When I calculate the median of even numbers for e.g 1,2,3,4,5,6,7,8.. i want the return value to be 5 i.e. the higher value of the two middle values and not the average of 4 & 5. Please help
I don't have Excel, so I can't try it, but I think you should be able to accomplish this with a combination of the LARGE function the COUNT function, and the TRUNC function. For example, if the numbers you are working with are in cells A1 through A8, you should be able to find the answer you want, though technically it's not the median, with the formula
=LARGE(A1:A8,TRUNC(COUNT(A1:A8)/2))
Edit
=LARGE(A1:A8,TRUNC((1+COUNT(A1:A8))/2))
If you know you will always be working with an even number of entries, the call to TRUNC could be omitted.
With data in column A:
=IF(ISODD(COUNT(A:A)),MEDIAN(A:A),ROUNDUP(MEDIAN(A:A),0))
EDIT#1:
Consider the array formula:
=IF(ISODD(COUNT(A:A)),MEDIAN(A:A),MIN(IF(A:A>MEDIAN(A:A),A:A)))
If the number of values is odd, return the median. If the number of value is even, return the smallest value greater than the median.
Array formulas must be entered with Ctrl + Shift + Enter rather than just the Enter key.
This approach does not require the data to be sorted.
EDIT#2:
This array formula appears to handle Ron's case:
=IF(ISODD(COUNT(A:A)),MEDIAN(A:A),MIN(IF(A:A>=MEDIAN(A:A),A:A)))
It returns the smallest value greater than or equal to the median (for the even case)
But I don't know if this is what the Poster wants.
=LARGE(A1:A9,INT((COUNT(A1:A9)/2)+0.5))
you can use Roundafter the Median
in your case you can type: =ROUND(MEDIAN(B4:I4),0)
lets say that the given range is from B4 to I4 from 1 to 8.
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.
How to get this in excel
excel table
F column is the result column
For the following answer I am going to assume you only ever have two numbers in any row, but they can be in any cell along the row and they are always greater than 0.
If you just wish to find the difference between the two numbers without worrying about which number is bigger, a simple equation using maximum and minimum can be used, eg in Cell F1 you would have
=MAX(A1:E1)-MIN(A1:E1)
However, from your example, it seems more likely that you want to know the difference between the first number and the second number.
The difficulty here, is that the cells in columns B, C and D could contain either the first number, the second number, or no number! The solution is to use the following equation in Cell F1
=(MAX(A1:E1)-MIN(A1:E1))*IF(MAX(A1:E1)=INDEX(A1:E1,MATCH(0,A1:E1,-1)),-1,1)
This formula works as follows:
We still start off with the simple difference between the max and min, and then this is multiplied by 1 or -1 depending on which way around the numbers are.
MATCH(0,A1:E1,-1)
This part of the equation looks along the row for a 0, and assumes they are in descending order, so it will return the position of the second number.
This is then inserted into the INDEX function and checked to see if it is the same as the maximum number and the IF function returns either -1 or 1 as required.
Paste this formula on F1, then copy to F2 and F3
=INDEX(A1:E1,MATCH(TRUE,INDEX(A1:E1<>"",),0)) - LOOKUP(9.99E+307,A1:E1)
I am trying to get the max value of a column based on the Left function
What I am doing is the following :
These are the results I get when i write this into column C :
=MAX(LEFT(A:A, 2))
But what I truly want is to get in column C the max value of all column A not for each cell.
So the result should be in this case 90 for all rows.
What should be the formula here?
Just another option that gets entered normally:
=AGGREGATE(14,6,--LEFT($A$1:INDEX(A:A,MATCH("ZZZ",A:A)),2),1)
Array formulas will calculate the enitre referenced array. So care should be taken to limit the number of iterations to only the data set.
The $A$1:INDEX(A:A,MATCH("ZZZ",A:A)) part of the formula does that. It finds the last cell in column A with data in it and sets that as the upper bound. So in this instance the reference range is A1:A3. But, it will grow dynamically as data in Column A is added, so no need to change the formula each time data is added.
Update 2
Here is another solution which I think is better than my original (below)
=INT(SUMPRODUCT(MAX(SUBSTITUTE(A:A,"-",".")*1)))
it can be entered as normal (just Enter)
Orignal Answer
You need numbers and arrays
=MAX(IFERROR(LEFT(A:A,2)*1,0))
Let's break this down. Multiplying by turns your strings into numbers - since Left only returns a string
LEFT(A:A,2)*1
Unfortunately this method returns #Value if you multiply an empty string by 1. You will definitely have some empty strings in the range A:A so we wrap the whole thing with an IFERROR function.
