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I have a Excel file with two columns. One has a name other has the corresponding mass to it. I have used the corresponding lines to read it and find the position of the name. But when I am trying to find the mass to the corresponding name as shown below it is not able to store it in the memory. In the Excel file, I have the mass values as 1.989*10^30. This seems to affect the code as the same code works fine when the cells in the excel has just numeric values.
majbod = 'Sun';
minbod = 'Earth';
majbodin = readtable("Major_and_Minor_Bodies.xlsx","Sheet",1);
minbodin = readtable("Major_and_Minor_Bodies.xlsx","Sheet",2);
MAJORBODY = table2array(majbodin(:,"Major_Body"));
MINORBODY = table2array(minbodin(:,"Minor_Body"));
mmaj = table2array(majbodin(:,"Mass"));
mmin = table2array(minbodin(:,"Mass"));
selected_majbody = find(strcmp(MAJORBODY,majbod));
selected_minbody = find(strcmp(MINORBODY,minbod));
M = mmaj(selected_majbody);
m = mmin(selected_minbody);
disp([M ;m])
Is there a better way to write the code compared to the way which I wrote?
Thanks.
Excel does it's best to figure out what kind of data is in each cell. Since your data has something besides just numbers, Excel treats it like a string. You have a couple of options for getting around that:
If you put an equals sign in front of it, it will treat it like an equation, and calculate the value of 1.989*10^3 for you. this will be a number.
Since scientific notation is so common, programmers have created a shortcut for representing it. They often use the character 'E' where you use "*10^". This means that if you type "1.989E30", excel will recognize that as a number.
If keeping the current string format is very important, you could probably modify the string during extraction - replace '*10^' with E, and then whatever language you are using will have a string to number parser you can use.
If the real problem is that the real numbers are just too long in Excel, you can always format the cell that they are in. (right click the cell, select format cells, then select scientific.)
Good luck
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've tried lots of searches for this but I'm still not coming up with anything that works.
I have a range of strings in Column A
Amend.Clause_1.1.AddMCQ
Amend.Clause_1.1.AddNo
Amend.Clause_1.1.AddRepeat
Amend.Clause_1.13.AddRepeat
Amend.Clause_1.13.AddTitle
Amend.Clause_1.13.AddUTQ
Amend.Clause_2.8.Heading_Edit
Amend.Clause_2.8.MCQ
Amend.Clause_2.8.Remove
Amend.Clause_4.26.AddUTQ
Amend.Clause_4.26.Heading_Edit
Amend.Clause_4.26.MCQ
Amend.Clause_5.15.AddMCQ
Amend.Clause_5.15.AddNo
Amend.Clause_5.15.AddRepeat
As you can see, the numbers always start in the same place, after the underscore "_" at position 13.
I need to extract the decimal numbers from these strings into a new column so I'm left with 1.1, 1.13, 1.14, 4.26 etc.
I've tried all sorts of combos of MID, LEFT, LEN, RIGHT but to no avail, trying to find the position of the last period.
Could anyone explain how to accomplish this? Ideally I'd like to do this without VBA.
Thanks
Here you are:
=VALUE(MID(A1,SEARCH("_",A1)+1,SEARCH(".",A1,SEARCH(".",A1,SEARCH("_",A1)+1)+1)-(SEARCH("_",A1)+1)))
Here's what inside =VALUE(MID(...)):
A1 - the whole string itself
SEARCH("_",A1)+1 - find the number starting position - right after "_".
SEARCH(".",A1,SEARCH(".",A1,SEARCH("_",A1)+1)+1)-(SEARCH("_",A1)+1) - find number length - position of second "." after first "." minus number starting position.
Try with three functions:
=MID(A1,14,FIND("#",SUBSTITUTE(A1,".","#",3))-14)
Try this - If the position of _ is not necessarily 13.
