I have a set of values as following in a row:
(9,888,000) (88,410,205) (76,030,786) (62,712,494) (48,416,610) (33,102,893) (16,729,517) 746,979 19,371,753 39,191,722 43,755,624 66,114,081 89,819,671 114,926,989 141,492,724
Each value represents some amount in a specific year. e.g.(16,729,517) is the cash outflow in year 7.
Now I've an amount of 110,000,000 as the initial payment in another cell.
Now i try to calculate payback year of initial payment. For me payback period should be: 7.96 as in that year amount becomes positive.
But I'm unable to calculate this 7.96 value. can anyone please help?
i've done it via indirect method:
as first step I check value in each column and store true or false against positive and negative values.
2nd Step: =MATCH(TRUE,Complete Array,0). This gives me 8 as Digit, where it finds first positive value.
Then, as third step, i use this formula:
=Year-OFFSET(CFC,0,Year)/OFFSET(Investment,0,Year) .
in this case Year is what i got in step 2. and CFC cell is empty (Cumulative Cash Flow in 0 year). Out put of this 7.96, which is my required value.
But this is a lengthy and indirect way. I want to have a one line formula for all these calculations.
You can combine your formula into a single line. Here's a slight variation on what you are doing, which is a straight line extrapolation between the values below and above zero:
=FORECAST(0,{-1,0},
OFFSET(CashFlows,,MATCH(TRUE,CashFlows>0,0)-2,1,2))
+MATCH(TRUE,CashFlows>0,0)
You may need to change the offset formula if your CashFlows range is in a column instead of a row. The formula is entered as an array formula by holding down ctrl+shift while hitting enter
EDIT This formula ignores the very first cash flow whether positive or negative. If you want to include it, change the > operator to >= and test for an error.
Your formula approach seems natural. I am not sure that you can do much better then what you are doing, although I'm sure that an array-formula guru could find a way to wrap it all into a single, albeit not-quite readable formula.
Using VBA you can write a function which can be called directly from the spreadsheet. In a standard code module put:
Function PayBack(R As Range) As Variant
'R is a range, assumed to be 1-dimensional
'consisting of negative numbers which
'at some point transition to positive
'returns the cross-over point
Dim i As Long, n As Long
Dim x As Double, y As Double
n = R.Cells.Count
x = R.Cells(1).Value
If x >= 0 Then
PayBack = CVErr(xlErrNA)
Exit Function
End If
'x < 0 in the following loop:
For i = 2 To n
y = R.Cells(i).Value
If y >= 0 Then
x = Abs(x)
PayBack = i - y / (x + y)
Exit Function
End If
x = y
Next i
'if the code reaches here -- still negative, so return error
PayBack = CVErr(xlErrNA)
End Function
Then it could be used like:
I have your sample data in row 1 (so A1:O1) and in A3 I just entered the formula
=PayBack(A1:O1)
and it computes as expected. It returns #N/A error if there is no negative to positive transition in the data -- though that can of course be tweaked in various ways.
Related
I have a polynomial equation that i want to solve: L^3-4043L-60647=0 using goal seek in the vba.
This equation gives 3 roots : L1=70.06, L2, -54.04 and L3=-16.02 according to my calculator. But i only want my L in my excel cell to show the first positive root as my answer.
However when i do the goalseek using vba, it only gives me -16.02. How do i tell in my code to only solve for positive value?
I already tried using Do until and if statement. However Do until statement kept crashing and If statement is giving me wrong values.
Sub GoalSeek()
'GoalSeek Macro
Dim Length As Double
Dim i As Long
Range("Length") = i
If i > 0 Then
Application.CutCopyMode = False
Application.CutCopyMode = False
Range("GS").GoalSeek Goal:=0.1, ChangingCell:=Range("Length")
Else
End If
End Sub
I tried using this if statement. However my L or "Length" comes up only to be 0. I am very very beginner level in VBA. I don't know what i am doing wrong.
GoalSeek gets the nearest solutions to the starting value.
You can use the following code:
Sub GoalSeek()
Dim i As Double
'Set the initial value to a very high number
Range("Result").Value = 9999
'Ask GoalSeek to get the neares solution to that high value
Range("Formula").GoalSeek Goal:=0, ChangingCell:=Range("Result")
If Range("Result").Value > 0 Then
'If the value is positive, we need to make sure that it is the first positive solution
i = -1
Do
i = i + 1
'Set a new inital value. This time, a small one (starting from 0)
Range("Result").Value = i
'Ask GoalSeek to get the neares solution to the small initial value
Range("Formula").GoalSeek Goal:=0, ChangingCell:=Range("Result")
'If the result is negative, loop (increase the initial value and try again till you find the first positive one
Loop While Range("Result").Value < 0
Else 'If the nearest result to the high value is negative, keep it & show a message box.
