I would like to count '01' sequence in 5760 binary bits.
First, I would like to combine several binary numbers then count # of '01' occurrences.
For example, I have 64 bits integer. Say, 6291456. Then I convert it into binary. Most significant 4 bits are not used. So I'll get 60 bits binary 000...000011000000000000000000000
Then I need to combine(just put bits together since I only need to count '01') first 60 bits + second 60 bits + ...so 96 of 60 bits are stitched together.
Finally, I want to count how many '01' appears.
s = binToString(5760 binary bits)
cnt = s.count('01');
num = 6291226
binary = format(num, 'b')
print(binary)
print(binary.count('01'))
If I use number given by you i.e 6291456 it's binary representation is 11000000000000000000000 which gives 0 occurrences of '01'.
If you always want your number to be 60 bits in length you can use
binary = format(num,'060b')
It will add leading 0 to make it of given length
Say that nums is your list of 96 numbers, each of which can be stored in 64 bits. Since you want to throw away the most 4 significant bits, you are really taking the number modulo 2**60. Thus, to count the number of 01 in the resulting string, using the idea of #ShrikantShete to use the format function, you can do it all in one line:
''.join(format(n%2**60,'060b') for n in nums).count('01')
I am trying to add in cell data validation for a string length to be between 8 and 16 and the max byte length less than 40(UTF8 encoding).
I created a data validation using the excel active support:
Data validation(data tab -> Data Validation (between Remove Duplicates and Consolidate in excel 2016 mac)), In Settings tab, there is validation
criteria:
Validation Criteria:
Allow: Text Length
Data : between
Min : 8 & Max : 16
Though the above validation satisfies all the restrictions i have(8
For other languages(say Japanese), though the string length is being counted though physical length(Eg : "こんにちはこんにちはこんにちは", hellohellohello in Japanese), the UTF8 byte value is 45 bytes, which is the violation of the 40 bytes, thought the length is only 15.
I found "LENB" function in excel, but it is giving the value as 30(instead of 45). I think it is based on different encoding(ansi maybe)
I found the UNICODE function which gives the unicode number of the first character(12371) in the above case. But i don't see how can i get the byte value from this number(3 bytes is the value for the first character(こ)).
Any help in this regard will be greatly appreciated.
I faced the same issue and here is the solution without VBA based on the answer above + this article. Assuming you have a string in A1:
=SUM(
IF(UNICODE(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))<128, 1,
IF(UNICODE(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))<2048, 2,
IF(UNICODE(MID(A1,ROW(INDIRECT("1:"&LEN(A1))),1))<65536, 3, 4
))))
Don't forget to use array function (CTRL+SHIFT+ENTER) when leaving the cell :)
With the Unicode value, you can compute how many bytes a particular one will take. <128 is 1 byte, else <2048 is 2, else <65536 is 3, else 4.
I have some large numbers in an Excel sheet and I want to convert them to binary.
e.g.
12345678
965321458
-12457896
If we are talking positive number between 0 and 2^32-1 you can use this formula:
=DEC2BIN(MOD(QUOTIENT($A$1,256^3),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^2),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^1),256),8)&DEC2BIN(MOD(QUOTIENT($A$1,256^0),256),8)
NOTE: =DEC2BIN() function cannot handle numbers larger than 511 so as you see my formula breaks your number into four 8-bit chunks, converts them to binary format and then concatenates the results.
Well, theoretically you can extend this formula up to six 8-bit chunks. Maximum precision you can get in Excel is 15 (fifteen) decimal digits. When exceeded, only the most significant 15 digits remain, the rest is rounded. I.e. if you type 12345678901234567 Excel will store it as 12345678901234500. So since 2^48-1 is 15 decimal digits long the number won't get rounded.
