Essentially, my code returns a string which only contains a 1 or 2 digit integer, such as "8" or "10". Is there any way to convert this to an actual number, such as one that could be then stored in a list or matrix?
Assuming the string is in Str1, you can simply do this:
expr(Str1->X
This will evaluate whatever crazy math is inside of Str1 (in this case, it's just a straight up number) and store it into X.
Related
I have some data which is badly formatted ( inherited) after some manipulation and some concatenation I have something resembling the following in a string
"SIGNAGE -- 11 Requires door signage. "
My table has to cross reference a some data from a VLOOKUP and then tries to get the first chars in a pattern after the --
My formula is this
=IF(VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)<>"",LEFT((RIGHT((VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)), LEN((VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)))-SEARCH("--", (VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)),1)-2)),2),"")
This successfully gives me the number 11.
My problem is that the number is being treated as a string and not as a numeric value.
What am I missing?
Use NUMBERVALUE() function to convert a string into a number.
I often use the trick when getting numbers out of text with LEFT(), MID() or RIGHT() to do a "*1" as the final step.
=mid(....) *1
for example,
So, yours would be :
=IF(VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)<>"",LEFT((RIGHT((VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)), LEN((VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)))-SEARCH("--", (VLOOKUP($C3,DoorCheck!$D3:$AD79,19,FALSE)),1)-2)),2),"")*1
I'm a newbie on Excel.
So I have a list of some names ending with Hexa decimals. And some names, that doesn't have any.
My mission is to see only those names with Hexa decimals. (Mabye somehow filter them out)
Column:
BFAXSPOINTDEVBAUHOFLAN2AD
BFAXSQLBAUHOFLAN207
BFAXSQLDEVBAUHOFLAN27A
BFREPDEVBAUHOFLAN258
BFREPORTINGBAUHOFLAN20B
COBALTSEA02900
COBALTSEAVHOST900
DIRECTO8000
DIRECTO9000
DIRECTODCDIRECTOLA009
DYNAMAEBSSISE006
SURVEYEBSSISE006
KVMSRV00",
KVMSRV01",
KVMSRV02",
ASR
CACTI
DBSYNC",
DTV
and so on...
The Function HEX2DEC will help you achieve what you want - it attempts to convert a number as a hexidecimal, into a decimal. If it is not a valid Hex input, it will produce an error.
The key is understanding how many digits you expect your decimal to be - is it the last 5 characters; the last 10; etc. Also note that there is a risk that random text / numbers will be seen as hexidecimal when really that's not what it represents [but that's a problem with the question as you have laid it out; going solely based on the text provided, all we can see is whether a particular cell creates a valid Hexidecimal].
The full formula would look like this[assuming your data starts in A1, and that your Hexidecimal numbers are expected to be 6 characters long, this goes in B1 and is copied down]:
=ISERROR(HEX2DEC(RIGHT(A1,6)))
This takes the 6 rightmost characters of a cell, and attempts to convert it from Hex to Decimal. If it fails, it will produce TRUE [because of ISERROR]; if it succeeds, it will produce FALSE.
Then simply filter on your column to see the subset of results you care about.
Consider the following UDF:
Public Function EndsInHex(r As Range) As Boolean
Dim s As String, CH As String
s = r(1).Text
CH = Right(s, 1)
If CH Like "[A-F]" Or CH Like "[0-9]" Then
EndsInHex = True
Else
EndsInHex = False
End If
End Function
For the string to end in a hex, the last character must be a hex.
I would like to use the command text to type numbers within 57 hexagons. I want to use a loop:
for mm=1:57
text(x(m),y(m),'m')
end
where x(m) and y(m) are the coordinates of the text .
The script above types the string "m" and not the value of m. What am I doing wrong?
Jubobs pretty much told you how to do it. Use the num2str function. BTW, small typo in your for loop. You mean to use mm:
for mm=1:57
text(x(mm),y(mm),num2str(mm));
end
The reason why I've even decided to post an answer is because you can do this vectorized without a loop, which I'd also like to write an answer for. What you can do place each number into a character array where each row denotes a unique number, and you can use text to print out all numbers simultaneously.
m = sprintfc('%2d', 1:57);
d = reshape([m{:}], 2, 57).';
text(x, y, d);
The (undocumented!) function sprintfc takes a formatting specifier and an array and creates a cell array of strings where each cell is the string version of each element in the array you supply. In order to ensure that the character array has the same number of columns per row, I ensure that each string takes up 2 characters, and so any number less than 10 will have a blank space at the beginning. I then convert the cell array of strings into a character array by converting the cell array into a comma-separated list of strings and I reshape the matrix into an acceptable form, and then I call text with all of the pairs of x and y, with the corresponding labels in m together on the screen.
I am trying to convert 14 bit hex numbers to decimal.
I have this VBA code.
Option Explicit
Public Function HexadecimalToDecimal(HexValue As String) As Double
Dim ModifiedHexValue As String
ModifiedHexValue = Replace(HexValue, "0x", "&H")
HexadecimalToDecimal = CDec(ModifiedHexValue)
End Function
With numbers like this to convert to decimal
0x047B1142591E80
0x044A81325A1E80
0x047B7542591E80
I keep getting random results across large amounts of data. Sometimes spot on other times the numbers are off by 6 or 2.
Try changing the return type of the function from Double to Variant. Double has only about 15 decimal digits of precision, so can't, for example, capture the value 1261213964639872 (which has 16 digits) exactly. The closest it can get is 1261213964639870. By changing the return type to Variant, the full precision returned by CDec will be preserved. You can't use a Decimal return type, because VBA for some reason does not support this.
The problem isn't with VBA. Excel cells can only hold 15 digits in number format. So the "number" 1234567891234567 will always display 1234567891234560. This can be avoided by converting items to text AND/OR changing the cell format to text.
But this doesn't always work.
The only surefire way to make sure it will retain all digits is to append something onto the string that isn't a number.
This code will append an apostrophe before the number, but return the entire string.
Public Function HexadecimalToDecimal(HexValue As String) As String
Dim ModifiedHexValue As String
ModifiedHexValue = Replace(HexValue, "0x", "&H")
HexadecimalToDecimal = "'" & CDec(ModifiedHexValue)
End Function
Unfortunately, not a perfect solution.
I am trying to use the Mid function in Excel, but I am stuck.
In column A, there is a string that looks like this: _string_unwanted
where _ equals a space, "string" is a string of variable length that i want to extract, _ is the 2nd space, and "unwanted" is a string of variable length that i don't care about.
Here's my function:
= Mid(A1, 2, __)
I put "2" to start at the 2nd character spot since I want to ignore the 1st space. What do I put in _ such that it finds the string of variable length? Meaning, it just isolates the "string" part of the above string..and ignores the 2 spaces, and the "unwanted" string?
Are you looking for this:-
=MID(A1,2,LEN(A1)-1)
Result:
string_unwanted
EDIT:
This is the formula:-
=MID(A1,2,FIND("_",A1,2)-2)
Result:
string
How about:
=MID(A1,2,LEN(A1)-LEN("_unwanted")-1)