I have a set of data with 3 column (example shown in the figure). The second column (height) is starting from 1 and increased to ~80 and then starting from 1 again. I need to select the row with the second column most close to 50. As shown in the figure below, I will need to select 3 rows. Can any one suggest how to find out those rows in excel?
Many thanks!
=match(50, B1:B80, 1)
will find the element that has the largest value less than or equal to 50 in that range (choose your ranges accordingly). I believe B1:B80 must be in ascending order.
Related
I have this (example):
Luffy 320
Coby 350
Zoro 180
Now I want to show the max from this info, with number and text (in seperate cells) like this:
col 1 col 2 col 3
1st 350 Coby
2nd 320 Luffy
3rd 180 Zoro
The 2nd Column no problem with the MAX() formula.
For the 3rd column to get the text I've tried the MAX(...) and INDEX(...) formulas but nothings working ...
Can anyone help me?
You first need to get which value is the largest, second largest and so on.
You can use the function LARGE(range, n) for this.
So in your col 2 use this formula:
=LARGE(B:B,1)
=LARGE(B:B,2)
=LARGE(B:B,3)
Assuming B is the column with the values.
Then we need to match this value and get the name
=INDEX(A:A,MATCH("the above calculated cell",B:B,0))
With the above calculated cell I mean the LARGE function cell. And assuming column A is the column with the names.
This should give you a dynamic table that will update when values or names change.
I'm not sure how you manage to get that column 2 using MAX formula since it only outputs the largest number of the inputs and thus can't output 2nd and 3rd position.
I've got a table in excel with 10 rows and 10 columns.
The table contains 100 different values between 1 and 3.
I want to find the highest sum of 10 values using only 1 value from each row and 1 from each column.
Do u guys know a function that finds the highest sum? - I've tried to do i manually, but there are to many combinations!
Hope it makes sense.
Thanks in advance:)
My solution builds on what I wrote in the comment, i.e. you first take the maximum value in the 10x10 array, then the maximum in the 9x9 array (excluding the row/column of the first maximum), etc. My solution tries not to do everything in one formula, but I add a few helper columns, and a bit more helper rows (it is fast and dirty, but it works and is easily audited/understandable). You always can do this on a separate worksheet which you could hide if needed.
The screenshot above goes from cell A1 till Y31.
The key formulas:
3.55 is the result of =MAX(B2:K11)
The first gray cell is =IFNA(MATCH($M12;B2:B11;0);""), and you drag this 9 cells to the left. This tries to find a match with the max result in each column of the table;
The 10 left of the 3.55 is =MATCH(TRUE;INDEX(ISNUMBER(P12:Y12);0);0) , and gives the column number of the max value.
The 2 next to the 10 is =INDEX(P12:Y12;N12) and gives the row number of the max value.
The 1 in cell B12 is =IF(OR(B$1=$N12;$A12=$O12);0;1), and creates a 10x10 matrix with a row and column with zeroes where the previous max value was found.
Then you multiply this with the preceding matrix and create a new 10x10 matrix below (enter {=B2:K11*B12:K21} array formula (ctrl+shift+enter) in B22-K31
You then copy/paste rows 12 till 31 9 times below
The 23.02 is the total sum =SUM($M$12:$M$211) from all 10 maximum values and is the result you are looking for. The 10 is just a check with =COUNT($M$12:$M$211)
I have 7 columns to choose from and I need to pick 4 of those columns and generate a total for each row. I also need every combination of 4, which means I'll have 35 new columns with the totals for each of those combinations showing in each row. I need the code for this and if it can be done only using Excel. Here is an image of the columns and the grayed ones are the 7 columns I'm talking about. My knowledge of Excel is very limited. There are over 1,500 rows if that matters.
multi step approach that is going to use some helper rows. there may be a more elegant formula that will do this, and much slicker options in VBA, but this is a formula only approach.