MsgBox "No +ve solution found"
End If
End Sub
In your example, you have three solutions 70.06, -54.04 & -16.02
The nearest to 0 is -16.02, to 9999 is 70.6 and to -9999 is -54.04
What if the solutions are -5, 7 & 12?
The nearest to 9999 is 12, but you want 7, right?
So we ask for the nearest to 0 (-5) then, we keep increasing the initial value till the nearest solution becomes 7.
Please note that this assumes that you have an idea about what the results would be.
For example, if the solutions are -1 & 1,000,000, this code will not work because -1 is nearer to 9999 than 1,000,000.
In this case, you will need to change the initial high value more.
AND if you set it to a too high value that exceeds the limit of double data type 1.79E+308 or even to a value that makes the result of the formula exceed it, you will get an error.
Please take a look at the attached image. I have a long list of items and I've created a common keywords to search in that list. I'm using this formula:
=INDEX(A:A,MATCH((("*"&B2&"*")&("*"&C2&"*")&("*"&D2&"*")&("*"&E2&"*")&("*"&F2&"*")),A:A,0))
The problem that the search is going through the same sequence that I entered.
It gives error if the sequence of the words in the cell is different than the sequence in my formula which make sense.
Is there a way I can search for 3 or more words that are existing in any cell in any sequence?
I am open to using VBA if necessary.
My search results:
Here is the user defined function:
Public Function indexMX(rng As Range, pat1 As Range, pat2 As Range, pat3 As Range, pat4 As Range, pat5 As Range) As Variant
Dim r As Range, rngx As Range, s(1 To 5) As String, Kount As Long, j As Long
s(1) = pat1.Value
s(2) = pat2.Value
s(3) = pat3.Value
s(4) = pat4.Value
s(5) = pat5.Value
Set rngx = Intersect(rng, rng.Parent.UsedRange)
For Each r In rngx
v = r.Value
Kount = 0
For j = 1 To 5
If InStr(1, v, s(j)) > 0 Or s(j) = "" Then Kount = Kount + 1
Next j
If Kount = 5 Then
indexMX = v
Exit Function
End If
Next r
indexMX = "no luck"
End Function
Here is an example of its usage:
As you see, we give the UDF() the address of the column and the addresses of the five keywords and the UDF() finds the first item containing all five words.
If a keyword is blank, it is not used. (so if you want to search for only two keywords, leave the other three blank). If no matches are found the phrase no luck is returned.
User Defined Functions (UDFs) are very easy to install and use:
ALT-F11 brings up the VBE window
ALT-I
ALT-M opens a fresh module
paste the stuff in and close the VBE window
If you save the workbook, the UDF will be saved with it.
If you are using a version of Excel later then 2003, you must save
the file as .xlsm rather than .xlsx
To remove the UDF:
bring up the VBE window as above
clear the code out
close the VBE window
To use the UDF from Excel:
=myfunction(A1)
To learn more about macros in general, see:
http://www.mvps.org/dmcritchie/excel/getstarted.htm
and
http://msdn.microsoft.com/en-us/library/ee814735(v=office.14).aspx
and for specifics on UDFs, see:
http://www.cpearson.com/excel/WritingFunctionsInVBA.aspx
Macros must be enabled for this to work!
EDIT#1:
to remove case sensitivity, replace:
If InStr(1, v, s(j)) > 0 Or s(j) = "" Then Kount = Kount + 1
with:
If InStr(1, LCase(v), LCase(s(j))) > 0 Or s(j) = "" Then Kount = Kount + 1
Yes, it is possible for a single cell to return three matching words from a different cell. The answer in this example uses a formula to return 6 matches. VBA and special array functions are not used.
This is the formula:
=SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUMPRODUCT((IFERROR(SEARCH(FIRST,TARGET),0)>0)*100000+(IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+(IFERROR(SEARCH(THIRD,TARGET),0)>0)*3000+(IFERROR(SEARCH(FOURTH,TARGET),0)>0)*400+(IFERROR(SEARCH(FIFTH,TARGET),0)>0)*50+(IFERROR(SEARCH(SIXTH,TARGET),0)>0)*6),"1",FIRST&DELIM),"2",SECOND&DELIM),"3",THIRD&DELIM),"4",FOURTH&DELIM),"5",FIFTH&DELIM),"6",SIXTH&DELIM),"0","")
Did you notice the named ranges? FIRST, SECOND, THIRD, etc are individual cells and each one holds a word. We are trying to find those words inside TARGET. If we find the words, then we will write them in this cell holding this formula and each word will be separated by DELIM
The ranges are optional. In the picture below, you'll see cell A2 contains the word "named". This is the first of the six words we're tying to find and it can be expressed as FIRST = "A2" = "named" Inside the formula you'll see that FIRST appears twice. You could replace it with "named" and the cell A2 would become blank but the functionality of the formula would not change.
