I am finding a formula which can convert a 16 bit binary number into two separate decimal number
0000000110010000 -> 0x0190
I want the decimal number to be 1 and 144
I have 50 columns(say M1 to M50) of binary numbers so need to make a generic formula for this
If M1 contains your binary number (as a text string), then use
=BIN2DEC(LEFT(M1, 8))
to extract the left part
and
=BIN2DEC(RIGHT(M1, 8))
to extract the right part.
If you want the result in the same cell then use something like
=BIN2DEC(LEFT(M1, 8)) & "|" & BIN2DEC(RIGHT(M1, 8))
where the | is an arbitrary separator, which you can change or omit to suit personal taste.
are they all exactly 16 characters long? You could do:
=BIN2DEC(RIGHT(M1,8))
=BIN2DEC(LEFT(M1,8))
Related
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)
Somewhat simple problem:
I need to turn a column A, which contains numbers with up to 1 decimal (20, 142, 2.5, etc.) to a string with a specific format, namely 8 whole digits and 6 decimal digits but without the actual decimal period, like so:
1 = 00000001000000
13 = 00000013000000
125 = 00000125000000
46.5 = 00000046500000
For what it's worth, the input data from column A will never be more than 3 total digits (0.5 to 999) and the decimal will always be either none or .5.
I also need for Excel to leave the zeroes alone instead of auto-formatting as a number and removing the ones at the beginning of the string.
As a makeshift solution, I've been using =CONCATENATE("'",TEXT(A1,"00000000.000000")), then copying the returning value and "pasting as value" where I actually need it.
It works fine, but I was wondering if there was a more direct solution where I don't have to manually intervene.
Thanks in advance!
=TEXT(A1*1000000,"0000000000000") I think that's what you mean.
I'm trying DEC2HEX(1000000000050000000) but it comes out as #NUM! as the number is too large for this function.
Is there another function I could use to turn this number into hexadecimal?
If you want to convert a decimal number to a 64 bit hex string, assuming that the decimal number is in cell A1 you can use the following:
=CONCATENATE(DEC2HEX(A1/2^32),DEC2HEX(MOD(A1,2^32),8))
This will work up to decimal value of 18,446,744,073,709,500,000 or hex value of 0xfffffffffffff800.
Bonus:
To convert from hex string to decimal, assuming that the 64bit hex string is in cell A1 and contains 16-characters then you can use the following:
=HEX2DEC(LEFT(A1,8))*2^32+HEX2DEC(RIGHT(A1,8))
You can adjust the number of characters in the LEFT(text,[num_chars]) to better suit your needs.
If your hex string has a 0x then you can use the following:
=HEX2DEC(MID(A1,3,8))*2^32+HEX2DEC(RIGHT(A1,8))
I found a simple solution for converting HEX to DEC and vice versa without the limits of characters.
HEX to DEC: use DECIMAL(input number or cell coordinates, input base number)
Case 1: I want to convert hex value "3C" to decimal, the formula is DECIMAL(3C, 16).
Case 2: I want to convert binary value "1001" to decimal, the formula is DECIMAL(1001, 2).
DEC to HEX: use BASE(input number or cell coordinates, output base number)
Case 1:I want to convert number value "1500" to hexadecimal, the formula is BASE(1500, 16)
Case 2:I want to convert number value "1500" to binary, the formula is BASE(1500, 2)
The DEC2HEX function has a limit of 549,755,813,887, try this formula it works for numbers up to 281,474,976,710,655.
=DEC2HEX(A7/(16^9),3)&DEC2HEX(MOD(A7,16^9),9)
There is a free add-in available that will handle that: Xnumbers
Seems to work OK:
=cvDecBase("1000000000050000000",16) --> DE0B6B3AA5EF080
Long formula but it is working for 64-HEX characters:
=HEX2DEC(MID(A24,1,8))*2^512 *(4) +HEX2DEC(MID(A24,9,8))*2^512 *(2) +HEX2DEC(MID(A24,17,8))*2^512+HEX2DEC(MID(A24,25,8))*2^256+HEX2DEC(MID(A24,33,8))*2^128+HEX2DEC(MID(A24,41,8))*2^64+HEX2DEC(MID(A24,49,8))*2^32+HEX2DEC(MID(A24,57,8))
please note: (*4) = *4 (remove brackets) and: (*2) = *2 (remove brackets)
also note: all 64 character must be present like the following example:
0000000000000000000000000000000000000000000000000000000000000fd1
I have an Excel spreadsheet with over 2000 entries:
Field B1: CustomerID as 000012345
Field B2: CustomerID as 0000432
Field C1: CustomerCountry as DE
Field C2: CustomerCountry as IT
I need to build codes 13 digits long including "CustomerCountry" + "CustomerID" without leading 0 + random number (can be 6 digits, more or less, depends in length of CustomerID).
The results should be like this: D1 Code as DE12345967895 or D2 Code as IT43274837401
How to do it with Excel functions?
UPDATED:
I tried this one. My big problem is to say that random number should be long enough to get 13 characters in all. Sometimes CustomerID is just 3 or 4 digits long, and concatenation of three variables can be just 10 or 9 characters. But codes have to be always 13 characters long.
Use & to concatenate strings.
Use VALUE(CustomerID) to trim the leading zeroes from the ID
Use RAND() to add a random number between 0 and 1 or RANDBETWEEN(x,y) to create one between x and y.
Combine the above and there you are!
If you always want 13 digits you can use LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID)))) for the random number to ALWAYS be the right length.
total formula
= CustomerCountry
& VALUE(CustomerID)
& LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID))))
=C1 & TEXT(B1,"0") & RIGHT(TEXT(RANDBETWEEN(0,99999999999),"00000000000"),11 - LEN(TEXT(B1,"0")))
that should do it
I don’t understand what is where and OP has accepted answer so have not bothered testing:
=LEFT(RIGHT(C1,2)&VALUE(MID(B1,15,13))&RANDBETWEEN(10^9,10^10),13)
(but I might revert to this if no one else picks the flaws in it first!)
I am putting a string into excel. The string is often only numeric digits but can have alpha characters or hypens etc.
When I don't set the number format or set it like this
(Where xlSheet(0) is Excel.Worksheet)
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "#"
It outputs in scientific notation.
When I use this code:
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "0"
It rounds up the number to the nearest 100,000 so that the last five digits are 0's when they shouldn't be.
Should be: 1539648751235678942
But is: 1539648751235600000
The cells that have a hyphen or a letter aren't affected and work fine.
Any help would be greatly appreciated.
EDIT:
I add the data like this:
I loop through and put in xlSheet(0).Cells(i, 14) = rs!value_number
Where rs is my ADODB.Recordset
EDIT2: Herbert Sitz got it by adding an apostrophe before the text! Thanks everyone.
I think problem is that the number you're trying to enter can't be accommodated exactly by Excel. Excel has limitations on what numbers it display/represent because of the way numbers are stored internally. In Excel's case numbers are limited to 15 digit precision (see http://office.microsoft.com/en-us/excel-help/excel-specifications-and-limits-HP010073849.aspx ), which is not enough to represent your number.
You can enter the number as a string ("152..42") and all digits will be displayed, but you won't be able to perform exact mathematical operations with it.
For numbers, Excel can only handle 15 significant digits.
If you want to store a number that is more than 15 digits long without losing data, you have to store the data as text.
Doing what you've been doing will resolve the issue:
You can do either of the following to add your numbers as text:
xlSheet(0).Cells(i, 14).Numberformat = "#"
xlSheet(0).Cells(i, 14) = rs!value_number
Or
xlSheet(0).Cells(i, 14) = "'" & rs!value_number