How do I get a random number in Lua to the eighth decimal?
Example : 0.00000001
I have tried the following and several variations of this but can not get the format i need.
math.randomseed( os.time() )
x = math.random(10000000,20000000) * 0.00000001
print(x)
i would like to put in say 200 and get this 0.00000200
Just grab a random number from 0-9, and slide it down 6 places. You can use format specifiers to create the string representation of the number that you desire. For floats we use %f, and indicate how many decimal places we want to have with an intermediate .n, where n is a number.
math.randomseed(os.time())
-- random(9) to exclude 0
print(('%.8f'):format(math.random(0, 9) * 1e-6))
--> '0.00000400'
string.format("%.8f",math.random())
to help anyone else. my question should have been worded a bit better. i wanted to be able to get random numbers and get it to the 8th decimal place.
but i wanted to be able to have those numbers from 1-10,000 so he is updated how i wanted it and the help of Oka got me to this
math.randomseed(os.time())
lowest = 1
highest = 7000
rand=('%.8f'):format(math.random(lowest, highest) / 100000000)
print(rand)
Hope this helps someone else or if it can be cleaned up please let me know
Related
This question already has an answer here:
Dealing with big numbers in Lua
(1 answer)
Closed 1 year ago.
When using lua to handle floating point numbers I found that lua can handle very limited precision, for example:
print(3.14159265358979)
output:
3.1415926535898
The result will be missing a few decimal places, which will lead to calculation bias. How can I deal with such a lack of precision
By default, Lua only displays 14 digits of a number. A float can require 15 to 17 digits to be represented exactly as a base-10 string. We can use a loop to find the right number of digits. Note that %g will drop the trailing zeros, so we can start our search at 15 digits, not 1. This is the function I use:
local function floatToString(x)
for precision = 15, 17 do
-- Use a 2-layer format to try different precisions with %g.
local s <const> = ('%%.%dg'):format(precision):format(x)
-- See if s is an exact representation of x.
if tonumber(s) == x then
return s
end
end
end
print(floatToString(3.14159265358979))
Output: 3.14159265358979
def digits(n):
total=0
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))=>1:
total+=i
else:
total+=0
return total
I want to find the number of digits in 13 so I do the below
print digits(13)
it gives me $\0$ for every number I input into the function.
there's nothing wrong with what I've written as far as I can see:
if a number has say 4 digits say 1234 then dividing by 10^4 will make it less than 1: 0.1234 and dividing by 10^3 will make it 1.234
and by 10^3 will make it 1.234>1. when i satisfies BOTH conditions you know you have the correct number of digits.
what's failing here? Please can you advise me on the specific method I've tried
and not a different one?
Remember for every n there can only be one i which satisfies that condition.
so when you add i to the total there will only be i added so total returning total will give you i
your loop makes no sense at all. It goes from 0 to exact number - not what you want.
It looks like python, so grab a solution that uses string:
def digits(n):
return len(str(int(n))) # make sure that it's integer, than conver to string and return number of characters == number of digits
EDIT:
If you REALLY want to use a loop to count number of digits, you can do this this way:
def digits(n):
i = 0
while (n > 1):
n = n / 10
++i
return i
EDIT2:
since you really want to make your solution work, here is your problem. Provided, that you call your function like digits(5), 5 is of type integer, so your division is integer-based. That means, that 6/100 = 0, not 0.06.
def digits(n):
for i in range(0,n):
if n/float(10**(i))<1 and n/float(10**(i-1))=>1:
return i # we don't need to check anything else, this is the solution
return null # we don't the answer. This should not happen, but still, nice to put it here. Throwing an exception would be even better
I fixed it. Thanks for your input though :)
def digits(n):
for i in range(0,n):
if n/(10**(i))<1 and n/(10**(i-1))>=1:
return i
HI I know there have been may question about this here but I wasn't able to find a detailed enough answer, Wikipedia has two examples of ISIN and how is their checksum calculated.
The part of calculation that I'm struggling with is
Multiply the group containing the rightmost character
The way I understand this statement is:
Iterate through each character from right to left
once you stumble upon a character rather than digit record its position
if the position is an even number double all numeric values in even position
if the position is an odd number double all numeric values in odd position
My understanding has to be wrong because there are at least two problems:
Every ISIN starts with two character country code so position of rightmost character is always the first character
If you omit the first two characters then there is no explanation as to what to do with ISINs that are made up of all numbers (except for first two characters)
Note
isin.org contains even less information on verifying ISINs, they even use the same example as Wikipedia.
I agree with you; the definition on Wikipedia is not the clearest I have seen.
There's a piece of text just before the two examples that explains when one or the other algorithm should be used:
Since the NSIN element can be any alpha numeric sequence (9 characters), an odd number of letters will result in an even number of digits and an even number of letters will result in an odd number of digits. For an odd number of digits, the approach in the first example is used. For an even number of digits, the approach in the second example is used
The NSIN is identical to the ISIN, excluding the first two letters and the last digit; so if the ISIN is US0378331005 the NSIN is 037833100.