Even TARGET is optional. It could be written as E1 or typed out word for word.
I don't know why anyone would do that... but it is possible.
DELIM is at cell B2, it is a double space
Now to explain how it works
SEARCH(search for what?, search where?) This is responsible for determining if a match exists or not. If you understand what the named ranges then you've already figured out the syntax is. The location of first letter that of match in TARGET is returned. In this formula it is always 1 If it is not found then the number is 0
IFERROR(value,value) tries to perform the operation. If successful then the result is displayed. If there's an error the second result is displayed. Every IFERROR in this formula is practically the same: IFERROR(SEARCH(FIRST,TARGET),0) It searches inside TARGET trying to find the FIRST word. Result if found is 1 and if not found is 0
It gets a little more complicated from here so lets recap. We're calling SEARCH 6 times. Once for each word we want to find and we're always looking in TARGET. Result will be a 1 if match is found or a 0 if not. Ironically, us humans can put it together and see the match but the formula can't determine which words have been matched without more information
SUMPRODUCT takes the sum (addition) of the product (multiplication) of two or more arrays.
multiply two arrays to get the product
a, b, c * e, f, g = ae, bf, cg
takethe sum of the product to get the SUMPRODUCT
ae + bf + cg`
This is easiest when thought of a price and quantity. If one array is the price of a group of items and the other is quantity of the same group of items, then multiplying the two arrays will create a new array where each element is the cost to buy all items of that type in the group the total, and adding all those numbers give you the total cost you'd pay for all of the items
Here we multiply two arrays:
Qty Price
12.0 0.3
70.0 0.1
20.0 0.4
Multiply them to get the product:
Qty Price Total
12.0 0.3 3.8
70.0 0.1 7.0
20.0 0.4 8.0
Take the sum of the product:
Qty Price Total
12.0 0.3 3.8
70.0 0.1 7.0
20.0 0.4 8.0
18.8 SUMPRODUCT
Lets look at part of the formula:
SUMPRODUCT((IFERROR(SEARCH(FIRST,TARGET),0)>0)*100000+IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+...
It is easy to see this segment is looking for two words. We know SUMPRODUCT want's to multiply and add arrays. If you're thinking (IFERROR(SEARCH(FIRST,TARGET),0)>0) is an array, you'd be right! It's not an array in the technical sense of the word, but it does evaluate into a single value which can thought of as a 1x1 array, or, a cell. The sharp eyed and quick witted, may have noticed there is something on this array we have mentioned. It's the inequality at the end! Many of you know that you can take numerical values and turn them into boolean by testing them with an inequality. So lets evaluate.... SEARCH for FIRST inside TARGET = 1 because FIRST = "named" which is inside TARGET waaaaaaay in the back and because it wasn't an error we get to keep the 1. Next we do the inequality 1 > 0 = TRUE One is greater than zero and evaluates to TRUE
This is what we have right now
SUMPRODUCT((TRUE*100000+IFERROR(SEARCH(SECOND,TARGET),0)>0)*20000+....
Can you identify the arrays now? We know TRUE is an array, a 1x1. You know the IFERROR bit all the way to the inequality is also an array. Lets evaluate that IFERROR .... Mathematically we should still be working from left to right but trust me, we're ok if we let it slide this once.
IFERROR(SEARCH(SECOND,TARGET),0)>0) SECOND = "array" = 1 = TRUE
Did you follow my short hand? It's ok if you didn't just back up and practice on FIRST until you understand.
Plugging in the value gives us something like this
SUMPRODUCT(TRUE*100000+TRUE*20000+...
SUMPRODUCT is the SUM(addition) of the PRODUCT(multiplication)
So we're adding the stuff we multiply
SUMPRODUCT = (TRUE * 100000) + (TRUE * 20000)
Remember how easy it was to go from 1 > 0 to TRUE. We're "getting all up into that boolean TRUE that" equals 1 Here's a fun fact, -1 is also equal to TRUE. If you've ever seen a formula with a double negative in it like this STUFF(--(MORESTUFF( that's just some Excel wizard person making sure they get a +1 instead of a -1 ... ok, so lets get back on track and evalute
SUMPRODCUT = 1 * 100000 + 1 * 20000+....
SUMPRODCUT = 100000 + 20000+....
SUMPRODCUT = 120000+.....
