Somewhat simple problem:
I need to turn a column A, which contains numbers with up to 1 decimal (20, 142, 2.5, etc.) to a string with a specific format, namely 8 whole digits and 6 decimal digits but without the actual decimal period, like so:
1 = 00000001000000
13 = 00000013000000
125 = 00000125000000
46.5 = 00000046500000
For what it's worth, the input data from column A will never be more than 3 total digits (0.5 to 999) and the decimal will always be either none or .5.
I also need for Excel to leave the zeroes alone instead of auto-formatting as a number and removing the ones at the beginning of the string.
As a makeshift solution, I've been using =CONCATENATE("'",TEXT(A1,"00000000.000000")), then copying the returning value and "pasting as value" where I actually need it.
It works fine, but I was wondering if there was a more direct solution where I don't have to manually intervene.
Thanks in advance!
=TEXT(A1*1000000,"0000000000000") I think that's what you mean.
I have unique identifiers for each row. For example 19Jan187938 or 19Jan206414 but there are some which are like 19Jan17333. I need to add a 0 before the number if it's 5 digits, so it becomes 19Jan017333.
I tried,
=TEXT(CONCATENATE(19,AB2,C2),"000000")
even with 11 0's, since the total length is 11. Nothing changes.
Try the following:
=CONCATENATE(LEFT(AB2,5),TEXT(RIGHT(AB2,LEN(AB2)-5),"000000"))
It will basically, take the first 5 characters and concatenate that with the remaining characters formatted as a six digit number with leading zeroes
If your identifier is on A1, you can try this:
=IF(LEN(A1)<11;CONCATENATE(LEFT(A1;5);RIGHT("000000"&MID(A1;6;5);6));A1)
See what happens.
I need to search an Excel range for specific text, and return how many are found, but want to search only on the first 4 characters. I have
=SUMPRODUCT(-- ISNUMBER(FIND("V",E4:E40)))
which does search the range, but it finds all V's and not those just in the first 4 characters. The data starts with a 4 character acronym and then has free form text.
I'm unable to get LEFT to work in this formula.
=SUMPRODUCT(-- (LEFT(E4:E40,4) = "VOID"))
If you only want to see if V is in the first 4 letters:
=SUMPRODUCT(--(FIND("V",E4:E40 & "XXXXV")<=4))
concatenating the & "XXXXV" on the end removes the possiblity of the error. So the Find will return a number. We then test whether that V is found in the first 4 characters.
We added fourX to the string to ensure that if the value has a length less than 4 we do not get false positives.
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.
I have an Excel spreadsheet with over 2000 entries:
Field B1: CustomerID as 000012345
Field B2: CustomerID as 0000432
Field C1: CustomerCountry as DE
Field C2: CustomerCountry as IT
I need to build codes 13 digits long including "CustomerCountry" + "CustomerID" without leading 0 + random number (can be 6 digits, more or less, depends in length of CustomerID).
The results should be like this: D1 Code as DE12345967895 or D2 Code as IT43274837401
How to do it with Excel functions?
UPDATED:
I tried this one. My big problem is to say that random number should be long enough to get 13 characters in all. Sometimes CustomerID is just 3 or 4 digits long, and concatenation of three variables can be just 10 or 9 characters. But codes have to be always 13 characters long.
Use & to concatenate strings.
Use VALUE(CustomerID) to trim the leading zeroes from the ID
Use RAND() to add a random number between 0 and 1 or RANDBETWEEN(x,y) to create one between x and y.
Combine the above and there you are!
If you always want 13 digits you can use LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID)))) for the random number to ALWAYS be the right length.
total formula
= CustomerCountry
& VALUE(CustomerID)
& LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID))))
=C1 & TEXT(B1,"0") & RIGHT(TEXT(RANDBETWEEN(0,99999999999),"00000000000"),11 - LEN(TEXT(B1,"0")))
that should do it
I don’t understand what is where and OP has accepted answer so have not bothered testing:
=LEFT(RIGHT(C1,2)&VALUE(MID(B1,15,13))&RANDBETWEEN(10^9,10^10),13)
(but I might revert to this if no one else picks the flaws in it first!)