=MID(A1,FIND("_",A1,1)+1,FIND("¬¬",SUBSTITUTE(A1,".","¬¬",LEN(A1)-LEN(SUBSTITUTE(A1,".",""))))-FIND("_",A1,1)-1)
Or this if the _ is always 13
=MID(A1,14,FIND("¬¬",SUBSTITUTE(A1,".","¬¬",LEN(A1)-LEN(SUBSTITUTE(A1,".",""))))-14)
Use This:
=VALUE(TRIM(LEFT(SUBSTITUTE(RIGHT(A1;LEN(A1)-FIND("_";A1));".";REPT(" ";LEN(A1));2);LEN(A1))))
assuming value is in A1
Far from ideal, but with a shorter formula than the solutions offered so far:
=SUBSTITUTE(A1,".","_",3)
Catch is that formulae would then need to be converted to values, parsed with delimiter _ (being careful to ensure Column data format is Text) and surplus columns deleted.
When the string Amend.Clause_1.1.AddMCQ is in A1
=Find(".",A1,Find(".",A1)+1)
will give the position of the second decimal point, then you should be able to extract the decimal number.
The syntax is
FIND(find_text, within_text, [start_num])
I have a sheet with a list of names in Column B and an ID column in A. I was wondering if there is some kind of formula that can take the value in column B of that row and generate some kind of ID based on the text? Each name is also unique and is never repeated in any way.
It would be best if I didn't have to use VBA really. But if I have to, so be it.
Solution Without VBA.
Logic based on First 8 characters + number of character in a cell.
= CODE(cell) which returns Code number for first letter
= CODE(MID(cell,2,1)) returns Code number for second letter
= IFERROR(CODE(MID(cell,9,1)) If 9th character does not exist then return 0
= LEN(cell) number of character in a cell
Concatenating firs 8 codes + adding length of character on the end
If 8 character is not enough, then replicate additional codes for next characters in a string.
Final function:
=CODE(B2)&IFERROR(CODE(MID(B2,2,1)),0)&IFERROR(CODE(MID(B2,3,1)),0)&IFERROR(CODE(MID(B2,4,1)),0)&IFERROR(CODE(MID(B2,5,1)),0)&IFERROR(CODE(MID(B2,6,1)),0)&IFERROR(CODE(MID(B2,7,1)),0)&IFERROR(CODE(MID(B2,8,1)),0)&LEN(B2)
Sorry, I didn't found a solution with formula only even if this thread might help (trying to calculate the points in a scrabble game) but I didn't find a way to be sure the generated hash would be unique.
Yet, here is my solution, based on a UDF (Used-Defined Function):
Put the code in a module:
Public Function genId(ByVal sName As String) As Long
'Function to create a unique hash by summing the ascii value of each character of a given string
Dim sLetter As String
Dim i As Integer
For i = 1 To Len(sName)
genId = Asc(Mid(sName, i, 1)) * i + genId
Next i
End Function
And call it in your worksheet like a formula:
=genId(A1)
[EDIT] Added the * i to take into account the order. It works on my unit tests
May be OTT for your needs, but you can use a call to CoCreateGuid to get a real GUID
Private Declare Function CoCreateGuid Lib "ole32" (ID As Any) As Long
Function GUID() As String
Dim ID(0 To 15) As Byte
Dim i As Long
If CoCreateGuid(ID(0)) = 0 Then
For i = 0 To 15
GUID = GUID & Format(Hex$(ID(i)), "00")
Next
Else
GUID = "Error while creating GUID!"
End If
End Function
Test using
Sub testGUID()
MsgBox GUID
End Sub
How to best implement depends on your needs. One way would be to write a macro to get a GUID populate a column where names exist. (note, using it as a udf as is is no good, since it will return a new GUID when recalculated)
EDIT
See this answer for creating a SHA1 hash of a string
Do you just want an incrementing numeric id column to sit next to your values? If so, and if your values will always be unique, you can very easily do this with formulae.
If your values were in column B, starting in B2 underneath your headers for example, in A2 you would type the formula "=IF(B2="","",1+MAX(A$1:A1))". You can copy and paste that down as far as your data extends, and it will increment a numeric identifier for each row in column B which isn't blank.
If you need to do anything more complicated, like identify and re-identify repeating values, or make identifiers 'freeze' once they're populated, let me know. Currently, when you clear or add values to your list the identifers will toggle themselves up and down, so you need to be careful if your data changes.
