I am trying to write a formula which will return a positive value to a cell G3 if the content of D3 is a partial match to anything found in the same column excluding itself. I.e.
Red - No Match
Blue - Partial Match
Light Blue - No Match
No match because "Red" is not a partial match with any of the other results, Partial match because "Blue" IS a partial match for "Light Blue".
I am using the current formula, but it returns false when I expect it to return true
=IF(ISNUMBER(SEARCH(D3,$D$1:D2,D4:$D$10000)),"Yes","No")
My original formula:
Suggested formula:
a formula which will return a positive value is too broad. But I did a formula that counts your partial matches. If the result of the formula is 0, means there is no matches or partial matches:
The formula I've used:
=COUNTIFS($A$1:$A$3;"*" &A1&"*")-1
I added a -1 because with this formula you will get always at least 1 match (the cell you want to exclude). So with -1 you make sure you get the other partial matches.
You cam combine this with a normal IF. If result of this formula is 0, means there is no partial matches (excluding the own in same row, of course). If is not zero, it means there are partial matches. Then do whatever you need to do to get that positive value.
UPDATE: Now I understand what you mean with positive value. You can combine the formula above with an IF as I said. Try something like this:
=IF(COUNTIFS($A$1:$A$3;"*" &A1&"*")-1=0;"NO";"YES")
Or try:
=IF(COUNTIFS($A$1:$A$3;"*" &A1&"*")-1=0;FALSE;TRUE)
Related
I am trying to make lookups with Multiple Non-Exact Criteria using INDEX-MATCH.
The formual looks like this:
=INDEX(C314:C318;MATCH(1;(D314:D318>=G313)*(E314:E318>=G314);0))
Criterias are: greater or equal to amount X.
Formula works fine, however when using a long list of values, it does not find the best matching value, it finds the first value that matches the criteria.
For example.
Condition 1 is: code "find code equal to 2055516"
Condition 2 is: numerical "find value equal or above 77"
Condition 3 is: alphabetical "find letter equal or greater than H"
In a large dataset where I´ve got many values, it finds only the next best value that matches this criteria. First value that meet that condition would be "80" and "R", however, following values in my dataset, way below, meet much better those criteria with "78" and "I". Problem here is clear I guess.
How can I adapt my formula to look for those much more fitting values that meet my condtions?
Dataset table looks like this:
The formula should return the name "A, B, C, D, E" of the best maching product.
I used a helper column called Helper to rank the letters in Condition 2 alphabetically first using the following formula (drag it down to apply to all rows):
=COUNTIF(Condition_2,"<="&Condition_2)
then use the following formula to find the best match (although it is an array formula it does not need to be confirmed by Ctrl+Shift+Enter):
=INDEX(Product1,MATCH(AGGREGATE(15,6,Helper/((Condition_1>=77)/(Condition_2>="H")),1),Helper,0))
Replace the named ranges in the above formula with the actual ranges in your worksheet.
Replace , with ; as the delimiter in all formulas to suit your system.
EDIT #2
Based on the new scenario, the problem can be solved by AGGREGATE function solely given that the look up value is a number (EAN)
The formula in Cell J2 of my above example is:
=AGGREGATE(15,6,EAN/((DIMENSION=F2)/(LOAD_INDEX>=G2)/(SPEED_INDEX>=H2)),1)
Please note the following are all named ranges which needs to be replaced with the actual range on your workbook:
DIMENSION being B2:B8
LOAD_INDEX being C2:C8
SPEED_INDEX being D2:D8
EAN being A2:A8
If you do not want to show the error #NUM! for no matching result, you can use IFERROR to return a blank cell as shown in Cell J3 of my example. The formula is:
=IFERROR(AGGREGATE(15,6,EAN/((DIMENSION=F3)/(LOAD_INDEX>=G3)/(SPEED_INDEX>=H3)),1),"")
EDIT #3
Please use the following array formula (need to confirm by pressing Ctrl+Shift+Enter) to find the closest match of LOAD INDEX and SPEED INDEX with the help of a Helper column.
