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I'm out of idea how I could format consecutive same (respectively only even) values in Excel tables without using VBA.
The conditional formatting shall color only consecutive values and only
all 0s or all even values, when there are not more than 2.
A: ID
B: binary
C: counting
1
1
1
2
0
2
3
0
2
4
1
3
5
0
4
6
0
4
7
0
4
8
1
5
9
1
5
I tried to format with: =COUNTIF(C1:C9, C1) < 3, but then it also colors the 1s and C6:C7, eventho there are more than 2.
I also tried =AND( COUNTIF(C1:C9,C1) < 3, ISEVEN(C1:C9) ) but then it colors nothing.
I could replace the 0s with empty cells so I could check ISEMPTY(B1:B9) but it again colors nothing. Using $ to set absolute changes nothing as well.
Formating duplicates also colors triplets, which also doesn't work for me.
=OR(COUNTIF($C$1:$C$9,C1) = 1, COUNTIF($C$1:$C$6,C1) = 2) works so far, but also colors the 1s (uneven).
=AND(OR(COUNTIF($C$1:$C$9,C1) = 1, COUNTIF($C$1:$C$6,C1) = 2), ISEVEN($C$1:$C$9)) doesn't work.
=AND(OR(COUNTIF($C$1:$C$9,C1) = 1, COUNTIF($C$1:$C$6,C1) = 2), $B$1:$B$9 <> 1) doesn't work as well.
My only solution so far is using 2 formating rules:
color =OR(COUNTIF($C$1:$C$9,C1) = 1, COUNTIF($C$1:$C$6,C1) = 2)
do not color =$B$1:$B$9 = 1
but I think it is terrible.
I worked on it for some hours, maybe I'm missing something really obvious.
I'm not allowed to use VBA, therefore this is ot an option.
EDIT: My 2.rule-solution can be simplificed with:
color =COUNTIF($C$1:$C$9,C1) < 3
do not color =$B$1:$B$9 = 1
I'm still confused why combining both doesn't work:
AND(COUNTIF($C$1:$C$9,C1) < 3; $B$1:$B$9 <> 1)
EDIT2: I know why it didn't work. Don't check <>1 with absolute value-range $B$1$:$B$9
Solution: B1 <> 1 then it loops through.
Now combining both works:
=AND( COUNTIF($C$1:$C$9, C1) < 3, B1 <> 1)
I can't see an easy answer for the binary numbers. You have two cases:
(1) Current cell is zero, previous cell is 1, next cell is zero and next cell but one is 1.
(2) Current cell is zero, previous cell is zero, previous cell but one is 1, next cell is 1.
But then the first pair of numbers is a special case because there is no previous cell.
Strictly speaking the last pair of numbers is a special case as well because there is no following cell.
=OR(AND(ROW()=1,B$1=0,B$2=0,B$3=1),AND(ROW()=2,B$1=0,B$2=0,B$3=1),AND(B1=0,B1048576=1,B2=0,B3=1),AND(B1=0,B1048576=0,B1048575=1,B2=1))
where I have used the fact that you are allowed to wrap ranges to the end of the sheet (B1048576) in conditional formatting.
