 10 MANUFACT 5 $52,813 $56,924 $109,736
        SURNAME 10 MANUFACT 9 $37,999 $51,480 $89,479
        SURNAME 10 MANUFACT 2 $49,523 $1,475 $50,998
        SURNAME 10 MANUFACT 4 $5,385 $5,385
        SURNAME 10 MANUFACT 10 $257 $257
        SURNAME 10 MANUFACT 7 $77 $34 $111
        SURNAME 10 MANUFACT 6 $42 $42
        SURNAME 11 MANUFACT 5 $529,114 $414,904 $944,018
        SURNAME 11 MANUFACT 3 $145,983 $28,705 $174,687
        SURNAME 11 MANUFACT 2 $56,614 $11,275 $67,889
        SURNAME 11 MANUFACT 4 $40,020 $40,020
        SURNAME 11 MANUFACT 9 $8,049 $13,264 $21,313
        SURNAME 11 MANUFACT 11 $4,818 $6,972 $11,790
        SURNAME 11 MANUFACT 8 $1,444 $7,909 $9,353
        SURNAME 12 MANUFACT 3 $409,018 $402,342 $811,360
        SURNAME 12 MANUFACT 11 $7,544 $7,544
        SURNAME 12 MANUFACT 9 $6,696 $6,696
        SURNAME 12 MANUFACT 1 $353 $353
        SURNAME 13 MANUFACT 3 $419,845 $335,662 $755,507
        SURNAME 13 MANUFACT 5 $28,130 $11,276 $39,406
        SURNAME 13 MANUFACT 2 $14,728 $525 $15,253
        SURNAME 13 MANUFACT 9 $4,800 $4,878 $9,678
        SURNAME 13 MANUFACT 8 $4,303 $1,200 $5,503
        SURNAME 14 MANUFACT 5 $661,634 $661,634
        SURNAME 14 MANUFACT 3 $392,000 $392,000
        SURNAME 14 MANUFACT 4 $101,677 $101,677
        SURNAME 14 MANUFACT 2 $55,390 $55,390
        SURNAME 14 MANUFACT 8 $21,579 $21,579
        SURNAME 14 MANUFACT 9 $6,099 $6,099
        SURNAME 14 MANUFACT 7 $210 $210
        SURNAME 14 MANUFACT 6 $92 $92
        SURNAME 15 MANUFACT 3 $32,761 $3,976 $36,737
        SURNAME 15 MANUFACT 9 $577 $832 $1,410
        SURNAME 15 MANUFACT 2 $975 $975
        SURNAME 15 MANUFACT 11 $334 $334
        SURNAME 16 MANUFACT 8 $930,731 $930,731
        SURNAME 16 MANUFACT 11 $1,052 $1,052
        SURNAME 16 MANUFACT 9 $165 $165
        SURNAME 16 MANUFACT 7 $12 $12
        SURNAME 17 MANUFACT 3 $31,007 $28,153 $59,160
        SURNAME 17 MANUFACT 9 $429 $4,787 $5,216

          • Hello!! Yes! That worked and I could see exactly what I did wrong before I tried it. I wasn't aware I need to convert back to the other MATCH syntax when doing the lateral lookup. Thanks so much!!

  10. Hi,
    This works perfectly for pulling out a single number in a data set that I have where I have multiple criteria in my columns & months in the rows (i.e. following your example, as if I had region, vendor & item in the columns, with a list of months in the for the full year).

    Is there a way in which I can continue to do this where I can sum all of the figures that come before a chosen date to get a year-to-date view? For example, if I wanted to know the total sales of Apples by Vendor 1 in the North up to and including September 2024?

    I have tried this with the offset function but quite can't get there with it...

    Thanks in advance!

  11. This was SO helpful - thank you!! I was beginning to think what I wanted to do did not exist!

  12. ACCOUNT CODE Total Jan-24 Feb-24 Mar-24 Apr-24 May-24 Jun-24 Jul-24 Aug-24 Sep-24 Oct-24 Nov-24 Dec-24
    USD USD USD USD USD USD USD USD USD USD USD USD USD
    518101-00-153-0000 4,000 4,000
    518102-00-153-0000 24,657 24,657
    518103-00-153-0000 23,020 23,020
    518105-00-153-0000 2,000 2,000
    515001-00-153-0000 127,500 63,750 63,750
    514004-00-153-0000 14,052 14,052
    528001-00-153-0000 1,000 80 80 80 80 80 80 80 80 80 80 80 120
    528003-00-153-0000 7,000 3,500 3,500
    525001-00-153-0000 3,000 100 100 100 100 150 150 150 150 200 200 200 1,400
    524001-00-153-0000 2,500 1,000 1,000 500

    how to find the value in the below format

    Jan-24 518101-00-153-0000
    Feb-24 518101-00-153-0000
    Mar-24 518101-00-153-0000
    Apr-24 518101-00-153-0000
    May-24 518101-00-153-0000

