I'm using the below formula to find the Maximum income in a data set (entered as an array):
{=MAX((DATA!$B$2:$B$10000=1)*(DATA!$M$2:$M$10000=MAXINCOME!$D6)*(DATA!$A$2:$A$10000))}
Column B is the product, M is an agent and A is the range of values. However, any negative values return as 0. For some agents, all their values will be negative but I'm struggling to adapt this formula so if that scenario occurs the number closest to 0 is returned. Can anyone please help?
By multiplying the conditions with the values you get zeroes when the conditions don't apply for any row, so you always get a minimum of zero as MAX - try using nested IFs so that doesn't happen, i.e.
=MAX(IF(DATA!$B$2:$B$10000=1,IF(DATA!$M$2:$M$10000=MAXINCOME!$D6,DATA!$A$2:$A$10000)))
confirmed with CTRL+SHIFT+ENTER
That version can give you a negative result
....also MAX will return zero if there are no rows where the conditions are satisfied. That may be confusing if MAX values can include negative numbers (and possibly zero) so it might be better to use LARGE function here instead - that will give an error if there are no matching rows (but the same results otherwise), so you can then "error trap" the error and return an appropriate text value, e.g.
=IFERROR(LARGE(IF(DATA!$B$2:$B$10000=1,IF(DATA!$M$2:$M$10000=MAXINCOME!$D6,DATA!$A$2:$A$10000)),1),"No matches")
Related
I currently have a column of data with numbers. I want to create a second column which returns TRUE when a set of multiple conditions are TRUE. This column should return FALSE when one or more conditions are FALSE. The conditions are the following:
The number is negative (<= 0)
The number in this list is the closest to zero
The number in this list is not #N/A
The big trouble I'm having is because of these #N/A's in my dataset. I can't get my formula to ignore it. This is the current formula I use in column B, but for some reason I can't get this to work:
=IF(AND(A1<=0;A1=MAX(IF(NOT(IFNA($A$1:$A$300;TRUE));TRUE;FALSE)));TRUE;FALSE)
I suppose something's wrong in my interpretation of the IF(NOT(IFNA(...)) part.
Sorry if it's obvious, but can anyone help me out here?
I would use aggregate because there is an option to exclude NA's:
=IF(ISNA(A1),FALSE,A1=AGGREGATE(14,6,A$1:A$10/(A$1:A$10<0),1))
(this is assuming the number should be equal to the nearest negative number to zero, in other words the highest negative number. In this case you don't need a separate test for the number being negative)
I would use the array formula:
=IF(ISNA(A1),FALSE,IF(A1>0,FALSE,IF(A1=MAX(IF(A:A<0,A:A)),TRUE,FALSE)))
Array formulas must be entered with Ctrl + Shift + Enter rather than just the Enter key. If this is done correctly, the formula will appear with curly braces around it in the Formula Bar.
I have two Excel lists:
One extensive with 20 thousand lines. In which:
Two columns are important: First: Unique ID, Second: a value (number formatted).
It can be a value that appears several times, or only once.
I have to create the second list. In this list I have only one column of values that I would like to have.
I need a formula that will look for values from List 2 in List 1 and then match a Unique ID to each value.
It is important that, when no direct value exist. In this case it has to search for a sample which is in about 3-5% value deviation.
Example: there was no value 127, but within 3%, 125 was found.
I've tried indexing and comparison, but it does not seem to work.
VLOOKUP worked, but without 3-5% deviation
I am very grateful for the help.
Example: http://www.filedropper.com/excellist1and2
If the value exists in the list, you can use VLOOKUP or INDEX(MATCH to find it - that's the easy part. If the value is not in the list, then you need to find the nearest value.
The nearest "low" value will be the MAX value ≤ our input, and the nearest "high" value will be the MIN value ≥ our input.
If you have Office 365, you can use MINIFS($D$1:$D$6,$D$1:$D$6,">="&B1,$D$1:$D$6,"<="&(B1*1.05)) and MAXIFS($D$1:$D$6,$D$1:$D$6,"<="&B1,$D$1:$D$6,">="&(B1*0.95)) )` here. If not, you'll need an Array Formula, we can build that "±5%" in early, to simplify the formula.