Adding the condition for the case where there there are two zeroes at the end of the range:
=OR(AND(ROW()=1,B$1=0,B$2=0,B$3=1),
AND(ROW()=2,B$1=0,B$2=0,B$3=1),
AND(B1=0,B1048576=1,B2=0,OR(B3=1,B3="")),
AND(B1=0,B1048576=0,B1048575=1,OR(B2=1,B2="")))
Even this could go wrong if there was something in the very last couple of rows of the sheet, so I suppose to be absolutely safe:
=OR(AND(ROW()=1,B$1=0,B$2=0,B$3=1),
AND(ROW()=2,B$1=0,B$2=0,B$3=1),
AND(Row()>1,B1=0,B1048576=1,B2=0,OR(B3=1,B3="")),
AND(Row()>2,B1=0,B1048576=0,B1048575=1,OR(B2=1,B2="")))
Shorter:
=OR(AND(ROW()<=2,B$1+B$2=0,B$3=1),
AND(B1+B2=0,B1048576=1,OR(B3=1,B3="")),
AND(B1+B1048576=0,B1048575=1,OR(B2=1,B2="")))
Not the cleanest wat but it works. You only need to move your data 1 row below, so headers would be in row 2 and data in row 3 for this formula to work:
=IF(AND(B3=B4,B3<>B5),IF(AND(B4=B3,B4<>B2),TRUE,FALSE),IF(AND(B3=B2,B3<>B1),IF(AND(B3=B4,B3<>B5),FALSE,TRUE),FALSE))
How about this approach (Office 365):
=LET(range,B$1:B$9,
s,IFERROR(TRANSPOSE(INDEX(range,ROW()+SEQUENCE(5,,-2))),1),
t,TEXTJOIN("",,(s=INDEX(range,ROW()))*ISEVEN(s)),
IFERROR(SEARCH("0110",t)<4,IFERROR(SEARCH("010",t)=2,FALSE)))
It creates an array s of 5 values starting point is the current row of the range, adding the 2 values above and below. If the value is out of range it will replace the error with a 1.
The array s is checked for being even (TRUE/FALSE, IFERROR created values are uneven) and the values to equal the value of the current row of the range (TRUE/FALSE).
These two booleans are multiplied creating 1 for both values being TRUE, else 0.
These values are joined and checked for 2 consecutive 1's (surrounded by 0) to be found in the 2nd or 3rd position of the range (this would be the case if two even consecutive equal numbers are found),
if it errors it will look if a unique even number is found (1 surrounded by 0 in 2nd position).
PS I'm unable to test if conditional formatting allows you to type the range as B:B instead of B$1:B$9 (working from a mobile) but that would make it more dynamical, because that way you can easily expand the conditional range.
One of my Excel column of my board have to store numbers of 9 digits.
I'm looking for a solution to keep only the 9 last digits of any bigger number past in this specific column. It's only entire number.
Also if after formatting the number it appear that the number starts with 0 the 0 have to be kept. Is there another solution than adding an '0 at first ?
Here is what I already done : (i is the row number / Range01 is Range("A14:O400"))
If Len(Range01.Cells(i,5).value) = 9 Then
Range01.Cells(i,5).Interior.color = vbGreen
ElseIf Len(Range01.Cells(i,5).value) = 8 Then
Range01.Cells(i,5).value = "'0" & Range01.Cells(i,5).value
ElseIf Len(Range01.Cells(i,5).value) > 9 Then
????
Else
Range01.Cells(i,5).Interior.color = vbRed
End If
Thanks for the help.
The simplest way to get the last nine numbers of an integer is:
=MOD(A1,1000000000)
(For your information, that's one billion, a one with nine zeroes.)
If you're interested in showing a number with leading zeroes, you can alter the cell formatting as follows: (the format simply contains nine zeroes)
If you're interested in keeping the zeroes, you might need to use your number as a string, and precede it with a good number of repeated zeroes, something like:
=REPT("0",9-LEN(F8))&F8
Take the length of your number (which gets automatically converted into a string)
Subtract that from 9 (so you know how many zeroes you need)
Create a string, consisting of that number of zeroes
Add your number behind it, using basic concatenation.
You can simply use the math operator of modulus. If you want the last 9 digit you can write:
n % 10000000000
Where n is the number in the column.
In VBA:
MOD(n,1000000000)
HI.
how can i come up with return value of "company name" (column H) at Column B IF any of the "PrefiX" (Column G) found at "con no" (Column A).
Sample of outcome needed as in column B.
