I want to add decimal to alpha numeric values of my variable ucod. UCOD values are as follows:
I want to add decimal point after 2 digits in a 3 digit alpha-numeric values. Kindly help whether it is possible to do this.
ucod
.B182 >>>B18.2
.I251 >>>I25.1
.F03 >>>F03
.C55 >>>C55
.J449 >>>J44.9
.N390 >>>N39.0
If all values are exactly EITHER one letter followed by two digits OR one letter followed by three digits, and only the latter should have a decimal point after the first two digits, then this code will work for you:
* Example generated by -dataex-. For more info, type help dataex
clear
input str6 varA
".B182"
".I251"
".F03"
".C55"
".J449"
".N390"
end
* Remove leading decimal point
gen varB = substr(varA,2,.)
* If string length is 4, take the first 3 charachters, then add a decimal,
* and then add the remaining part of the string and
* then replace the value with the results
replace varB = substr(varB,1,3) + "." + substr(varB,4,.) if strlen(varB) == 4
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)
I am trying to understand the maths in this code that converts binary to decimal. I was wondering if anyone could break it down so that I can see the working of a conversion. Sorry if this is too newb, but I've been searching for an explanation for hours and can't find one that explains it sufficently.
I know the conversion is decimal*2 + int(digit) but I still can't break it down to understand exaclty how it's converting to decimal
binary = input('enter a number: ')
decimal = 0
for digit in binary:
decimal= decimal*2 + int(digit)
print(decimal)
Here's example with small binary number 10 (which is 2 in decimal number)
binary = 10
for digit in binary:
decimal= decimal*2 + int(digit)
For for loop will take 1 from binary number which is at first place.
digit = 1 for 1st iteration.
It will overwrite the value of decimal which is initially 0.
decimal = 0*2 + 1 = 1
For the 2nd iteration digit= 0.
It will again calculate the value of decimal like below:
decimal = 1*2 + 0 = 2
So your decimal number is 2.
You can refer this for binary to decimal conversion
The for loop and syntax are hiding a larger pattern. First, consider the same base-10 numbers we use in everyday life. One way of representing the number 237 is 200 + 30 + 7. Breaking it down further, we get 2*10^2 + 3*10^1 + 7*10^0 (note that ** is the exponent operator in Python, but ^ is used nearly everywhere else in the world).
There's this pattern of exponents and coefficients with respect to the base 10. The exponents are 2, 1, and 0 for our example, and we can represent fractions with negative exponents. The coefficients 2, 3, and 7 are the same as from the number 237 that we started with.
It winds up being the case that you can do this uniquely for any base. I.e., every real number has a unique representation in base 10, base 2, and any other base you want to work in. In base 2, the exact same pattern emerges, but all the 10s are replaced with 2s. E.g., in binary consider 101. This is the same as 1*2^2 + 0*2^1 + 1*2^0, or just 5 in base-10.
What the algorithm you have does is make that a little more efficient. It's pretty wasteful to compute 2^20, 2^19, 2^18, and so on when you're basically doing the same operations in each of those cases. With our same binary example of 101, they've re-written it as (1 *2+0)*2+1. Notice that if you distribute the second 2 into the parenthesis, you get the same representation we started with.
What if we had a larger binary number, say 11001? Well, the same trick still works. (((1 *2+1 )*2+0)*2+0)*2+1.
With that last example, what is your algorithm doing? It's first computing (1 *2+1 ). On the next loop, it takes that number and multiplies it by 2 and adds the next digit to get ((1 *2+1 )*2+0), and so on. After just two more iterations your entire decimal number has been computed.
Effectively, what this is doing is taking each binary digit and multiplying it by 2^n where n is the place of that digit, and then summing them up. The confusion comes due to this being done almost in reverse, let's step through an example:
binary = "11100"
So first it takes the digit '1' and adds it on to 0 * 2 = 0, so we
have digit = '1'.
Next take the second digit '1' and add it to 1* 2 =
2, digit = '1' + '1'*2.
Same again, with digit = '1' + '1'*2 +
'1'*2^2.
Then the 2 zeros add nothing, but double the result twice,
so finally, digit = '0' + '0'*2 + '1'*2^2 + '1'*2^3 + '1'*2^4 = 28
(I've left quotes around digits to show where they are)
As you can see, the end result in this format is a pretty simple binary to decimal conversion.
I hope this helped you understand a bit :)
I will try to explain the logic :
Consider a binary number 11001010. When looping in Python, the first digit 1 comes in first and so on.
To convert it to decimal, we will multiply it with 2^7 and do this till 0 multiplied by 2^0.
And then we will add(sum) them.
Here we are adding whenever a digit is taken and then will multiply by 2 till the end of loop. For example, 1*(2^7) is performed here as decimal=0(decimal) +1, and then multiplied by 2, 7 times. When the next digit(1) comes in the second iteration, it is added as decimal = 1(decimal) *2 + 1(digit). During the third iteration of the loop, decimal = 3(decimal)*2 + 0(digit)
3*2 = (2+1)*2 = (first_digit) 1*2*2 + (seconds_digit) 1*2.
It continues so on for all the digits.
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.
In matlab, how can I turn a string or cell of digits into a vector of numbers, where each digit in the string is an element in the vector.
That is, for eg., how to turn this:
A=3141592;
(where class(A)=char)
into this:
A=[3 1 4 1 5 9 2];
(where class(A)=double)
This is related to this question
Subtract ascii value of '0' from each of the ascii characters that constitute the string in A to get the double array -
A-'0'
Straight away plugging in the ascii value would work too -
A-48
Output -
ans =
3 1 4 1 5 9 2
In excel I am using the left,mid and Right functions to pull the 'suffix' of a string.
Example:
1234-1234567-1234
The prefix is 4 digits long
The Base is 7 or 8 digits long
and the Suffix is either 3 or 4 digits long.
I have the right formula as: =RIGHT(A6,LEN(A6)-FIND("-",A6)-8) to handle the varying lengths of the suffix
I need the MID formula that pulls the base section that can handle the varying lengths of the base and suffix.
Given
The prefix is 4 digits long The Base is 7 or 8 digits long....
then you can use this formula
=MID(A1,6,8-ISERR(MID(A1,13,1)+0))
Please try:
=MID(A1,1+FIND("-",A1),FIND("-",MID(A1,1+FIND("-",A1),9))-1)
(just for the part between the hyphens).
But Text to Columns with - as delimiter might be more convenient.
You could try:
=MID(A1,6,FIND("-",A1,6)-FIND("-",A1,1)-1)