  13. Is there a way to find a number in a matrix (or a number close to it) and return either or both the row header and/or column header. Example:

    0 Mth 1 Mth 2 Mth
    1 Day 6 174 348
    2 Days 12 180 354
    3 days 17 186 359

    I have the number 186 that I need to find in the matrix and return the number of months and days -- should be 1 mth and 3 days.

    I hope this is clear. Thanks!

    • Table does not align, hopefully this will work:

      ------------0 Mth ----- 1 Mth ----- 2 Mth
      1 Day ------ 6 ----------174 -------- 348
      2 Days --- 12 --------- 180 -------- 354
      3 Days ----17 --------- 186 -------- 359

    • Hi! If your matrix of numbers is written in the range B2:F5 and the number to look up is written in H1, try these formulas to get the row header and column header:

      =INDEX(A1:F1,MIN(IF(B2:F5=H1,COLUMN(B2:F2))))
      =INDEX(A1:A5,MIN(IF(B2:F5=H1,ROW(A2:A5))))

      • That works perfectly...if the number is in the matrix. However, and this is my fault for not making it clear, if I have a number not in the matrix, I would look for the next higher number to determine the month/day. What change would need to be made? With the same sample matrix, if I had the number 181, it would go to the 186 and return the same answer.

        • Hi! To find the nearest number, find the minimum variance of the numbers in the matrix from the number to look up. Use the ABS function to read the absolute value of the variance.

          =INDEX(A1:F1,MIN(IF(ABS(B2:F5-H1)=MIN(ABS(B2:F5-H1)),COLUMN(B2:F2))))
          =INDEX(A1:A5,MIN(IF(ABS(B2:F5-H1)=MIN(ABS(B2:F5-H1)),ROW(A2:A5))))

          I hope my advice will help you solve your task.

          • Alex,

            I appreciate all the help on this. It almost works. 181 - 183 would return 1 Mth and 1 Day and 184 - 185 would return 1 Mth and 2 Days, where 181-185 should all be 1 Mth and 2 Days.

              • The way the formulas work now is finding the range around a number, if it doesn't match exactly to a number in the matrix. If the number is 181, which is not listed in the matrix, it finds 180 (listed) and returns 1 mth and 1 Day. However, what I need it to do is find the next higher number in the matrix (186) and return 1 Mth and 2 Days. The same would go for 182, 183, 184, and 185.

                So, in the above matrix, my return should be:

                179 -- 1 Mth 1 Day
                180 -- 1 Mth 1 Day
                181 -- 1 Mth 2 Days
                182 -- 1 Mth 2 Days
                183 -- 1 Mth 2 Days
                184 -- 1 Mth 2 Days
                185 -- 1 Mth 2 Days
                186 -- 1 Mth 2 Days
                187 -- 1 Mth 3 Days

                I hope this makes it clearer. Again, thank you so much. I appreciate it. :)

            • I’ll try to guess and offer you the following formula:

              =INDEX(A1:F1,MIN(IF(IF(B2:F5-H1<0,9999999,B2:F5-H1)=MIN(IF(B2:F5-H1<0,9999999,B2:F5-H1)),COLUMN(B2:F2))))
              =INDEX(A1:A5,MIN(IF(IF(B2:F5-H1<0,9999999,B2:F5-H1)=MIN(IF(B2:F5-H1<0,9999999,B2:F5-H1)),ROW(A2:A5))))

              It finds the nearest number that is greater than the number to look up, and returns the row and column headers.

              • PERFECT!!!!!!! Thank you for all the help. I truly appreciate it. Have a wonderful holiday time. :)

  14. I need to return the value from within the table below using two different criteria, but the criteria is a number that falls within a number range.
    For example, I have 13skus and 1300boxes, the returning value should be $160.
    The top two rows are the range of box counts for each column (ie. between 501-1000).
    The first two columns are the range for the number of skus.
    How do I set up the formula to return the value of $160 for 13skus and 1300boxes?