I know you've been asking about those numbers. One hundred thousand? What's one hundred thousand for? I've been purposefully ignoring until it became convenient to talk about it. And now it's convenient. Go look at the whoooole formula and you'll find a pattern. Those numbers are in a decreasing sequence. Anyone who's ever done bitwise logic can see where this is going.I'm running short on time so I'll cut to the chase. Assume a hypothetical situation where every word was matched. You'd end up with
SUMPRODUCT = 100000 + 20000 + 3000 + 400 + 50 + 6
SUMPRODCUT = 123456
123456 are you pulling my leg? No, I am not. We're almost done so if you're still with me then you're gonna drive it home.
We have a large group of SUBSTITUTE teachers at the front of the line and we gotta get rid of them.
We also have this to contend with :"1",FIRST&DELIM),"2",SECOND&DELIM),"3",THIRD&DELIM),"4",FOURTH&DELIM),"5",FIFTH&DELIM),"6",SIXTH&DELIM),"0","")
Thankfully for us, they are part of the same problem. We worked our way from the middle out.
SUBSTITUTE(SUBSTITUTE(text, old text, new text)
SUBSTITUTE(SUBSTITUTE("123456","1", FIRST & DELIM),"2",SECOND & DELIM)....
Remember at top DELIM was pointing to a cell holding a double space. Each DELIM can be replaced with " " or any other delimiter you want.
SUBSTITUTE(SUBSTITUTE("123456","1", "named" & " "),"2",SECOND & DELIM)...
SUBSTITUTE("named 23456","2",SECOND & DELIM)...
SUBSTITUTE("named 23456","2","array"& " ")...
("named array 3456")... and so on.
Any questions?
Ok, class is dismissed!
I want to do an INDEX-MATCH-like lookup between two documents, except my MATCH's index array doesn't stay in one column.
In Vague-English: I want a value from a known column that matches another value that may be found in any column.
Refer to the image below. Let's call everything to the left of the bold vertical line on column H doc1, and the right side will be doc2.
Doc2 has a column "Find This", which will be the INDEX's array. It is compared with "ID1" from doc1 (Note that the values in "Find This" will not be in the same order as column ID1, but it's easier to undertsand this way).
The "[Result]" column in doc2 will be the value from doc1's "Want This" column from the row that matches "FIND THIS" ...However, sometimes the value from "FIND THIS" is not in the "ID1" column, and is instead in "ID2","ID3", etc.
So, I'm trying to generate Col K from Col J. This would be like pressing Ctrl+F and searching for a value in Col J, then taking the value from Col D in that row and copying it to Col K.
I made identical values from a column the same color in the other doc to make it easier to visualize where they are coming from.
Note also that in column F of doc1, the same value from doc2's "Find This" can be found after some other text.
Also note that the column headers are only there as examples, the ID columns are not actually numbered.
I would simply hard-code the correct column to search from, but I'm not in control of doc1, and I'm worried that future versions may have new "ID" columns, with other's being removed.
I'd prefer this to be a solution in the form of a formula, but VB will do.
To generate column K based on given values of column J then you could use the following:
=INDEX(doc1!$D$2:$D$14,SUMPRODUCT((doc1!$B$2:$H$14=J2)*ROW(doc1!$B$2:$H$14))-1)
Copy that formula down as far as you need to go.
It basically only returns the row of the where a matching column J is found. we then find that row in the index of your D range to get your value in K.
Proof of concept:
UPDATE:
If you are working with non unique entities n column J. That is the value on its own can be found in multiple rows and columns. Consider using the following to return the Last row where there J value is found:
=INDEX(doc1!$D$2:$D$14,AGGREGATE(14,6,(doc1!$B$2:$H$14=J2)*ROW(doc1!$B$2:$H$14),1)-1)
UPDATE 2:
And to return the first row where what you are looking in column J is found use:
=INDEX($D$2:$D$14,AGGREGATE(15,6,1/($B$2:$H$14=J2)*ROW($B$2:$H$14)-1,1))
Thanks to Scott Craner for the hint on the minimum formula.
To determine if you have UNIQUE data from column J in your range B2:H14 you can enter this array formula. In order to enter an array formula you need to press CTRL+SHFT+ENTER at the same time and not just ENTER. You will know you have done it right when you see {} around your formula in the formula bar. You cannot at the {} manually.
=IF(MAX(COUNTIF($B$2:$H$14,J2:J14))>1,"DUPLICATES","UNIQUE")
UPDATE 3
AGGREGATE - A relatively new function to me but goes back to Excel 2010. Aggregate is 19 functions rolled into 1. It would be nice if they all worked the same way but they do not. I think it is functions numbered 14 and up that will perform the same way an array formula or a CSE formula if you prefer. The nice thing is you do not need to use CSE when entering or editing them. SUMPRODUCT is another example of a regular formula that performs array formula calculations.