Perhaps a simpler option:
For positive numbers only, just use BASE (as in BASE2) for numbers between 0 to 2^53 in Excel. Here are some examples:
=BASE(3,2) # returns 11
=BASE(11,2) # returns 1011
Credit for answer goes here:
https://ask.libreoffice.org/en/question/69797/why-is-dec2bin-limited-to-82bits-in-an-32-and-64-bits-world/
Negative numbers: Come to think of it, negative numbers could be handled as well by building upon howy61's answer. He shifts everything by a power of two (2^31 in his case) to use the 2's complement:
=BASE(2^31+MyNum, 2)
so (using 2^8 for only 8 bits):
=BASE(2^8+(-1),2) # returns 11111111
=BASE(2^8+(-3),2) # returns 11111101
The numbers given by the OP requires more bits, so I'll use 2^31 (could go up to 2^53):
=BASE(2^31+(-12457896),2) # returns 11111111010000011110100001011000
For either positive or negative, both formulas could be coupled in a single IF formula. Here are two ways you could do it that give the same answer, where MyNum is the decimal number you start with:
=IF(MyNum<0, BASE(2^31+MyNum,2), BASE(MyNum, 2))
or
=BASE(IF(MyNum<0, MyNum+2^32, MyNum), 2)
See VBA posted here
' The DecimalIn argument is limited to 79228162514264337593543950245
' (approximately 96-bits) - large numerical values must be entered
' as a String value to prevent conversion to scientific notation. Then
' optional NumberOfBits allows you to zero-fill the front of smaller
' values in order to return values up to a desired bit level.
Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As Variant) As String
DecToBin = ""
DecimalIn = CDec(DecimalIn)
Do While DecimalIn <> 0
DecToBin = Trim$(Str$(DecimalIn - 2 * Int(DecimalIn / 2))) & DecToBin
DecimalIn = Int(DecimalIn / 2)
Loop
If Not IsMissing(NumberOfBits) Then
If Len(DecToBin) > NumberOfBits Then
DecToBin = "Error - Number too large for bit size"
Else
DecToBin = Right$(String$(NumberOfBits, "0") & _
DecToBin, NumberOfBits)
End If
End If
End Function
I just tried the formula above, and found that Microsoft screwed up the DEC2BIN function in another way that keeps the formula from working correctly with negative numbers. Internally, DEC2BIN uses a ten bit result; leading zeroes are dropped from the text result, unless the optional length parameter is used, in which case the required number of leading zeroes are left in the string. But here's the rub: a negative number always starts with a one, so there are no leading zeroes to drop, so DEC2BIN will always show all ten bits! Thus, DEC2BIN(-1,8), which should show 11111111 (eight ones) will instead show 1111111111 (ten ones.)
To fix this, use RIGHT to trim each eight bit chunk to eight bits, dumb as that sounds.
=RIGHT(DEC2BIN(QUOTIENT(A1,256^3),8),8) & RIGHT(...
(I read through the VBA, and it does not have the same problem, but it doesn't look like it will handle negatives at all.)
To add easier to read formatting to Taosique's great answer, you can also break it up into chunks of 4 bits with spaces in between, although the formula grows to be a monster:
=DEC2BIN(MOD(QUOTIENT($A$1,16^7),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^6),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^5),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^4),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^3),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^2),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^1),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^0),16),4)
1101 0100 1111 0110 0011 0001 0000 0001
Of course, you can just use the right half of it, if you're just interested in 16 bit numbers:
=DEC2BIN(MOD(QUOTIENT($A$1,16^3),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^2),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^1),16),4)&" "&DEC2BIN(MOD(QUOTIENT($A$1,16^0),16),4)
0011 0001 0000 0001
While I didn't write this for negatives or decimals, it should be relatively easy to modify. This VBA will convert any super large (or not so large if you want, but that wasn't the point) decimal up to the converted binary result containing up to 32767 digits (maximum string length in VBA).
Enter decimal in cell "A1" as a string, result will be in "B1" as a string.