    Boxes 01 501 1,001 1,501 2,001
    500 1,000 1,500 2,000 2,500
    Skus
    01-10 $100 $120 $140 $160 $180
    11- 20 $120 $140 $160 $180 $200
    21-30 $140 $160 $180 $200 $220
    31-40 $160 $180 $200 $220 $240

  15. I have 3 columns of data having (thickness),(grade) & (speed) along with tons/hr matrix based on different widths(horizontal).
    how do i make a index..match criteria to select tons/hr for given thk, grade, speed and width

    Width ton/h
    thk grade speed 900 1000 1050 1100 1200 1250 1300 1350
    110 lc 6 247 329 245 360 462 432 470 476
    110 mc 5.8 256 329 245 360 473 470 470 462
    112 mc 6 297 329 245 360 450 432 432 476
    114 lc 6.1 280 329 245 360 432 470 470 462
    114 hc 6 301 329 245 360 470 470 432 476
    117 ap 5.5 297 329 245 360 432 470 462 476
    124 gh 4.7 297 329 245 360 470 470 470 462

  16. Hi,
    I have this data special number and total claim(money).
    How I want to pull data total claim (money) but ignore zero because some of the special no has duplicate?

    SPECIAL no Total claim
    231362 48.36
    231361 0
    231361 4200.83
    231363 852
    231364 54.4
    231281 21.92
    231327 21.92
    231328 21.92

    • Hi! I'm not quite clear what you want to count. If you are counting the sum, then zero is irrelevant. Explain in detail what result you want. You can find the sum for each special number using the SUMIF function.

      • Thank you very much! That solved my problem! It's really simple solution.

  17. How can I get "column1" and "column2" from the first three column? I need to get latest "etd date" and itsitem names of "modal". Also, if etd date has the value of "11/11/2000", I need to know in a same cell like column 1 an 2.

    Modal Item name etd date column1 column2
    1 x 11/11/2000 1 (y 20/01/2021 + x no etd)
    1 y 20/01/2024 2 (x 10/12/2023)
    1 z 10/01/2024 3 (a,b,c 09/01/2024 + e no etd)
    2 x 10/12/2023 4 (b 20/01/2024)
    2 b 09/12/2023
    3 a 09/12/2023
    3 b 09/01/2024
    3 c 09/01/2024
    3 y 09/01/2024
    3 e 11/11/2000
    4 a 10/01/2024
    4 b 20/01/2024

  18. IF any duplicate row will given in this example, and i want to add all same row, then what formula will required?

  19. This page helped to be able to pull one value from a table, but how could I modify to concatenate and pull all values that meet the criteria into one cell?

    *doing a two criteria index match based on a name and "yes" in two different columns. Need it to return all values with this name and "yes", not just the first one*

  20. Column A Column B
    5715363263461 271F0345028709
    5715363263461 271F0345028890

    Need a formula where in if i put column a value in column c then i get 2 different values in column d

  21. Can I combined partial text match and if greater than zero match and return with text?

    Example:
    Column A:
    Row1: Apple
    Row2: Banana
    Row3: Grapes

    Column B:
    Row1: $100
    Row2: $0
    Row3: $1

    Formula: match if "apple", >1, return "valid"

    Answer: Valid

    Is this possible?

    Thank you

  22. I having issues figure this out. Below is a sample of data. Here the logic I want to apply

    1) For the first row match to a row that date equals month +1 and that has the same division number. Example 2022 07 match 2202 08 and division are both 1001.
    2) subtract count 172 - 64 = 108

    Any suggest for how to do this?

    Division Date Year Month count
    1001 2022 07 2022 July 172
    1002 2022 07 2022 July 3
    1003 2022 07 2022 July 1
    1005 2022 07 2022 July 7
    1006 2022 07 2022 July 27
    1007 2022 07 2022 July 33
    1009 2022 07 2022 July 1
    1001 2022 08 2022 August 64
    1002 2022 08 2022 September 22
    1003 2022 08 2022 October 11
    1005 2022 08 2022 November 40
    1006 2022 08 2022 December 1214
    1007 2022 08 2022 January 76
    1009 2022 08 2022 February 9

    • Hi! I hope you have studied the recommendations in the tutorial above. It contains answers to your question. Try this formula:

      =F1-INDEX(F1:F14, MATCH(1,(A1=$A$1:$A$14)*(B1=$B$1:$B$14)*((C1+1)=$C$1:$C$14),0))

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