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$9 (working from a mobile) but that would make it more dynamical, because that way you can easily expand the conditional range.
I'm trying to complete division for a dynamic range with two variables and it keeps dividing the dynamic range by the last number in the range. Below is my VBA code.
For i = 2 To 8
For r = 13 To 19
If ThisWorkbook.Sheets("Sheet1").Cells(i, 28) = "" Then
ThisWorkbook.Sheets("Sheet1").Cells(r, 28) = ""
Else
ThisWorkbook.Sheets("Sheet1").Cells(r, 28) = ThisWorkbook.Sheets("Sheet1").Cells(i, 28) / Range("$AB$8")
End If
Next r
Next i
Essentially it is dividing the last i value (Cell row 8) by the Range("$AB$8") (cells row 19).
What I would like to happen is the values in rows i to divide by Range("$AB$8")....in other words the value in cell (2,28)/ab8, (3/28)/ab8, (4,28)/ab8 etc etc.
It current is taking the value in cell (8,28) dividing it by ab8...and applying it to all defined r rows.
There are a number of issues here - all of which are small tweaks but end up with the wrong result you are seeing.
Your example code is not a dynamic range. You have hardcoded Cells(AB2:AB8) and Cells(AB13:AB19). You just did it in a way that is not obvious.
Also not very obvious is that you are writing the results to a single column. See the pattern here:
Loop 1: i = 2, results may be writing to Cells(AB13:AB19)
[…]
Loop 7: i = 8, results may be writing to Cells(AB13:AB19)
I said "may" because you have the If statement.
Depending on what you really want to happen, the code can be amended.
Instead off the first loop put a conditional there (e.g. If all cells
in that range are blank then …, or if any cells are blank then ...)
Use an Exit For after fixing the first blank loop
Also address the column (i.e. results spread across multiple columns)
Use a single loop (For i = 2 to 8 … and then adjust r according to i… r = i+12)
I have a list of numbers (that will be changed weekly) and I also have a list of ranges (that also change weekly). I need to check whether each number falls between each range.
Eg. My list of numbers on the left and my list of ranges on the right.
4 1 3
10 67 99
54 120 122
155
So what I need is to return a value if 4 falls between 1-3, then check if it falls between 67-99 and so on. Then return a value if 10 falls between 1-3 or 67-99 etc.
I have tried array and vba but I'm noob and I cant find much in the way of examples for this issue. I have had success with the following nested if;
=IF(OR(AND(G2>$L$2,G2<$M$2),AND(G2>$L$3,G2<$M$3),AND(G2>$L$4,G2<$M$4),G2,"")
=IF(OR(AND(G3>$L$2,G3<$M$2),AND(G3>$L$3,G3<$M$3),AND(G3>$L$4,G3<$M$4),G3,"")
However, once my number of ranges gets above a certain number it says i have too many characters.
Any help would be appreciated.
Regards,
Will.
So what I need is to return a value if 4 falls between 1-3, then check if it falls between 67-99 and so on. Then return a value if 10 falls between 1-3 or 67-99 etc.
If you want to match each value in col G with the list in L and M then, rearrange the G column values in a row as shown below so that you can get the entire view in one go.
Put this formula =(AND($N$1>L2,$N$1<M2)) in N2 and drag it down. Similarly put the formula =(AND($O$1>L2,$O$1<M2)) in O2 and pull it down and so on...
Thanks Siddarth. This works but the number of rows will be changing weekly and your method would be too tedious to do weekly.
Ended up working it out on my own. Pretty simple but took me forever!
Sub Subtract_Start()
Set rng1 = Range(Range("G2"), Range("G2").End(xlDown))
Set rng2 = Range(Range("L2"), Range("L2").End(xlDown))
For i = 2 To rng1.Rows.count
For j = 2 To rng2.Rows.count
If Cells(i, "G").Value > Cells(j, "L").Value Then
If Cells(i, "G").Value < Cells(j, "M").Value Then
Cells(i, "G").Copy Cells(i, "J")
End If
End If
Next j
Next i
End Sub
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.
I have 1 column with about 60 cells with values or different length. Each (or at least most) of the values have a numeric characters in the value. I want to split the columns cells into more columns which I normally would do with the 'tekst to columns' function of excel.
But this function does not have an advanced option of splitting the value at the first numeric character. splitting based on spaces, comma etc. is possible but this does not help me.
Is there any way to divide the cells into 2 columns at the first number in the cell value?
I have looked at numerous other questions but none of them (or other internet fora) have helped me to split the value at the first number of the cell value.
Thanks #quantum285 for the answer. This routine works if the string contains one number. I changed the teststring to firstpart323secondpart.
then part1 returns 'firstpart32' and part2 return secondpart.
I tried to understand what happens in this code, please correct me if I'm wrong:
First, the lenght of the string is determined.
Secondly, for each position in this string is checked if it is numeric or not. But this check is dan from right to left? So in case of firstpart323secondpart: the length is 22.
then isnumeric() checks for every position from 1 to 22 if it is numeric and stops when it finds a number?
If so, part 1 is the the tekst before the value i, where i is the first position from right to left in the string where a number is found.
and part 2 is then the string on the right from this same position.
However, I am looking for a routine which find the first position from left to right (or the last position from right to left) where a number is, ...
So I changed the routine now, simply adjusting the for i = 1 to line:
Sub test()
For j = 4 To Cells(Rows.Count, 4).End(xlUp).Row
For i = Len(Cells(j, 4)) To 1 Step -1
If IsNumeric(Mid(Cells(j, 4), i, 1)) Then
Cells(j, 5) = Left(Cells(j, 4), i - 1)
Cells(j, 6) = (Right(Cells(j, 4), Len(Cells(j, 4)) - i + 1))
End If
Next i
Next j
End Sub
this almost perfectly works (except for a few cells which have multiple number combinations in the value (like: soup 10g 20boxes). But as these are only a few, I can adjust them by hand.
Thanks!
Sub test()
testString = "firstpart3secondpart"
For i = 1 To Len(testString)
If IsNumeric(Mid(testString, i, 1)) Then
part1 = Left(testString, i - 1)
part2 = (Right(testString, Len(testString) - i))
End If
Next i
MsgBox (part1)
MsgBox (part2)
End Sub
Use something like this within your loop.