{=INDEX(EAN,MATCH(AGGREGATE(15,6,Helper/((LOAD_INDEX/((DIMENSION=G2)/(LOAD_INDEX>=H2)/(SPEED_INDEX>=I2)))=AGGREGATE(15,6,LOAD_INDEX/((DIMENSION=G2)/(LOAD_INDEX>=H2)/(SPEED_INDEX>=I2)),1)),1),Helper/((LOAD_INDEX/((DIMENSION=G2)/(LOAD_INDEX>=H2)/(SPEED_INDEX>=I2)))=AGGREGATE(15,6,LOAD_INDEX/((DIMENSION=G2)/(LOAD_INDEX>=H2)/(SPEED_INDEX>=I2)),1)),0))}
The logic is to first find the closest matches to LOAD INDEX and then find the closest match to SPEED LIMIT from the range with the closest matches to LOAD INDEX.
Again if you do not want to show #NUM! error for no matching result, you can use IFERROR to return the desired result.
Let me know if there is any question. Cheers :)
This might be done through something much smarter, but the following worked for me:
Formula in G2, following the above sample data to return the row with the best match:
=INDEX(A2:A6,MATCH(SMALL(IF(B2:B6-E2>-1,B2:B6-E2+IF(CODE(C2:C6)-CODE(F2)>-1,CODE(C2:C6)-CODE(F2),""),""),1),IF(B2:B6-E2>-1,B2:B6-E2+IF(CODE(C2:C6)-CODE(F2)>-1,CODE(C2:C6)-CODE(F2),""),""),0))
Note Enter as array formula through Ctrl+Shift+Enter
When no criteria matches, it will return an error, you could catch through IFERROR().
I have a spreadsheet that I track my hours. Each cell initially is populated with a formula, i.e. =IF(WORKDAY(B24-1,1,holidays2019)=B24,OFFSET(C24,-1,2),0)
and then as the month progresses I enter my actual time.
In the following excerpt all values through 5/10/2019 are entered.
The formula =SUMIF(C5:C19,NOT(ISFORMULA(C5:C19))) shows zero. I do not understand why this does not work.
I appreciate any help! Column B in my spreadsheet corresponds to the dates shown below and Column C to the time entries.
Expected Result: 48.9
=SUMPRODUCT(J6:J20,--NOT(ISFORMULA(J6:J20)))
The key to this solution is the -- in front of the NOT(). A boolean that is processed by a math operator gets converted to 1 or 0. --, +0, -0, *1, /1 would have all worked to do the conversion. So now you wind up with an array of values you may want to sum being multiplied by an array of 1 and 0 to indicate the ones you want. The 1 are manual entry and the 0 are your formulas entries.
Now SUMPRODUCT performs array like calculations. As a result avoid using full column/row references inside it or you will wind up with a lot of excess calculations. Adjust the ranges in the answer to suit your needs.
Here's the MSDN definition of the Criteria in =SUMIF
criteria Required. The criteria in the form of a number, expression,
a cell reference, text, or a function that defines which cells will be
added. For example, criteria can be expressed as 32, ">32", B5, "32",
"apples", or TODAY().
Important: Any text criteria or any criteria that includes logical or
mathematical symbols must be enclosed in double quotation marks (").
If the criteria is numeric, double quotation marks are not required.
So, the reason, why your SUMIF returns 0 is, because none of the cells match the criteria, as they return a number and meanwhile they expect FALSE
Another issue here being, that the ISFORMULA will return TRUE, even when a range contains a single formula while all the rest has none. So basically you need to drag the formula down for each cell individually and sum them up only when a value is TRUE
Starting from cell D1:
=ISFORMULA(B1)
And then you can simply sum them up with the formula you provided.
=SUMIF(D1:D16,TRUE,C1:C16)
Obviously, you can hide the column D to make it more aesthetically pleasing.
Your formula fails because the criteria you're matching against, is TRUE/FALSE. Obviously the values in C5:C19 don't contain any booleans, so the sum is 0.
To solve this, you can add the correct criteria in cell D5 and below: =ISFORMULA(C5)
Then use =SUMIF(D5:D19,FALSE,C5:C19) to sum the values in column C.
I have this function:
MATCH(1,(PositionParameter[[#All],[Position Revised]]=$C94)asterisk(PositionParameter[[#All],[Campus Type Short]]=G$3)asterisk(PositionParameter[[#All],[Campus Num Arbitrary]]=G$1),0))
and I can't figure out what it does. I don't know what the asterisks are for. PositionParameter is the name of the worksheet, Position Revised is the name of a column, Campus Type Short is the name of a column, and Campus Num Arbitrary is the name of a column. There is suppose to be an asterisk between the first PositionParameter() and the second PositionParameter(). There is supposed to be another asterisk between the second PositionParameter() and the third PositionParameter(), but it is rendered as an italic. I took the asterisk out and spelled it out. The tooltip tells me this is suppose to return some sort of array, but I can't figure out its components. Can someone explain the asterisks to me? I would appreciate it.
