I'm receiveing in a lua script a hash from redis with numbers in a string format.
1) "30"
2) "30.7"
3) "12.7"
4) "15.7"
5) "20.7"
6) "19.7"
7) "20.5"
8) "21.5"
9) "22.3"
10) "30.7"
I know that Lua does not differentiate between floating point numbers and integers.
This is the script:
local sum = "0.0"
local matches = redis.call('KEYS', 'sdid:*')
for _,key in ipairs(matches) do
local val = redis.call('HGET', key ,'data')
sum = sum + val
end
return sum
I receive the result in integer format ((integer) 224
) how could I receive de result in a real format string "224.5" ?
In arithmetic operations, Lua automatically converts strings containing numbers to numbers.
So, just add the strings with a+b+c and you'll get the number 6.6.
If you want the result with 1 decimal place even if its an integer, use string.format("%.1f",sum).
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)
I'm trying to create a list of integers, similar to python where one would say
x = input("Enter String").split() # 1 2 3 5
x = list(map(int,x)) # Converts x = "1","2",3","5" to x = 1,2,3,5
Here's my code asking for the input, then splitting the input into a table, i need help converting the contents of the table to integers as they're being referenced later in a function, and i'm getting a string vs integer comparison error. I've tried changing the split for-loop to take a number but that doesn't work, I'm familiar with a python conversion but not with Lua so I'm looking for some guidance in converting my table or handling this better.
function main()
print("Hello Welcome the to Change Maker - LUA Edition")
print("Enter a series of change denominations, separated by spaces")
input = io.read()
deno = {}
for word in input:gmatch("%w+") do table.insert(deno,word) end
end
--Would This Work?:
--for num in input:gmatch("%d+") do table.insert(deno,num) end
Just convert your number-strings to numbers using tonumber
local number = tonumber("1")
So
for num in input:gmatch("%d+") do table.insert(deno,tonumber(num)) end
Should do the trick
I wrote a small script that creates Fibonacci sequence and returns a sum of all even integers.
function even_fibo()
-- create Fibonacci sequence
local fib = {1, 2} -- starting with 1, 2
for i=3, 10 do
fib[i] = fib[i-2] + fib[i-1]
end
-- calculate sum of even numbers
local fib_sum = 0
for _, v in ipairs(fib) do
if v%2 == 0 then
fib_sum = fib_sum + v
end
end
return fib_sum
end
fib = even_fibo()
print(fib)
The function creates the following sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55
And returns the sum of its even numbers: 44
However, when I change the stop index from 10 to 100, in for i=3, 100 do the returned sum is negative -8573983172444283806 because the values become too big.
Why is my code working for 10 and not for 100?
Prior to version 5.3, Lua always stored numbers internally as floats. In 5.3 Lua numbers can be stored internally as integers or floats. One option is to run Lua 5.2, I think you'll find your code works as expected there. The other option is to initialize your array with floats which will promote all operations on them in the future to floats:
local fib = {1.0, 2.0}
Here is a hack written in hindsight.
The code exploits the mathematical fact that the even Fibonacci numbers are exactly those at indices that are multiple of 3.
This allows us to avoid testing the parity of very large numbers and provides high-order digits that are correct when you do the computation in floating-point. Then we redo it looking only at the low-order digits and combine the results. The output is 286573922006908542050, which agrees with WA. Values of d between 5 and 15 work fine.
a,b=0.0,1.0
s=0
d=10
for n=1,100/3 do
a,b=b,a+b
a,b=b,a+b
s=s+b
a,b=b,a+b
end
h=string.format("%.0f",s):sub(1,-d-1)
m=10^d
a,b=0,1
s=0
for n=1,100/3 do
a,b=b,(a+b)%m
a,b=b,(a+b)%m
s=(s+b)%m
a,b=b,(a+b)%m
end
s=string.format("%0"..d..".0f",s)
print(h..s)
I am trying to return very long integer number but my result returns as
"7.6561197971049e+016".
How do I make it return 76561197971049296 ?
local id64 = 76561197960265728
Z = string.match("STEAM_0:0:5391784", 'STEAM_%d+:%d+:(%d+)')
Y = string.match("STEAM_0:0:5391784", 'STEAM_%d+:(%d+):%d+')
--For 64-bit systems
--Let X, Y and Z constants be defined by the SteamID: STEAM_X:Y:Z.
--Let V be SteamID64 identifier of the account type (0x0110000100000000 in hexadecimal format).
--Using the formula W=Z*2+V+Y
if Z == nil then
return "none"
else
return Z*2+id64+Y
end
I installed lbc arbitrary precision now with this code
return bc.add(bc.number(id64),bc.number(2)):tostring()
it returns 70000000000000002 but if I delete 3 digits from id64 it displays correctly.
How can I get correct result without deleting the digits?
You need to use strings for long numbers. Otherwise, the Lua lexer converts them to doubles and loses precision in this case. Here is code using my lbc:
local bc=require"bc"
local id64=bc.number"76561197960265728"
local Y,Z=string.match("STEAM_0:0:5391784",'STEAM_%d+:(%d+):(%d+)')
if Z == nil then
return "none"
else
return (Z*2+id64+Y):tostring()
end
check out this library for arbitrary precision arithmetics. this so post might be of interest to you as well.
Assuming your implementation of Lua supports that many significant digits in the number type, your return statement is returning that result.
You're probably seeing exponential notation when you convert the number to a string or printing it. You can use the string.format function to control the conversion:
assert( "76561197971049296" == string.format("%0.17g", 76561197971049296))
If number is an IEEE-754 double, then it doesn't work. You do have to know how your Lua is implemented and keep in mind the the technical limitations.
If you have luajit installed, you can do this:
local ffi = require("ffi")
steamid64 = tostring(ffi.new("uint64_t", 76561197960265728) + ffi.new("uint64_t", tonumber(accountid)))
steamid64 = string.sub(steamid64, 1, -4) -- to remove 'ULL at the end'
Hope it helps.
I am trying to get a random 16 digit number in Lua. What I have written isn't working out for me when logically it should. How does math.random work with exponents?
This is what I keep getting.
> return math.random(10^15, 10^16)
> -1637272360
If you want to have a 16 digit number, try generating them this way:
local fmt = "%d%07d%08d"
local random = math.random
local num = fmt:format(random(1, 9), random(0, 10^7), random(0, 10^8))
and then keep the variable num in string type. As a number, it converts the values to exponential form(because of the very large; in your case > 10^14; exponential value) or otherwise, you can store them as a(n) Hex string?