I'm trying to round up a list of numbers to the nearest 5 and nearest 10.
example:
1562
1706
1665
1378
1439
I created this code to round up/off to nearest 5:
:exe "%s/\\d\\d\\d\\d/\\=substitute(submatch(0).'\\.0', '.*', (round(submatch(0)/5)*5), 'g')/g"
In the first part of the substitution I want to make a float from the submatch value adding .0 to the submatch value.
Expected result:
1560
1705
1665
1380
1440
However, it gives a trailing character error.
What did I wrong?
Since / is present in the replacement string as the division operator, you need to use another delimiter character for the :s command. As per documentation, this can be any other single-byte character, but not an alphanumeric character, '\', '"'' or '|'.
To round a number n to the nearest multiple of k you can do:
(n + k/2) / k * k
Putting this all together, the command can be:
:%s!\v\d{4}!\=(submatch(0) + 2) / 5 * 5!
Related
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
Can someone explain this behavior?
Buffer.from('5d9RAjZ2GCob-86_Ql', 'base64url').toString('base64url')
// 5d9RAjZ2GCob-86_Qg
Please take a close look at the last character l - g
Your string is 18 characters long, With 6 bits encoded in each character it means the first 16 characters represent 96 bits (12 bytes) and the last two represent one byte plus 4 unused bits. Only the first two bits of the last character are significant here. g is 100000, l is 100101. As the last 4 characters are not used, g is just the first choice for the two bits 1 0.
So for any character in the range between g and v, you would get a g when you convert it back to Base64Url.
See https://en.wikipedia.org/wiki/Base64#Base64_table_from_RFC_4648
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.
In vim editor, when I try to divide numbers using the command
<C-r>= 320 / 1024
I get 0. How to fix it?
You have to have one of the numbers as a float one:
320 / 1024.0 will give what you want. More info here
What you're trying to do is integer division. While doing integer division some sample results are as follows:
1/3 = 0
5/3 = 1
but converting either the numerator/denominator to a floating point value will return a floating point result.
1.0/3 = 0.333333
5/3.0 = 1.666667
Therefore what you're looking for is 320/1024.0 or 320.0/1024 or 320.0/1024.0
The result of which is 0.3125
I've written a small function in C, which almost do the same work as standart function `fcvt'. As you may know, this function takes a float/double and make a string, representing this number in ANSI characters. Everything works ;-)
For example, for number 1.33334, my function gives me string: "133334" and set up special integer variable `decimal_part', in this example will be 1, which means in decimal part only 1 symbol, everything else is a fraction.
Now I'm curious about what to do standart C function `printf'. It can take %a or %e as format string. Let me cite for %e (link junked):
"double" argument is output in scientific notation
[-]m.nnnnnne+xx
... The exponent always contains two digits.
It said: "The exponent always contains two digits". But what is an Exponent? This is the main question. And also, how to get this 'exponent' from my function above or from `fcvt'.
The notation might be better explained if we expand the e:
[-]m.nnnnnn * (10^xx)
So you have one digit of m (from 0 to 9, but it will only ever be 0 if the entire value is 0), and several digits of n. I guess it might be best to show with examples:
1 = 1.0000 * 10^0 = 1e0
10 = 1.0000 * 10^1 = 1e1
10000 = 1.0000 * 10^4 = 1e4
0.1 = 1.0000 * 10^-1 = 1e-1
1,419 = 1.419 * 10^3 = 1.419e3
0.00000123 = 1.23 * 10^-5 = 1.23e-5
You can look up scientific notation off Google, but it is useful for expressing very large or small numbers like 1232100000000000000 would be 1.2321e24 (I didn't actually count, exponent may be inaccurate).
In C, I think you can actually extract the exponent from the top 12 bits (the first being the sign which you will have to ignore). See: IEEE758-1985 Floating Point
The exponent is the power 10 is raised to then multiplied by the base.
SI is explained at wikipeida. http://en.wikipedia.org/wiki/Scientific_notation
m.nnnnnne+xx is logically equal to m.nnnnnn * 10 ^ +xx
In scientific notation, the exponent is the ten to the XX power, so 1234.5678 can be represented as 1.2345678E03 where the normalized form is multiplied by 10^3 to get the "real" answer.
400 = 4 * 10 ^ 2
2 is the exponent.
If you write a number in scientific notation then the exponent is part of that notation.
You can see a full description here http://en.wikipedia.org/wiki/Scientific_notation, but basically its just another way to write a number, typically used for very large or very small numbers.
Say you have the number 300, that is equal to 3 * 100, or 3 * 10^2 in scientific notation.
If you use %e it will be printed as 3.0e+02