Unique identifier based on the number of specific characters in text. I used an identifier based on vowels and numbers.
=LEN($J$14)-LEN(SUBSTITUTE($J$14;"a";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"e";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"i";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"j";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"o";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"u";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"y";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"1";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"2";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"3";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"4";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"5";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"6";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"7";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"8";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"9";""))&LEN($J$14)-LEN(SUBSTITUTE($J$14;"0";""))
You say you are confident that there are no duplicate values in your words. To push it further, are you confident that the first 8 characters in any word would be unique?
If so, you can use the below formula. It works by individually taking each character's ASCII code - 40 [assuming normal characters, this puts numbers at between 8 & 57, and letters at between 57 & 122], and multiplying that characters code by 10 ^ [that character's digit placement in the word]. Basically it takes that character code [-40], and concatenates each code onto the next.
EDIT Note that this code no longer requires that at least 8 characters exist in your word to prevent an error, as the actual word to be coded has 8 "0"'s appended to it.
=TEXT(SUM((CODE(MID(LOWER(RIGHT(REPT("0",8)&A3,8)),{1,2,3,4,5,6,7,8},1))-40)*10^{0,2,4,6,8,10,12,14}),"#")
Note that as this uses the ASCII values of the characters, the ID # could be used to identify the name directly - this does not really create anonymity, it just turns 8 unique characters into a unique number. It is obfuscated with the -40, but not really 'safe' in that sense. The -40 is just to get normal letters and numbers in the 2 digit range, so that multiplying by 10^0,2,4 etc. will create a 2 digit unique add-on to the created code.
EDIT FOR ALTERNATIVE METHOD
I had previously attempted to do this so that it would look at each letter of the alphabet, count the number of times it appears in the word, and then multiply that by 10*[that letter's position in the alphabet]. The problem with doing this (see comment below for formula) is that it required a number of 10^26-1, which is beyond Excel's floating point precision. However, I have a modified version of that method:
By limiting the number of allowed characters in the alphabet, we can get the max total size possible to 10^15-1, which Excel can properly calculate. The formula looks like this:
=RIGHT(REPT("0",15)&TEXT(SUM(LEN(A3)*10^{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}-LEN(SUBSTITUTE(A3,MID(Alphabet,{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15},1),""))*10^{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}),"#"),15)
[The RIGHT("00000000000000"... portion of the formula is meant to keep all codes the same number of characters]
Note that here, Alphabet is a named string which holds the characters: "abcdehilmnorstu". For example, using the above formula, the word "asdf" counts the instances of a, s, and d, but not 'f' which isn't in my contracted alphabet. The code of "asdf" would be:
001000000001001
This only works with the following assumptions:
The letters not listed (nor numbers / special characters) are not required to make each name unique. For example, asdf & asd would have the same code in the above method.
And,
The order of the letters is not required to make each name unique. For example, asd & dsa would have the same code in the above method.
I have price column in which the prices are displayed in 3 decimals and 4 decimals places, for example 123.456 or 123.4357.
So irrespective of the number of decimal places i want the value to be with only two decimals.
So i am selecting the column and in the VBA i am using Selection.NumberFormat = "0.00"
Which is resulting me the rounded value i.e when i format 123.456 and 123.4357 to 0.00 i am getting 123.46 and 123.44 but i want it to be 123.46 and 123.43.
So just wondering is there a way we can just trim the values instead of rounding.
Please give me some sample examples.
Thank you in advance.
Excel has a built-in function, trunc, that should do the trick for you.
This is what I placed in a1:b2.
123.456 =trunc(A1,2)
123.4357 =trunc(A2,2)
This will display
123.456 123.45
123.4357 123.43
You could treat the result as a string and extract the number of characters you need as in:
dim a as double
dim s as string
a = 123.4357
s = MID(a,1,FIND(".",a))&MID(a,FIND(".",a)+1,2)
I don't believe there is a built in mask that will truncate without rounding, you could instead use another column containing =TRUNC(A1, 2).