Starting with the Low values, we want the MAX value ≤ our input and ≥ 95% of our input. Putting an Array Formula in a SUMPRODUCT so that we can use it in a normal formula, we get =SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$1:$D$6>=(B1*0.95))))
The High values are slightly harder, because we can't just multiply be 0 to cancel out anything too low, or over 105% of the target. We need to add a huge number like 1E+99 (a 1 with ninty-nine 0s after it) instead, so that the MIN will ignore them: SUMPRODUCT(MIN($D$1:$D$6+1E+99*(--($D$1:$D$6<B1)+--($D$1:$D$6>(B1*1.05)))))
The last steps are to decide which of these numbers is closer to the target, and then to find the Unique ID to match. The %closeness calculations are (TARGET - LOW)/TARGET and (HIGH - TARGET)/TARGET), and subtracting one from the other gives you 2-(HIGH + LOW)/TARGET - a Positive number means "High" is closer, a Negative number means that "Low" is closer, and 0 means they are both the same distance (I'll default this to the Low number). We then use SIGN to change it to ±1, add 2 to get 1,2 or 3 and finish up with CHOOSE to output our number. In pseudo-code, CHOOSE(2+SIGN(2-(HIGH+LOW)/TARGET),LOW,LOW,HIGH), and the full thing:
CHOOSE(2+SIGN(2-(SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95)))+SUMPRODUCT(MIN($D$1:$D$6+1E+99*(--($D$1:$D$6<B1)+--($D$1:$D$6>(B1*1.05))))))/B1),SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95))),SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95))),SUMPRODUCT(MIN($D$1:$D$6+1E+99*(--($D$1:$D$6<B1)+--($D$1:$D$6>(B1*1.05))))))
Now, we have a number. All we need to do is either use VLOOKUP, or use MATCH to get the row it is on, and INDEX to pull the data for that row:
Office 365:
=IFERROR(VLOOKUP(B1,$D$1:$E$6,2,FALSE),VLOOKUP(CHOOSE(2+SIGN(2-(MAXIFS($D$1:$D$6,$D$1:$D$6,"<="&B1,$D$1:$D$6,">="&(B1*0.95))+MINIFS($D$1:$D$6,$D$1:$D$6,">="&B1,$D$1:$D$6,"<="&(B1*1.05)))/B1),MAXIFS($D$1:$D$6,$D$1:$D$6,"<="&B1,$D$1:$D$6,">="&(B1*0.95)),MAXIFS($D$1:$D$6,$D$1:$D$6,"<="&B1,$D$1:$D$6,">="&(B1*0.95)),MINIFS($D$1:$D$6,$D$1:$D$6,">="&B1,$D$1:$D$6,"<="&(B1*1.05))),$D$1:E$7,2,FALSE))
Otherwise:
=IFERROR(VLOOKUP(B1,$D$1:$E$6,2,FALSE),VLOOKUP(CHOOSE(2+SIGN(2-(SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95)))+SUMPRODUCT(MIN($D$1:$D$6+1E+99*(--($D$1:$D$6<B1)+--($D$1:$D$6>(B1*1.05))))))/B1),SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95))),SUMPRODUCT(MAX($D$1:$D$6*--($D$1:$D$6<=B1)*--($D$6>B1*0.95))),SUMPRODUCT(MIN($D$1:$D$6+1E+99*(--($D$1:$D$6<B1)+--($D$1:$D$6>(B1*1.05)))))),$D$1:E$7,2,FALSE))
(Obviously, change $D$1:$D$6 and $D$1:$E$6 to your actual data table ranges, and B1 to the input-value range)
I am trying to use index match functions to determine the appropriate rate for the below table.
So for example a consumer loan that is for a person that owns property, the car is 2 years or less in age and the total loan to value ratio is less than 140% should return a value of 5.15%
I believe this is what you wanted...
I would use a series of nested if functions to evaluate which column of LTV I would want the value to come from.
"That is what is done in the AND( ) part. If the value is greater than the 110% and smaller than 140% let's do the Index Match on the 110% Column, Otherwise do it on the 140% Column."
You could extend this for more columns with more IFs in the false condition.
Then it is a simple INDEX match with concatenation. It searches for the three parameters all concatenated in a single range of concatenations.
Hope it helped.
Proof of Concept
In order to achieve the above I had to make a minor edit to your header to be able to distinguish between the two 140% columns.
The functions used in this answer are:
AGGREGATE function
MATCH function
INDEX function
ROW function
IFERROR function
I placed the main part of the formula inside the IFERROR function as a way of dealing with things that may be out of range or when not all the input have been provided. I then assumed that what you were basing your search on would be provided in a series of cells. In my example I assumed the questions would be asked in the range H3 to K3 and I place the results in L3.
The main concept is centered around the INDEX function. I specified the index range as being the height of your table and the width of the percentage rates. Or for this example D2:F9.