The meat of this explanation I believe is what is happening inside of the AGGREGATE brackets. If you click on the link you will get an idea of what the first two arguments are. The first defines which function you are using, and the second tell AGGREGATE how to deal with Errors, hidden rows, and some other nested functions. That is the relatively easy part. What I believe you want to know is what is happening with this:
(doc1!$B$2:$H$14=J2)*ROW(doc1!$B$2:$H$14)
For illustrative purpose lets reduce this formula to something a little smaller in scale that does the same thing. I'll avoid starting in A1 as that can make life a little easier when counting since it the 1st row and first column. So by placing the example range outside of it you can see some more special considerations potentially.
What I want to know is what row each of the items list in Column C occurs in column B
| B | C
3 | DOG | PLATYPUS
4 | CAT | DOG
5 | PLATYPUS |
The full formula for our mini example would be:
{=($B$3:$B$5=C2)*ROW($B$3:$B$5)}
And we are going to look at the following as an array
=INDEX($B$3:$B$5,AGGREGATE(14,6,($B$3:$B$5=C2)*ROW($B$3:$B$5),1)-2)
So the first brackets is going to be a Boolean array as you noted. Every cell that is TRUE will TRUE until its forced into a math calculation. When that happens, True becomes 1 and False becomes 0.I that formula was entered as a CSE formula and place in D2, it would break down as follows:
FALSE X 3
FALSE X 4
TRUE X 5
The 3, 4 and 5 come from ROW() returning the value of the row number that it is dealing with at the time of the array math operation. Little trick, we could have had ROW(1:3). Just need to make sure the size of the array matches! This is not matrix math is just straight across multiplication. And since the Boolean is now experiencing a math operation we are now looking at:
0 X 3 = 0
0 X 4 = 0
1 X 5 = 5
So the array of {0,0,5} gets handed back to the aggregate for more processing. The important thing to note here is that it contains ONLY 0 and the individual row numbers where we had a match. So with the first aggregate formula, formula 14 was chosen which is the LARGE function. And we also told it to ignore errors, which in this particular case does not matter. So after providing the array to the aggregate function, there was a ,1) to finish off the aggregate function. The 1 tells the aggregate function that we want the 1st larges number when the array is sorted from smallest to largest. If that number was 2 it would be the 2nd largest number and so on. So the last row or the only row that something is found on is returned. So in our small example it would be 5.
But wait that 5 was buried inside another function called Index. and in our small example that INDEX formula would be:
=INDEX($B$3:$B$5,AGGREGATE(...)-2)
Well we know that the range is only 3 rows long, so asking for the 5th row, would have excel smacking you up side the head with an error because your index number is out of range. So in comes the header row correction of -1 in the original formula or -2 for the small example and what we really see for the small example is:
=INDEX($B$3:$B$5,5-2)
=INDEX($B$3:$B$5,3)
and here is a weird bit of info, That last one does not become PLATYPUS...it becomes the cell reference to =B5 which pulls PLATYPUS. But that little nuance is a story for another time.
Now in the comments Scott essentially told me to invert for the error to get the first row. And this is important step for the aggregate and it had me running in circles for awhile. So the full equation for the first row option in our mini example is
=INDEX($B$3:$B$5,AGGREGATE(15,6,1/($B$3:$B$5=C2)*ROW($B$3:$B$5),1)-2)
And what Scott Craner was actually suggesting which Skips one math step is:
=INDEX($B$3:$B$5,AGGREGATE(15,6,ROW($B$3:$B$5)/($B$3:$B$5=C2),1)-2)
However since I only realized this after writing this all up the explanation will continue with the first of these two equations
So the important thing to note here is the change from function 14 to function 15 which is SMALL. Think of it a finding the minimum. And this time that 6 plays a huge factor along with the 1/. So our array in the middle this time equates to:
1/FALSE X 3
1/FALSE X 4
1/TRUE X 5
Which then becomes:
1/0 X 3
1/0 X 4
1/1 X 5
Which then has excel slapping you up side the head again because you are trying to divide by 0:
#div/0 X 3
#div/0 X 4
1/1 X 5
But you were smart and you protected yourself from that slap upside the head when you told AGGREGATE to ignore error when you used 6 as the second argument/reference! Therefore what is above becomes:
{5}
Since we are performing a SMALL, and we passed ,1) as the closing part of the AGGREGATE, we have essentially said give me the minimum row number or the 1st smallest number of the resulting array when sorted in ascending order.
The rest plays out the same as it did for the LARGE AGGREGATE method. The pitfall I fell into originally is I did not use the 1/ to force an error. As a result, every time I tried getting the SMALL of the array I was getting 0 from all the false results.
SUMPRODUCT works in a very similar fashion, but only works when your result array in the middle only returns 1 non zero answer. The reason being is the last step of the SUMPRODUCT function is to all the individual elements of the resulting array. So if you only have 1 non zero, you get that non zero number. If you had two rows that matched for instance 12 and 31, then the SUMPRODUCT method would return 43 which is not any of the row numbers you wanted, where as aggregate large would have told you 31 and aggregate small would have told you 12.