So, if you want to verify the checksum digit of US0378331005, you'll have to use the "first algorithm" because there are 9 digits in the NSIN. Conversely, if you want to check AU0000XVGZA3 you're going to use the "second algorithm" because the NSIN contains 4 digits.
As to the "first" and "second" algorithms, they're identical, with the only exception that in the former you'll multiply by 2 the group of odd digits, whereas in the latter you'll multiply by 2 the group of even digits.
Now, the good news is, you can get away without this overcomplicated algorithm.
You can, instead:
Take the ISIN except the last digit (which you'll want to verify)
Convert all letters to numbers, so to obtain a list of digits
Reverse the list of digits
All the digits in an odd position are doubled and their digits summed again if the result is >= 10
All the digits in an even position are taken as they are
Sum all the digits, take the modulo, subtract the result from 0 and take the absolute value
The only tricky step is #4. Let's clarify it with a mini-example.
Suppose the digits in an odd position are 4, 0, 7.
You'll double them and get: 8, 0, 14.
8 is not >= 10, so we take it as it is. Ditto for 0. 14 is >= 10, so we sum its digits again: 1+4=5.
The result of step #4 in this mini-example is, therefore: 8, 0, 5.
A minimal, working implementation in Python could look like this:
import string
isin = 'US4581401001'
def digit_sum(n):
return (n // 10) + (n % 10)
alphabet = {letter: value for (value, letter) in
enumerate(''.join(str(n) for n in range(10)) + string.ascii_uppercase)}
isin_to_digits = ''.join(str(d) for d in (alphabet[v] for v in isin[:-1]))
isin_sum = 0
for (i, c) in enumerate(reversed(isin_to_digits), 1):
if i % 2 == 1:
isin_sum += digit_sum(2*int(c))
else:
isin_sum += int(c)
checksum_digit = abs(- isin_sum % 10)
assert int(isin[-1]) == checksum_digit
Or, more crammed, just for functional fun:
checksum_digit = abs( - sum(digit_sum(2*int(c)) if i % 2 == 1 else int(c)
for (i, c) in enumerate(
reversed(''.join(str(d) for d in (alphabet[v] for v in isin[:-1]))), 1)) % 10)
Ok...I think I'm missing something very obvious here but haven't been able to google myself through this solution. I have two simple rails methods that calculate the number of up votes and down votes. They will always return a fraction because i'm trying to show a percentage (up_vote_count / votal_vote_count). I open the rails console and run the following:
y = #somespecificrecord
then...
y.up_vote_count
This returns 1 as is expected
y.down_vote_count
This returns 1 as is expected
y.total_vote_count
This returns 2 as is expected.
However, when I run in the console...
y.up_vote_count / y.total_vote_count
This returns 0 when it should return .50. I've been reading about floats/integers/decimals, etc and I do see this in the schema on the model i'm working from:
t.float "value", default: 0.0
Is this my problem?...and if so what do I have to do to allow myself to do a simple formula like the one above in rails console that will return the correct decimal rounded to 2 digits (i.e, .50 in this case above). I don't know if I want to run any migrations to change data types because this is a gem (& as a beginner I tend to stay away from customizing code from any gems I'm using). Is there another way? something small i'm missing hopefully?
UPDATE:
I'm learning decimals are slower than floats also, so is there any way to accomplish this with continuing to use t.float "value", default: 0.0
thanks for any help.
1 / 2 = 0.5
With integers this will round down to 0
You can get around this by casting the divisor to a float, forcing it to do division with floating point accuracy.
y.up_vote_count / y.total_vote_count.to_f
Float
Float objects represent inexact real numbers using the native
architecture's double-precision floating point representation.
Floating point has a different arithmetic and is an inexact number.
its important to know if you divide 2 hole numbers you are going to get a hole number.
if you are looking a decimal you should first convert your numbers to a decimal or a "float" like this
up_votes = y.up_vote_count.to_f
total_vote = y.down_vote_count.to_f
(up_votes / total_vote) * 100.0
I hope that this helps
I have two cards which contains a hex value, I am struggling to find out what kind of algorithm is used ti get the decimal value.
8HEX from chip: 0b98c44a Printed on card: 3491308370
8HEX from chip: 0c96425c Printed on card: 812204602
does any one of you number experts in here find out how it is done, you helped me with this before :)
The relation between the numbers seems to be that the bits of every byte has been reversed.
When you print first pair in binary they are:
11010000 00011001 00100011 01010010 = 3491308370
00001011 10011000 11000100 01001010 = 0x0b98c44a
The second pair is:
00110000 01101001 01000010 00111010 = 812204602
00001100 10010110 01000010 01011100 = 0x0c96425c
If you want to know how to convert one number to the other, you should mention which programming language you are using.