IFERROR(LEFT(A:A,2)*1,0)
Now we still need excel to treat this as an array (i.e. a whole column of numbers, rather than just one number). So we put the MAX formula in and enter it with Ctrl+Shift+Enter rather than just Enter. The result is that the formula looks like this in the formula bar
{=MAX(IFERROR(LEFT(A:A,2)*1,0))}
which would return 90 in your example, as required
Update 1
If you are using Excel 2013 or later, you can also use the NUMBERVALUE function
=MAX(NUMBERVALUE(LEFT(A:A,2)))
again, enter it with Ctrl+Shift+Enter
How do I average a list of numbers whose values are greater than 0? I know I can use AVERAGEIF function in Excel
My data is located in A2, A5, A6, A10, A17.
I only want to average it if the data is greater than 0.
Since my data is not an range, I am not able to use AVERAGEIF Function range.
Need some help on this.
EDIT
For example,
I tried with three numbers:
1) 98.068 and 98.954 and 0 so my forumla looked like this:
=AVERAGE(IF(N(OFFSET(A2,{0,5,10},))>0,N(OFFSET(A2,{0,5,10},))))
The answer came out as 99.106. Not sure why.
A few options:
1)=SUM(SUMIF(INDIRECT({"A2","A5","A6","A10","A17"}),">0"))/SUM(COUNTIF(INDIRECT({"A2","A5","A6","A10","A17"}),">0"))
2)=AVERAGE(IF(N(INDIRECT({"A2","A5","A6","A10","A17"}))>0,N(INDIRECT({"A2","A5","A6","A10","A17"}))))
3)
=AVERAGE(IF(N(OFFSET(A2,{0,3,4,8,15},))>0,N(OFFSET(A2,{0,3,4,8,15},))))
2) and 3) must be committed as array formulas**
Regards
(0) A simple method
=SUM(A2*(A2>0),A5*(A5>0),A6*(A6>0),A10*(A10>0),A17*(A17>0))/SUM(A2>0,A5>0,A6>0,A10>0,A17>0)
(4) A more general method
=SUM((A1:A20>0)*A1:A20*(ADDRESS(ROW(A1:A20),1,4)={"A2","A5","A6","A10","A17"}))/
SUM((A1:A20>0)*(ADDRESS(ROW(A1:A20),1,4)={"A2","A5","A6","A10","A17"}))
The second one is an array formula and must be entered with CtrlShiftEnter
If it's possible to have text in the cells rather than numbers, then this should replace the first formula:-
=SUM(N(A2)*(A2>0),N(A5)*(A5>0),N(A6)*(A6>0),N(A10)*(A10>0),N(A17)*(A17>0))/SUM(N(A2)>0,N(A5)>0,N(A6)>0,N(A10)>0,N(A17)>0)
(I haven't used N in the > brackets in the numerator because I reason that if A2 etc. is text, the product will always be zero)
I can't persuade N to work with arrays in the second formula, so at the moment I have the rather lengthy
=SUM((IF(ISNUMBER(A1:A20),A1:A20,0)>0)*IF(ISNUMBER(A1:A20),A1:A20,0)*(ADDRESS(ROW(A1:A20),1,4)={"A2","A5","A6","A10","A17"}))/
SUM((IF(ISNUMBER(A1:A20),A1:A20,0)>0)*(ADDRESS(ROW(A1:A20),1,4)={"A2","A5","A6","A10","A17"}))
but I have tested it on text values and negative numbers and it does seem fine.
The only exception is if one of the cells contains TRUE. In this case the first formula will count it as 1, the second formula will ignore it.
I'm using Excel 2010 and I'm looking for a way to return the first negative number of a column. For instance, I have the following numbers distributed in a column:
1
4
6
-3
4
-1
-10
8
Which function could I use to return -3?
Thanks!
This could be interpreted two ways... If all the numbers are in a single cell (one column) as a string, the MID function can be used. If the numbers are in A1, a formula that could work is this:
=VALUE(MID(A1,SEARCH("-",A1),SEARCH(" ",A1,SEARCH("-",A1))-SEARCH("-",A1)))
If the numbers are each in their own columns (in my example, A3:H3), a different technique must be used:
{=INDEX(A3:H3,1,MATCH(TRUE,A3:H3<0,0))}
Don't type the { } - enter the equation using CTRL+SHIFT+ENTER.
In each case, the formula will return the number -3, which is the first negative number in the series.
Another possibility, avoiding the array formula (which are a big source of performance issues):
=LOOKUP(1;1/(M2:M15<0);M2:M15)
(I assume your numbers are in the M2:M15 range).
This will return the first number matching the "<0" condition. You may use any other condition, including text comparisons.
You may also extract the value of another array corresponding to the matching cell:
=LOOKUP(1;1/(M2:M15<>"OK");T2:T15)
In this example, the first cell containing another string than "OK" will be searched for in the m2:m15 array and the corresponding value in array t2:t15 will be returned.
Please note that the usage of the lookup function should be avoided whenever possible (but in this case, it's very handy !)
(I got the original inspiration for this answer from this post)