Something like this maybe, starting in K2 and copied down:
=IFERROR(INDEX(D:D,MAX(IFERROR(MATCH(J2,B:B,0),-1),IFERROR(MATCH(J2,E:E,0),-1),IFERROR(MATCH(J2,G:G,0),-1),IFERROR(MATCH(J2,H:H,0),-1))),"")
If you want to keep the positions of the columns for the Match variable, consider creating generic range names for each column you want to check, like "Col1", "Col2", "Col3". Create a few more range names than you think you will need and reference them to =$B:$B, =$E:$E etc. Plug all range names into Match functions inside the Max() statement as above.
When columns are added or removed from the table, adjust the range name definitions to the columns you want to check.
For example, if you set up the formula with five Matches inside the Max(), and the table changes so you only want to check three columns, point three of the range names to the same column. The Max() will only return one result and one lookup, even if the same column is matched several times.
I came up with a vba solution if I understood correctly:
Sub DisplayActiveRange()
Dim sheetToSearch As Worksheet
Set sheetToSearch = Sheet2
Dim sheetToOutput As Worksheet
Set sheetToOutput = Sheet1
Dim search As Range
Dim output As Range
Dim searchCol As String
searchCol = "J"
Dim outputCol As String
outputCol = "K"
Dim valueCol As String
valueCol = "D"
Dim r As Range
Dim currentRow As Integer
currentRow = 1
Dim maxRow As Integer
maxRow = sheetToOutput.UsedRange.Rows.Count
For currentRow = 1 To maxRow
Set search = Range("J" & currentRow)
For Each r In sheetToSearch.UsedRange
If r.Value <> "" Then
If r.Value = search.Value Then
Set output = sheetToOutput.Range(outputCol & currentRow)
output.Value = sheetToSearch.Range(valueCol & currentRow).Value
currentRow = currentRow + 1
Set search = sheetToOutput.Range(searchCol & currentRow)
End If
End If
Next
Next currentRow
End Sub
There might be better ways of doing it, but this will give you what you want. We assume headers in both "source" and "destination" sheets. You will need to adapt the "Const" declarations according to how your sheets are named. Press Control & G in Excel to bring up the VBA window and copy and paste this code into "This Workbook" under the "VBA Project" group, then select "Run" from the menu:
Option Explicit
Private Const sourceSheet = "Source"
Private Const destSheet = "Destination"
Public Sub FindColumns()
Dim rowCount As Long
Dim foundValue As String
Sheets(destSheet).Select
rowCount = 1 'Assume a header row
Do While Range("J" & rowCount + 1).value <> ""
rowCount = rowCount + 1
foundValue = FncFindText(Range("J" & rowCount).value)
Sheets(destSheet).Select
Range("K" & rowCount).value = foundValue
Loop
End Sub
Private Function FncFindText(value As String) As String
Dim rowLoop As Long
Dim colLoop As Integer
Dim found As Boolean
Dim pos As Long
Sheets(sourceSheet).Select
rowLoop = 1
colLoop = 0
Do While Range(alphaCon(colLoop + 1) & rowLoop + 1).value <> "" And found = False
rowLoop = rowLoop + 1
Do While Range(alphaCon(colLoop + 1) & rowLoop).value <> "" And found = False
colLoop = colLoop + 1
pos = InStr(Range(alphaCon(colLoop) & rowLoop).value, value)
If pos > 0 Then
FncFindText = Mid(Range(alphaCon(colLoop) & rowLoop).value, pos, Len(value))
found = True
End If
Loop
colLoop = 0
Loop
End Function
Private Function alphaCon(aNumber As Integer) As String
Dim letterArray As String
Dim iterations As Integer
letterArray = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
If aNumber <= 26 Then
alphaCon = (Mid$(letterArray, aNumber, 1))
Else
If aNumber Mod 26 = 0 Then
iterations = Int(aNumber / 26)
alphaCon = (Mid$(letterArray, iterations - 1, 1)) & (Mid$(letterArray, 26, 1))
Else
'we deliberately round down using 'Int' as anything with decimal places is not a full iteration.
iterations = Int(aNumber / 26)
alphaCon = (Mid$(letterArray, iterations, 1)) & (Mid$(letterArray, (aNumber - (26 * iterations)), 1))
End If
End If
End Function
I am using excel and i want to display a value to a certain number of significant figures.
I tried using the following equation
=ROUND(value,sigfigs-1-INT(LOG10(ABS(value))))
with value replaced by the number I am using and sigfigs replaced with the number of significant figures I want.
This formula works sometimes, but other times it doesn't.