=IFERROR(INDEX($D$2:$F$9,row number, column number),"Not Found")
That is the easy part. That more challenging part is determining the row and column number to look in. Lets start with the column number as it is the slightly easier of the two. I assumed the ratio to look for, or rather the header of the column to look in would be supplied. I basically used this equation to determine the column number:
=MATCH(K3,$D$1:$F$1,0)
which in layman's terms is which column between D and F, counting column D as 1, has the value equal to the contents of K3. So now that there is a formula to determine the column, we can drop that into our original formula and wind up with:
=IFERROR(INDEX($D$2:$F$9,row number,MATCH(K3,$D$1:$F$1,0)),"Not Found")
Now we just need to determine the row number. This is the most complex operation. We are going to basically make a bunch of logical checks and take the first row that matches all the logical checks. The premise here is that a logical check is either TRUE or FALSE. In excel 0 is false an every other integer is TRUE. So if we multiply a series of logical checks together, only the one that is true in all cases will be equal to 1. The first logical check is the loan type. it will be followed by the living status and then the vehicle age.
=(H3=$A$2:$A$9)*(I3=$B$2:$B$9)*(J3=C2:C9)
now if you put that into an array formula you will get a series of true false or 1/0. We are going to use it inside an AGGREGATE function with a special feature. The AGGREGATE function will perform array like calculation for some of its functions. We are going to use function 15 which will do this. We are also going to tell the aggregate function to ignore all errors, which is what the 6 does. So in the end what we wind up doing is dividing each row number by the logical check. If the logical check is false or 0, it will generate a Div/0! error which aggregate will choose to ignore. In the end we wind up with a list of row which match our logical check. We then tell the aggregate that we want the first result with the ,1. so we wind up with a formula that looks like:
=AGGREGATE(15,6,ROW($A$2:$A$9)/((H3=$A$2:$A$9)*(I3=$B$2:$B$9)*(J3=C2:C9)),1)
While this does provide us with the row number we want, we need to adjust it to make it an index number. In order to do this you need to subtract the number of header rows. In this case 1. So the index row number is given by this formula:
=AGGREGATE(15,6,ROW($A$2:$A$9)/((H3=$A$2:$A$9)*(I3=$B$2:$B$9)*(J3=C2:C9)),1)-1
And when we substitute that back into the earlier equation for the row number, we wind up with the final equation of:
=IFERROR(INDEX($D$2:$F$9,AGGREGATE(15,6,ROW($A$2:$A$9)/((H3=$A$2:$A$9)*(I3=$B$2:$B$9)*(J3=C2:C9)),1)-1,MATCH(K3,$D$1:$F$1,0)),"Not Found")
hello first of all this is my code which returns the error
=INDEX(Steel_table!A3:A151,LOOKUP(10^10,MATCH(H7,Steel_table!C3:C151,{1,0})+{1,0}))
i have based this code from this thread:
Use INDEX MATCH to find greater than/equal to value
This is the scenario
i have 2 sheets namely, stress analysis and steel table
the value that i would like to compare is located in the stress analysis sheet cell H7 and i would like to compare it to the steel table sheet from cell's values from C3 until C151.
please help me as it always returned an #N/A error
Stress Analysis Sheet
Steel Table sheet
Many thanks
The use of third argument to MATCH functions must be considered depending on whether the lookup range is sorted or not.
The only time the range sort need not be taken into account is when using 0 - exact match. But exact match is not what you need since you are looking up calculated results against a table of prefixed values which won't likely be exactly matched...
So in order to use third argument of 1, the looking range should be sorted in ASCENDING order. In your case teh second value of the lookup range is larger than your lookup value, and so is the first, so Excel shortcuts to N/A, assuming that no smaller value will be found further down.
The reverse logic is true for using -1.
You can simply use the difference of the lookup values and the range to obtain the smallest one. So we use the MIN function. But you only want positive differences , so you'll need to transform all negative numbers into a ridiculously large number(such as 10^10), so that they are not used as potential minimum values:
INDEX(Steel_table!$A$3:$A$151, MATCH(MIN(IF(Steel_table!C3:C151-H7>=0, Steel_table!C3:C151-H7, 10^10)), Steel_table!C3:C151-H7,0))
This will give you the smallest value in Steel_table!C3:C151) that is equal or greater then the value in cel H7:
=MIN(IF((Steel_table!C3:C151)>=H7,(Steel_table!C3:C151)))
It is an array formula, so confirm with [Control-Shift-Enter]. The output will be 1451.61. It will return 0 when there is no equal or greater value.
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).