Thanks,
Howard Hong
Your formula returns a single value - the relative position of the first row in the data where all three conditions are met.
It works like this:
Each of these three conditional statements:
PositionParameter[[#All],[Position Revised]]=$C94
PositionParameter[[#All],[Campus Type Short]]=G$3
PositionParameter[[#All],[Campus Num Arbitrary]]=G$1
.....returns an array of TRUE/FALSE values. Multiplying these three arrays together produces a single array of 1/0 values, 1 when all conditions are met in a row, 0 otherwise. This array forms the "lookup array" of the MATCH function
The "lookup value" is 1 so that value is looked up in the lookup array and the result of the MATCH function is the position of the first 1, which corresponds to the first row where all conditions are satisfied.
If there are no rows which meet all three conditions then the result is #N/A
Note that the zero at the end is the third parameter of the MATCH function - zero menas that an exact match must be found.
This is an "array formula" which needs to be confirmed with CTRL+SHIFT+ENTER
Often you would use this in conjunction with INDEX function to return a value from another column in the first row where conditions are satisfied, e.g. using normal cell references
=INDEX(A:A,MATCH(1,(B:B="x")*(C:C="y"),0))
That formula will return the value from column A in the first row where the two specified conditions are met (col B = "x"and col C = "y")
Well, asterisk could be a multiplication symbol or it could be a wildcard in Match. By the looks of the placement, I'd say it's multiplying data from an array or table.
And, um... I don't know what the asterisks are for but I took the asterisk out and spelled it out? Why would you do that? Was it working before you changed it? Where did you find this formula?
Please read [mcve]. Without sample data or other information about the purpose of the formula, I will take a wild guess:
Paste this into the cell:
=MATCH(1,(PositionParameter[[#All],[Position Revised]]=$C94)*(PositionParameter[[#All],[Campus Type Short]]=G$3)*(PositionParameter[[#All],[Campus Num Arbitrary]]=G$1),0))
. . . and assuming it's supposed to be an array, instead of hitting Enter on that cell:
hit: Ctrl+Shift+Enter to create an array formula.
Besides the link above, here is some other reading & practice for you:
Create an array formula
MATCH function
I think certains applications replace certain symbols (that aren't allowed in the application] with words when copying and pasting from Excel to them, but without more information about what happened, I can't say for sure what happened.
Assuming that the * are real and that the formula is entered as an array formula then it should return an array of 0s and 1s.
The formula is looking for Position Revised=C94 AND Campus Type Short =G3 AND Campus Num Arbitrary = G1
It will return a 1 for each row that matches all these conditions and a 0 for each row that does not.
If no rows match the conditions it will return #N/A
Hi there , I have this formula
=IFERROR(IF(MATCH(A2,G:G,0)*OR(MATCH(B2,G:G,0)),"Present",),"Absent")
What I want is to return Present if one of the email from Column A and B present in Column G.
The Formula Work with *And but its not working with *OR.
If there is no match, then Match() will return the #N/A error, which will be multiplied with the other Match() result and still return an error. Therefore, this formula will only have a non-error result if BOTH Match formulas return a proper value. That's not what you want, I assume.
You need a formula or function that does not resolve in an error if there is only one match for the two conditions.
One option is to encase each Match into an Iferror. Anonther option is to use Countif, along the lines of this:
=if(countif(G:G,A2)+countif(G:G,B2),"Present","Absent")
Countif returns 0 if nothing is found or a count of the items found. The zero will equal a FALSE in the IF statement, so if neither Countif finds anything, the FALSE argument of the IF function will fire. Bit if any of the two Countif functions find a match, the result will be a number greater than zero, so the TRUE part of the IF function will fire.
I made a formula using match and isnumber.
=IF(ISNUMBER(MATCH(A2;G:G;0));"Present";IF(ISNUMBER(MATCH(B2;G:G;0));"Present";"Absent"))
So if match number is value that means there are match in column G so then it will return Present, simple as that.
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)?