Its a very tiny bit of math and more a question of the most efficient, most elegant way possible.
If I give an integer such as
1.50
or
1.22
or
10.99
How can I get rid of the number to the left of the decimal and output the right side as an integer or float, for example
.50 or 50
.22 or 22
.99 or 99
It matters what the fastest way to do it is. I would rather not turn it into a string if possible.
Thanks for the help.
BR
First of all, 1.50 is not an integer, it's a float. If there's a decimal, it's not an integer. That said, you could get the two digits after the decimal like this:
Precision = 100.
Value = 1.50.
Decimal = trunc(Value * Precision) rem Precision.
If you need more digits, alter the precision variable.
One way is via trunc(Number):
1> X = 10.23 .
10.23
2> Y = X - trunc(X) .
0.23000000000000043
% To get an integer out of the float, you can use `trunc` twice, but you'll have
% to adjust your number first ...
3> Precision = 100 .
100
4> Z = trunc(Y * Precision) .
23
% In one statement:
5> Z = trunc((X - trunc(X)) * Precision) .
Related
I wanted to convert numbers greater than 64 bits, including up to 256 bits number from decimal to hex in lua.
Example:
num = 9223372036854775807
num = string.format("%x", num)
num = tostring(num)
print(num) -- output is 7fffffffffffffff
but if I already add a single number, it returns an error in the example below:
num = 9223372036854775808
num = string.format("%x", num)
num = tostring(num)
print(num) -- error lua54 - bad argument #2 to 'format' (number has no integer representation)
Does anyone have any ideas?
I wanted to convert numbers greater than 64 bits, including up to 256 bits number from decimal to hex in lua.
Well that's not possible without involving a big integer library such as this one. Lua 5.4 has two number types: 64-bit signed integers and 64-bit floats, which are both to limited to store arbitrary 256-bit integers.
The first num in your example, 9223372036854775807, is just the upper limit of int64 bounds (-2^63 to 2^63-1, both inclusive). Adding 1 to this forces Lua to cast it into a float64, which can represent numbers way larger than that at the cost of precision. You're then left with an imprecise float which has no "integer representation" as Lua tells you.
You could trivially reimplement %x yourself, but that wouldn't help you extend the precision/size of floats & ints. You need to find another number representation and find or write a bigint library to go with it. Options are:
String representation: Represent numbers as hex- or bytestrings (base 256).
Table representation: Represent numbers as lists of numbers (base 2^x where x is < 64)
Hi everyone i am trying compare integers by getting the amount between them
Lets say that i for example have a base integer
local i = 100
Then i have other integers that are for example 200 and 300.
I want to get the amount between i and the other to values two see which one is closest to the base integer.
To get the 'distance' between two integers, you can just compute the absolute difference:
local i = 100
local x = 200
print(math.abs(i - x))
print(math.abs(x - i))
The math.abs function gets rid of any negative numbers resulting from the subtraction.
Can somebody explain why multiplying by 100 here gives a less accurate result but multiplying by 10 twice gives a more accurate result?
± % sc
Loading development environment (Rails 3.0.1)
>> 129.95 * 100
12994.999999999998
>> 129.95*10
1299.5
>> 129.95*10*10
12995.0
If you do the calculations by hand in double-precision binary, which is limited to 53 significant bits, you'll see what's going on:
129.95 = 1.0000001111100110011001100110011001100110011001100110 x 2^7
129.95*100 = 1.1001011000010111111111111111111111111111111111111111011 x 2^13
This is 56 significant bits long, so rounded to 53 bits it's
1.1001011000010111111111111111111111111111111111111111 x 2^13, which equals
12994.999999999998181010596454143524169921875
Now 129.95*10 = 1.01000100110111111111111111111111111111111111111111111 x 2^10
This is 54 significant bits long, so rounded to 53 bits it's 1.01000100111 x 2^10 = 1299.5
Now 1299.5 * 10 = 1.1001011000011 x 2^13 = 12995.
First off: you are looking at the string representation of the result, not the actual result itself. If you really want to compare the two results, you should format both results explicitly, using String#% and you should format both results the same way.
Secondly, that's just how binary floating point numbers work. They are inexact, they are finite and they are binary. All three mean that you get rounding errors, which generally look totally random, unless you happen to have memorized the entirety of IEEE754 and can recite it backwards in your sleep.
There is no floating point number exactly equal to 129.95. So your language uses a value which is close to it instead. When that value is multiplied by 100, the result is close to 12995, but it just so happens to not equal 12995. (It is also not exactly equal to 100 times the original value it used in place of 129.95.) So your interpreter prints a decimal number which is close to (but not equal to) the value of 129.95 * 100 and which shows you that it is not exactly 12995. It also just so happens that the result 129.95 * 10 is exactly equal to 1299.5. This is mostly luck.
Bottom line is, never expect equality out of any floating point arithmetic, only "closeness".
Even though Lua does not differentiate between floating point numbers and integers, there are some cases when you want to use integers. What is the best way to covert a number to an integer if you cannot do a C-like cast or without something like Python's int?
For example when calculating an index for an array in
idx = position / width
how can you ensure idx is a valid array index? I have come up with a solution that uses string.find, but maybe there is a method that uses arithmetic that would obviously be much faster. My solution:
function toint(n)
local s = tostring(n)
local i, j = s:find('%.')
if i then
return tonumber(s:sub(1, i-1))
else
return n
end
end
You could use math.floor(x)
From the Lua Reference Manual:
Returns the largest integer smaller than or equal to x.
Lua 5.3 introduced a new operator, called floor division and denoted by //
Example below:
Lua 5.3.1 Copyright (C) 1994-2015 Lua.org, PUC-Rio
>12//5
2
More info can be found in the lua manual
#Hofstad is correct with the math.floor(Number x) suggestion to eliminate the bits right of the decimal, you might want to round instead. There is no math.round, but it is as simple as math.floor(x + 0.5). The reason you want to round is because floats are usually approximate. For example, 1 could be 0.999999996
12.4 + 0.5 = 12.9, floored 12
12.5 + 0.5 = 13, floored 13
12.6 + 0.5 = 13.1, floored 13
local round = function(a, prec)
return math.floor(a + 0.5*prec) -- where prec is 10^n, starting at 0
end
why not just use math.floor()? it would make the indices valid so long as the numerator and denominator are non-negative and in valid ranges.
I've been participating in a programming contest and one of the problems' input data included a fractional number in a decimal format: 0.75 is one example.
Parsing that into Double is trivial (I can use read for that), but the loss of precision is painful. One needs to be very careful with Double comparisons (I wasn't), which seems redundant since one has Rational data type in Haskell.
When trying to use that, I've discovered that to read a Rational one has to provide a string in the following format: numerator % denominator, which I, obviously, do not have.
So, the question is:
What is the easiest way to parse a decimal representation of a fraction into Rational?
The number of external dependencies should be taken into consideration too, since I can't install additional libraries into the online judge.
The function you want is Numeric.readFloat:
Numeric Data.Ratio> fst . head $ readFloat "0.75" :: Rational
3 % 4
How about the following (GHCi session):
> :m + Data.Ratio
> approxRational (read "0.1" :: Double) 0.01
1 % 10
Of course you have to pick your epsilon appropriately.
Perhaps you'd get extra points in the contest for implementing it yourself:
import Data.Ratio ( (%) )
readRational :: String -> Rational
readRational input = read intPart % 1 + read fracPart % (10 ^ length fracPart)
where (intPart, fromDot) = span (/='.') input
fracPart = if null fromDot then "0" else tail fromDot