**Please Note: **
This article is written for users of the following Microsoft Excel versions: 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. If you are using an earlier version (Excel 2003 or earlier), *this tip may not work for you*. For a version of this tip written specifically for earlier versions of Excel, click here: Finding the Number of Significant Digits.

Brenda is interested in knowing the number of significant digits in a value. She wonders if there is an Excel function or formula, she can use that would return the number of significant digits in the value shown in a cell.

This question is not as simple as it seems. For some people, finding the number of significant digits in a value means counting the number of digits, excluding any decimal points or negative signs. If that is all you need, then something like this formula will work just fine:

=IF(A1<0,IF(A1=INT(A1),LEN(A1)-1,LEN(A1)-2),IF(INT(A1)=A1,LEN(A1),LEN(A1)-1))

The reason that this isn't that simple, however, is because what constitutes the number of significant digits in a value depends on many things. The bottom line is that you can't always tell by looking at a value how many significant digits it has.

For instance, the value 100 could have 1, 2, or 3 significant digits. It is presumed that the value 1.00 has 3 significant digits, but that may not be the case if the value displayed is the result of formatting imposed by Excel—for instance, the value in the cell could be 1.0000437, which Excel formats as 1.00. You can discover more about the topic of significant digits here:

https://excelribbon.tips.net/T012083

There are some generally accepted ways to identify significant digits in a number, but any attempt to codify a set of rules is always open to debate. One such set of rules has been noted at Wikipedia, in the "Identifying Significant Digits" section of this article:

http://en.wikipedia.org/wiki/Significant_figures

With at least a rudimentary set of rules in mind (such as the one in the Wikipedia article) it is possible to develop a user-defined function that will give you the most likely number of significant digits for a value.

Function SigFigs(rng As Range, Optional iType As Integer = 1) 'iType = 1 is Min 'iType = 2 is Max Dim rCell As Range Dim sText As String Dim sText2 As String Dim iMax As Integer Dim iMin As Integer Dim iDec As Integer Dim i As Integer Application.Volatile Set rCell = rng.Cells(1) 'if not a number then error If Not IsNumeric(rCell) Or IsDate(rCell) Then SigFigs = CVErr(xlErrNum) Exit Function End If sText2 = Trim(rCell.Text) sText = "" 'find position of decimal point (it matters) iDec = InStr(sText2, ".") 'strip out any non-numbers (including decimal point) For i = 1 To Len(sText2) If Mid(sText2, i, 1) >= "0" And _ Mid(sText2, i, 1) <= "9" Then _ sText = sText & Mid(sText2, i, 1) Next 'remove any leading zeroes (they don't matter) While Left(sText, 1) = "0" sText = Mid(sText, 2) Wend iMax = Len(sText) 'strip trailing zeroes (they don't matter if no decimal point) sText2 = sText If iDec = 0 Then While Right(sText2, 1) = "0" sText2 = Left(sText2, Len(sText2) - 1) Wend End If iMin = Len(sText2) 'return Min or Max Select Case iType Case 1 SigFigs = iMin Case 2 SigFigs = iMax Case Else SigFigs = CVErr(xlErrNum) End Select End Function

You call this function by using the following in your worksheet:

=SigFigs(A1, x)

You can replace ** x** with either 1 or 2. If you specify 1, then the function returns the minimum number of significant digits. If you specify 2, then the function returns the maximum number of significant digits. In most cases the two possible return values will be the same, except with values that are whole numbers, without a trailing decimal point, that have trailing zeroes. In other words, if you use the function to evaluate the number 1234000, then the minimum (

The function takes into consideration how the number appears in the worksheet, meaning that it matters how the number is formatted. It strips out any formatting characters, such as negative signs, parentheses, and commas.

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This tip (10976) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. You can find a version of this tip for the older menu interface of Excel here: **Finding the Number of Significant Digits**.

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2018-06-02 12:27:29

Rick Rothstein

&HEAD ==> 3

&7654 ==> 4

123D45 ==> 5

($1,23,,3.4,,,5,,E67$) ==> 8

In this mini-blog article of mine, I explain why that is not necessarily a good thing to use IsNumeric for that purpose...

excelfox.com/forum/showthread.php/2004-Thinking-About-Using-VBA-s-IsNumeric-Function-Read-this-first

I also noticed the following evaluations by the SigFigs function which also produce answers that I do not agree with. First, numbers that Excel converts to E-notation when entered into a cell or the Formula Bar...

123456000000 ==> 8

0.00000012345 ==> 7

0.0000000012 ==> 5

Second, some floating point numbers when entered into a cell formatted as Text... I have not delved into the code to see why, I am simply noting the problem. Here are some examples all of which return a #NUM! error from the function... 1.2, 1.09, 12.34, 10.002, etc. I would note that there are lots of examples of floating point numbers entered into Text formatted cells that work correctly.

Now, of course, I have my own SigFigs function to offer which is more compact, uses no loops and handles the above noted problems correctly (as far as my limited testing is able to confirm)...

Function SigFigs(Rng As Range, Optional iType As Long = 1)

Dim S As String

Application.Volatile

S = Trim(Replace(Replace(Split(Rng.Text, "E", , vbTextCompare)(0), "$", ""), "%", ""))

If Not S Like "*[!0-9.+-]*" And Not S Like "*.*.*" And S <> "." Then

If iType = 1 And Not S Like "*.*" Then S = Replace(Trim(Replace(S, "0", " ")), " ", "0")

SigFigs = Len(Format(Replace(S, ".", ""), String(31, "#") & ";" & String(31, "#") & ";;"))

Else

SigFigs = CVErr(xlErrNum)

End If

End Function

2018-05-20 23:52:05

Don

2015-08-29 07:25:38

I hope there is an easy way out.

=LEN(A1)-LEN(SUBSTITUTE(A1,"Digit",""))

This formula basically counts the original length of the cell, replaces the desired number by null and gets the length again, finally reduces the new length from the original length and you get your number of instances of a specified digit.

Regards, Arun

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