Dim NBN As String
Dim Bin As String
5 Big = Range("A1")
AA = Len(Big)
For XX = 1 To AA
L1 = Mid(Big, XX, 1) + CRY
CRY = 0
If L1 = 0 Then
FN = "0"
GoTo 10
End If
If Int(L1 / 2) = L1 / 2 Then
FN = L1 / 2
GoTo 10
End If
If Int(L1 / 2) <> L1 / 2 Then
FN = Int(L1 / 2)
CRY = 10
GoTo 10
End If
10 NBN = NBN & FN
Next XX
If Left(NBN, 1) = "0" Then
NBN = Right(NBN, (Len(NBN) - 1))
End If
If CRY = 10 Then Bin = "1" & Bin Else Bin = "0" & Bin
Range("A1") = NBN
Range("A2") = Bin
If Len(NBN) > 0 Then
NBN = ""
CRY = 0
GoTo 5
End If
Someone can find binary shift operations more clear and relevant here
=DEC2BIN(BITRSHIFT($A$1,24),8) & DEC2BIN(MOD(BITRSHIFT($A$1,16),256),8) & DEC2BIN(MOD(BITRSHIFT($A$1,8),256),8) & DEC2BIN(MOD($A$1,256),8)
This formula is for 32-bit values
This vba function solves the problem of binary conversion of numbers greater than 511 that can not be done with WorksheetFunction.dec2bin.
The code takes advantage of the WorksheetFunction.dec2bin function by applying it in pieces.
Function decimal2binary(ByVal decimal2convert As Long) As String
Dim rest As Long
If decimal2convert = 0 Then
decimal2binary = "0"
Exit Function
End If
Do While decimal2convert > 0
rest = decimal2convert Mod 512
decimal2binary = Right("000000000" + WorksheetFunction.Dec2Bin(rest), 9) + decimal2binary
decimal2convert = (decimal2convert - rest) / 512
Loop
decimal2binary = Abs(decimal2binary)
End Function
=IF(Decimal>-1,BASE(Decimal,2,32),BASE(2^32+(Decimal),2))
Does both positive and negative numbers.
Took a bit LOL. Tech pun.
You're welcome.
Here's another way. It's not with a single formula, but I have tried and converted up to the number 2,099,999,999,999. My first intention was to build a 51 bit counter, but somehow it does not work with numbers beyond the one I mentioned. Download from
http://www.excelexperto.com/content/macros-production/contador-binario-de-51-bits/
I hope it's useful. Regards.
Without VBA and working with negative numbers as well (here: sint16), however, taking much more space:
You can download the excel file here: (sorry, didn't know where to put the file)
int16 bits to decimal.xlsx
or alternatively follow these steps (if your Excel is not in English, use Excel Translator to "translate" the formula into your MS Office language):
Enter the binary number in 4-bit nibbles (A4 = most significant to D4 = least significant) like shown in the screenshot. Enter all 4 digits (even if starting with 0) and format them as "text"
Enter formula in F4:
=IF(NUMBERVALUE(A4)>=1000,TRUE,FALSE)
Enter the letter "A" in G2-J2, "B" in K2-N2, "C" in O2-R2, "D" in S2-V2
Enter "1" in G3, K3, O3 and S3; "2" in H3, L3, P3 and T3; "3" in I3, M3, Q3 and U3; "4" in J3, N3, R3 and V3
In G4, enter:
=MID(INDIRECT(G$2&ROW()),G$3,1)
Copy the formula to H4-V4
In X4, enter:
=IF(G4="1",0,1)
Copy X4 to Y4-AM4
In BD3 enter "1"
In BC4, enter:
=IF((AM$4+BD3)=2,1,0)
IN BD4, enter:
=IF((AM$4+BD3)=2,0,IF((AM$4+BD3)=1,1,0))
Copy BD4 and BD4 and insert it 15 times diagonally one row further down and one column further left (like in the screenshot), i.e. insert it to BB5 and BC5, then BA6 and BB6, ..., AN19 and AO19.
In AO20, enter "=AO19"; in AP20, enter "=AP18" and so on until BD20 ("=BD4") - i.e. bring down the numbers into one line as seen in the screenshot
In BE20, enter (this is your result):
=IF(F4=FALSE,BIN2DEC(A4&B4)*2^8+BIN2DEC(C4&D4),-1*(BIN2DEC(AO20&AP20&AQ20&AR20&AS20&AT20&AU20&AV20)*2^8+BIN2DEC(AW20&AX20&AY20&AZ20&BA20&BB20&BC20&BD20)))
There maybe a simple solution. I have several 4.2 billion cells that are actually a negative Two's Complement and this works to get the correct value:
=SUM(2^31-(A1-2^31))
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!)