For instance, the value 18.036, will change to 18, which has 2 significant figures. The way around this is to change the source formatting to retain 1 decimal place. But that can introduce an extra significant figure. For instance, if the result was 182 and then the decimal place made it change to 182.0, now I would have 4 sig figs instead of 3.
How do I get excel to set the number of sig figs for me so I don't have to figure it out manually?
The formula (A2 contains the value and B2 sigfigs)
=ROUND(A2/10^(INT(LOG10(A2))+1),B2)*10^(INT(LOG10(A2))+1)
may give you the number you want, say, in C2. But if the last digit is zero, then it will not be shown with a General format. You have then to apply a number format specific for that combination (value,sigfigs), and that is via VBA. The following should work. You have to pass three parameters (val,sigd,trg), trg is the target cell to format, where you already have the number you want.
Sub fmt(val As Range, sigd As Range, trg As Range)
Dim fmtstr As String, fmtstrfrac As String
Dim nint As Integer, nfrac As Integer
nint = Int(Log(val) / Log(10)) + 1
nfrac = sigd - nint
If (sigd - nint) > 0 Then
'fmtstrfrac = "." & WorksheetFunction.Rept("0", nfrac)
fmtstrfrac = "." & String(nfrac, "0")
Else
fmtstrfrac = ""
End If
'fmtstr = WorksheetFunction.Rept("0", nint) & fmtstrfrac
fmtstr = String(nint, "0") & fmtstrfrac
trg.NumberFormat = fmtstr
End Sub
If you don't mind having a string instead of a number, then you can get the format string (in, say, D2) as
=REPT("0",INT(LOG10(A2))+1)&IF(B2-(INT(LOG10(A2))+1)>0,"."&REPT("0",B2-(INT(LOG10(A2))+1)),"")
(this replicates the VBA code) and then use (in, say, E2)
=TEXT(C2,D2).
where cell C2 still has the formula above. You may use cell E2 for visualization purposes, and the number obtained in C2 for other math, if needed.
WARNING: crazy-long excel formula ahead
I was also looking to work with significant figures and I was unable to use VBA as the spreadsheets can't support them. I went to this question/answer and many other sites but all the answers don't seem to deal with all numbers all the time. I was interested in the accepted answer and it got close but as soon as my numbers were < 0.1 I got a #value! error. I'm sure I could have fixed it but I was already down a path and just pressed on.
Problem:
I needed to report a variable number of significant figures in positive and negative mode with numbers from 10^-5 to 10^5. Also, according to the client (and to purple math), if a value of 100 was supplied and was accurate to +/- 1 and we wish to present with 3 sig figs the answer should be '100.' so I included that as well.
Solution:
My solution is for an excel formula that returns the text value with required significant figures for positive and negative numbers.
It's long, but appears to generate the correct results according to my testing (outlined below) regardless of number and significant figures requested. I'm sure it can be simplified but that isn't currently in scope. If anyone wants to suggest a simplification, please leave me a comment!
=TEXT(IF(A1<0,"-","")&LEFT(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00"),sigfigs+1)*10^FLOOR(LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00")),1),(""&(IF(OR(AND(FLOOR(LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00")),1)+1=sigfigs,RIGHT(LEFT(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00"),sigfigs+1)*10^FLOOR(LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00")),1),1)="0"),LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00"))<=sigfigs-1),"0.","#")&REPT("0",IF(sigfigs-1-(FLOOR(LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00")),1))>0,sigfigs-1-(FLOOR(LOG10(TEXT(ABS(A1),"0."&REPT("0",sigfigs-1)&"E+00")),1)),0)))))
Note: I have a named range called "sigfigs" and my numbers start in cell A1
Test Results:
I've tested it against the wikipedia list of examples and my own examples so far in positive and negative. I've also tested with a few values that gave me issues early on and all seem to produce the correct results.
I've also tested with a few values that gave me issues early on and all seem to produce the correct results now.
3 Sig Figs Test
99.99 -> 100.
99.9 -> 99.9
100 -> 100.
101 -> 101
Notes:
Treating Negative Numbers
To Treat Negative Numbers, I have included a concatenation with a negative sign if less than 0 and use the absolute value for all other work.
Method of construction:
It was initially divided into about 6 columns in excel that performed the various steps and at the end I merged all of the steps into one formula above.
Use scientific notation, say if you have 180000 and you need 4 sigfigs the only way is to type as 1.800x10^5
I added to your formula so it also automatically displays the correct number of decimal places. In the formula below, replace the digit "2" with the number of decimal places that you want, which means you would need to make four replacements. Here is the updated formula:
=TEXT(ROUND(A1,2-1-INT(LOG10(ABS(A1)))),"0"&IF(INT(LOG10(ABS(ROUND(A1,2-1-INT(LOG10(ABS(A1)))))))<1,"."&REPT("0",2-1-INT(LOG10(ABS(ROUND(A1,2-1-INT(LOG10(ABS(A1)))))))),""))
For example, if cell A1 had the value =1/3000, which is 0.000333333.., the above formula as-written outputs 0.00033.