Step 1 - Generate List of Column Combination
To generate the list 4 helper rows will need to be insert at the top of your data. either above or below you header row. These 4 rows will represent the column number you are going to pick. To keep the math simpler for me I just assumed the 1 for the first column and 7 for the last column. those numbers will get converted to later to account for column in between in your spreadsheet. For the sake of this example The first combination sum will occur in column AO and the first helper row will be row 1. The first combination will be hard coded and it will seed the pattern for the remainder of column combinations. Enter the following values in the corresponding cells:
AO1 = 1
AO2 = 2
AO3 = 3
AO4 = 4
In the adjacent column a formula will be placed and copied to the right. It will automatically augment the bottom value by 1 until it hits its maximum value at which point the value in the row above will increase by 1 and the the value of the current will be 1 more than the cell above. This will produce a pattern that covers all 35 combinations by the time column BW is reached. Place the formulas below in the appropriate cell and copy to the right:
AP1
=IF(AO2=5,AO1+1,AO1)
AP2
=IF(AO2=5,AP1+1,IF(AO3=6,AO2+1,AO2))
AP3
=IF(AO3=6,AP2+1,IF(AO4=7,AO3+1,AO3))
AP4
=IF(AO4=7,AP3+1,AO4+1)
Step2 - Sum The Appropriate Columns
I was hoping to use a some sort of array type operation to read through the column reference numbers above, but I could not get my head around it. Since it was just 4 entries to worry about I simply added each reference manually in a SUM function. Now the important thing to note is that we will be using the INDEX function over the 13 columns that cover the range of your columns so to convert the index number we figured out above, to something that will work to grab every second row, the number that was calculated will be multiplied by 2 and then 1 will be subtracted. That means 1,2,3,4 for the first column combination becomes 1,3,5,7. You can see this in the following formula. Place the following formula in the appropriate cell and copy down and to the right as needed.
AO5
=INDEX($AB5:$AN5,AO$1*2-1)+INDEX($AB5:$AN5,AO$2*2-1)+INDEX($AB5:$AN5,AO$3*2-1)+INDEX($AB5:$AN5,AO$4*2-1)
pay careful attention to the $ which will lock row or column reference and prevent them from changing as the formula is copied.
Now you may need to adjust the cell references to match your sheet.
how can I merge two columns of data into one like the following:
Col1 Col2 Col3
========================
A 1 A
B 2 1
C 3 B
2
C
3
You can use the following formula in column D as per my example. Keep in mind to increase the $A$1:$B$6 range according to your data.
=INDEX($A$1:$B$6,INT((ROWS(D$2:D2)-1)/2)+1,MOD(ROWS(D$2:D2)-1,2)+1)
Result:
Thank you to #Koby Douek for the answer. Just an addition--if you are using Open Office Calc, you replace the commas with semi-colons.
=INDEX($A$1:$B$6;INT((ROWS(D$2:D2)-1)/2)+1;MOD(ROWS(D$2:D2)-1;2)+1)
Expanding #koby Douek's answer to more columns and explaining some of the terms
Original Code for 2 columns to 1 alternating
=INDEX($A$1:$B$6,INT((ROWS(D$2:D2)-1)/2)+1,MOD(ROWS(D$2:D2)-1,2)+1)
$A$1:$B$6 Defines the columns and rows to source the final set of data from, the $s are only present to keep the formula from changing the columns and rows selects if it is copied and pasted or dragged.
To extend to work on any values you dump into the columns instead of having to expand the range every time it should be amended to $A:$B or A:B so you can easily copy it to other sets of columns and create new merges, but it will also give the 1st value in every column as one of the alternating values so if you instead have headers you would be able to do this by instead using a large number so $A$1:$B$99999 or A$1:B$99999 if you want to past and move the columns ymmv which is better by situation.
lets assume you are fine including the values in the 1st row
This changes the formula to
=INDEX($A:$B,INT((ROWS(D$2:D2)-1)/2)+1,MOD(ROWS(D$2:D2)-1,2)+1)
Now on to D$2:D2
This is the row that is being used to calculate the difference between the current row the formula is in (D2) and the reference row (D$2) The important thing to make sure you do is to set the reference row number to the 1st row you will be putting values in, so if your 1st row is a header in the sort column you will use the 2nd row as the reference, if your values in the combined column D begin on the 3rd row then the reference row would be D$3
Since I like the more general form where the 1st row isn't a header row I'll use D$1:D1 but you could still mix source rows without headers into a combined row with a header of as many rows as you like just by incrementing that reference row number to be the 1st row where your values should begin.
This changes the formula to
=INDEX($A:$B,INT((ROWS(D$1:D1)-1)/2)+1,MOD(ROWS(D$1:D1)-1,2)+1)
Now INT((ROWS(D$1:D1)-1)/2)+1 and MOD(ROWS(D$1:D1)-1,2)+1
INT returns an integer value so any decimal places are dropped, it essentially functions like rounding down to the nearest whole number
MOD functions by returning the remainder of a division, it's result will be a whole number between 0 and n-1 where n is the number we are dividing by. (eg: 0/3=0; 1/3=1; 2/3=2; 3/3=0; 4/3=1 ... etc)
So -1)/2)+1 and -1,2)+1
the first value is again the difference between the current row and the reference row. but D$1:D1 is going to be the count of the rows, which is 1 so we have to correct for the rows count starting at 1 instead of 0 which would throw off our calculations, so both are using the -1 to reduce the count of the rows by 1
in the case of /2 and ,2 both are because we are dividing by 2 in the first statement it's a normal division by 2 /2 in the modulus statement it's an argument of the Mod function so ,2
finally we need to add 1 using +1 to correct for the index's need to have a value series which begins at 1.