Sample:
620011113 = DD
CN1234 = BB
thanks
=INDEX($H:$H,AGGREGATE(15,6,ROW($G$1:$G$7)/(--(FIND($G$1:$G$7,$A2)=1)*--(LEN($G$1:$G$7)>0)),1),1)
Breaking this down, the INDEX retrieves the Nth item from Column H (Company name). To find the value of N, we are using the AGGREGATE function
AGGREGATE is a weird function - it lets us use things like MAX or LARGE or SUM while ignoring any error values. In this case, we will be using it for SMALL (first argument, 15), while Ignoring Error Values (second argument, 6). We will want the very smallest value, so the fourth argument will be 1. (If we wanted the second smallest, it would be 2, and so on)
=INDEX($H:$H,AGGREGATE(15,6, <SOMETHING> ,1),1)
So, all we need now is a list of values to compare! To make things slightly simpler, I'll break that bit of the code out for you here:
ROW($G$1:$G$7) / (--(FIND($G$1:$G$7,$A2)=1) * --(LEN($G$1:$G$7)>0))
There are 3 parts to this. The first, ROW($G$1:$G$7)is the actual value we want to retrieve - these will be the Row Numbers for each Prefix that matches your value. On its own, however, it will be all the row numbers. Since we are skipping errors, we want any Rows that don't match the prefix to throw an error. The easiest way to do this is to Divide by Zero
At the start of --(FIND($G$1:$G$7,$A2)=1) and --(LEN($G$1:$G$7)>0) we have a double-negative. This is a quick way to convert True and False to 1 and 0. Only when both tests are True will we not divide by 0, as this table shows:
A | B | A*B
1 | 1 | 1
1 | 0 | 0
0 | 1 | 0
0 | 0 | 0
Starting with the second test first (it's easier), we have LEN($G$1:$G$7)>0 - basically "don't look at blank cells".
The other test (FIND($G$1:$G$7,$A2)=1) will search for the Prefix in the Con No, and return where it is found (or a #VALUE! error if it isn't). We then check "is this at position 1" - in other words, "Is this at the start of the Con No, rather than in the middle". We don't want to say Con No CNQ6060 is part of Company AA instead of Company BB by mistake!
So, if the Prefix is at the Start of the Con No, AND it isn't Blank (because there is an infinite amount of Nothing Before, After, and Between every number and letter), then we get it added to our list of Rows. We then take the smallest row (i.e. closest to the top - change AGGREGATE(15 to AGGREGATE(14 if you want the closest to the bottom!), and use that to get the Company Name
You could try the below formal:
=VLOOKUP(IF(LEFT(A3,1)="6",LEFT(A3,4),IF(LEFT(A3,1)="C",LEFT(A3,2),IF(LEFT(A3,1)="E",LEFT(A3,7)))),$G$3:$H$7,2,0)
Have in mind that you have to use ' before the cell value of column A & G in order to convert cell value into text get the correct out comes using VLOOKUP
Result:
I would like to search for a specific string in matlab cell. For example my cell contains a column of strings like this
variable(:,5) = {'10';'10;20';'20';'10;20';'10';'10';'20'};
I would like to search for all cells that have only '10' and delete them.
I tried using this statement for searching
is10 = ~cellfun(# isempty , strfind (variable(:,5) , '10'));
But this returns all cells with '10' (including the ones with '10;20').
I would like to have just the cells with pure '10' values
What is the best way to do this?
It is not working as you expect because strfind allows for a partial string match. What you want is an exact match. You can do this using strcmp. Also, the input to strcmp can actually be a cell array of strings so you can use it the following way.
A = {'10';'10;20';'20';'10;20';'10';'10';'20'};
is10 = strcmp(A, '10');
%// 1 0 0 0 1 1 0
You could also use ismember to do the same thing.
is10 = ismember(A, '10');
%// 1 0 0 0 1 1 0
As a side note, most string functions (including strfind) can actually accept a cell array of strings as input. So in your initial post, the wrapping of strfind inside of cellfun is unnecessary.
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.