This is an old question, but I've modified sancho.s' VBA code so that it's a function that takes two arguments: 1) the number you want to display with appropriate sig figs (val), and 2) the number of sig figs (sigd). You can save this as an add-in function in excel for use as a normal function:
Public Function sigFig(val As Range, sigd As Range)
Dim nint As Integer
Dim nfrac As Integer
Dim raisedPower As Double
Dim roundVal As Double
Dim fmtstr As String
Dim fmtstrfrac As String
nint = Int(Log(val) / Log(10)) + 1
nfrac = sigd - nint
raisedPower = 10 ^ (nint)
roundVal = Round(val / raisedPower, sigd) * raisedPower
If (sigd - nint) > 0 Then
fmtstrfrac = "." & String(nfrac, "0")
Else
fmtstrfrac = ""
End If
If nint <= 0 Then
fmtstr = String(1, "0") & fmtstrfrac
Else
fmtstr = String(nint, "0") & fmtstrfrac
End If
sigFig = Format(roundVal, fmtstr)
End Function
It seems to work in all the use cases I've tried so far.
Rounding to significant digits is one thing... addressed above. Formatting to a specific number of digits is another... and I'll post it here for those of you trying to do what I was and ended up here (as I will likely do again in the future)...
Example to display four digits:
.
Use Home > Styles > Conditional Formatting
New Rule > Format only cells that contain
Cell Value > between > -10 > 10 > Format Number 3 decimal places
New Rule > Format only cells that contain
Cell Value > between > -100 > 100 > Format Number 2 decimal places
New Rule > Format only cells that contain
Cell Value > between > -1000 > 1000 > Format Number 1 decimal place
New Rule > Format only cells that contain
Cell Value > not between > -1000 > 1000 > Format Number 0 decimal places
.
Be sure these are in this order and check all of the "Stop If True" boxes.
The formula below works fine. The number of significant figures is set in the first text formula. 0.00 and 4 for 3sf, 0.0 and 3 for 2sf, 0.0000 and 6 for 5sf, etc.
=(LEFT((TEXT(A1,"0.00E+000")),4))*POWER(10,
(RIGHT((TEXT(A1,"0.00E+000")),4)))
The formula is valid for E+/-999, if you have a number beyond this increase the number of the last three zeros, and change the second 4 to the number of zeros +1.
Note that the values displayed are rounded to the significant figures, and should by used for display/output only. If you are doing further calcs, use the original value in A1 to avoid propagating minor errors.
As a very simple display measure, without having to use the rounding function, you can simply change the format of the number and remove 3 significant figures by adding a decimal point after the number.
I.e. #,###. would show the numbers in thousands. #,###.. shows the numbers in millions.
Hope this helps
You could try custom formatting instead.
Here's a crash course: https://support.office.com/en-nz/article/Create-a-custom-number-format-78f2a361-936b-4c03-8772-09fab54be7f4?ui=en-US&rs=en-NZ&ad=NZ.
For three significant figures, I type this in the custom type box:
[>100]##.0;[<=100]#,##0
You could try
=ROUND(value,sigfigs-(1+INT(LOG10(ABS(value)))))
value :: The number you wish to round.
sigfigs :: The number of significant figures you want to round to.
My formula is giving me unexpected responses.
=IF(I5+H5=0,"Paid","Due")
see below
H I J k
-£34.40 £34.40 £0.00 Due
Cell H is calculated with this
=(SUM(F5+G5))*-1
See correct output with exact same formula on same worksheet
=IF(I3+H3=0,"Paid","Due")
H I J K
-£205.44 £205.44 £0.00 Paid
Cell H is calculated he same
=(SUM(F3+G3))*-1
Any ideas why the top calculation not correct but the bottom one is.
This is most likely the floating point issue. You should not compare floating point numbers directly with = because computers can't store the full decimal places. Just like if you divide 1 dollar by 3, you end up with .3333333333333 cents, well if you add 3 of those you don't necessarily get back 1 dollar, but slightly less due to the "lost" 3333's at the end. The proper way to compare is using a Delta threshold, meaning "how close" it needs to be.
so instead of
if (a+b=c,"paid", "due")
you would do
if(ABS(c-(a+b))<.01, "paid", "due")
so in that case .01 is the delta, or "how close" it has to be. It has to be within 1 cent. the formula literally means "if the absolute value of the difference between c and (a+b) is less than 1 cent, return paid, else return due. (of course, this will say due if they overpaid, so keep that in mind)
you should always do this.