INT((ROWS(D$2:D2)-1)/2)+1 is finding the row number to select the value from.
MOD(ROWS(D$1:D1)-1,2)+1 is finding the column number to select the value from
Thus we can change /2 and ,2 to /3 and ,3 to do this with 3 columns
This yields:
=INDEX($A:$B,INT((ROWS(D$1:D1)-1)/3)+1,MOD(ROWS(D$1:D1)-1,3)+1)
So maybe that's the confusing way to look at it but it's closer to how my mind works on it. Here is an alternative view:
=INDEX([RANGE],[ROW_#],[COLUMN_#]) returns the value from a range of rows and columns
Using the example:
=INDEX($A:$B,INT((ROWS(D$1:D1)-1)/3)+1,MOD(ROWS(D$1:D1)-1,3)+1)
[RANGE] = $A:$B this is the range of source columns.
[ROW_#] = INT((ROWS(D$1:D1)-1)/3)+1
INT([VALUE_A])+1 returns an integer value so any decimal places are dropped. Then adds one to it. we add one to the value because the result of the next steps will be 1 less than the value we need.
[Value_A] = (ROWS(D$1:D1)-1)/3
ROWS(D$1:D1) returns the number of rows in the Range to the current row in the results column, we use D$1 to designate the row number where the values in the results column begin. D1 is the current row in the results column giving us a range from the source row, allowing us to count the rows. we have to subtract 1 from this value using -1 to get the difference between the source and current. This is then divided by /3 because we have three columns we want to look through in this example so we only change rows when the result is divisible by 3. the INT drops any decimal places as mentioned so it only increments when cleanly divisible by 3.
[COLUMN_#] = MOD(ROWS(D$1:D1)-1,3)+1
MOD([VALUE],[Divisor])+1 returns the remainder of the value when divided by the divisor.
Using the example:
MOD(ROWS(D$1:D1)-1,3)+1
In this case we still divide by 3 but it's an argument to the MOD function, we still need to count the number of rows and subtract 1 before dividing it, this will return a 0, 1, or 2 for the column, but as above we are shifted backwards by 1 as the column numbers begin with the number 1, so as before we must add 1
And here we add column A and D
two different formulas depending on if you add the formula to an odd row or an even row.
https://1drv.ms/x/s!AncAhUkdErOkguUaToQkVkl5Qw-l_g?e=5d9gVM
Odd Start row
=INDEX($A$2:$D$9;ROUND(ROW(A1)/2;0);IF(MOD(ROW()-ROW($A$2);2)=1;4;1))
Even Start row
=INDEX($A$2:$D$9;ROUND(ROW(A1)/2;0);IF(MOD(ROW()-ROW($A$1);2)=1;4;1))
What is A1 in the picture is the cell directly above your first data cell.
If you want to place it on a different sheet you just add the sheet name:
=INDEX(MySheet!$A$2:$D$9;ROUND(ROW(MySheet!A1)/2;0);IF(MOD(ROW()-ROW(MySheet!$A$2);2)=1;4;1))
=INDEX(MySheet!$A$2:$D$9;ROUND(ROW(MySheet!A1)/2;0);IF(MOD(ROW()-ROW(MySheet!$A$1);2)=1;4;1))
I have searched through many similar topics but could find nothing that will do what I need.
I am trying to create a worksheet that will track scores for a darts game.
On Sheet 1 I have two columns that simply tracks each players throws from 501 down to 0
Row 25 is the amount remaining for each player.
In Sheet 2 I have 2 columns. The Column A contains scores that you can check out on, and Column B contains the checkout e.g. (T20, T20, D18). So if the value in row 25 of Sheet 1 matches any of the values in Column A of sheet 2, the I want to display the Value of Column B in the matching row on Sheet 2 Underneath the remaining score on Sheet 1.
Can anyone point me in the right direction?
not sure what you mean exactly, but this formula in row 26 should do the trick:
=index('Sheet 2'!$B:$B;match(A25;'Sheet 2'!$A:$A;0))
if your list separator is comma ,, use that instead of semicolon ;
you might want to use 1 as the third argument of match function, if you want to display the checkout according to the nearest match that is bigger than the number in row 25 and the column A in Sheet 2 is sorted in ascending order (1-9)
or -1 if you want the nearest match that is smaller and column A is sorted in descending order (9-1)
You can use this:
=IFERROR(VLOOKUP(A4, Sheet2!$C$2:$E$65535, 3, FALSE),0)