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
Related
I am working on programming a Markov chain in Lua, and one element of this requires me to uniformly generate random numbers. Here is a simplified example to illustrate my question:
example = function(x)
local r = math.random(1,10)
print(r)
return x[r]
end
exampleArray = {"a","b","c","d","e","f","g","h","i","j"}
print(example(exampleArray))
My issue is that when I re-run this program multiple times (mash F5) the exact same random number is generated resulting in the example function selecting the exact same array element. However, if I include many calls to the example function within the single program by repeating the print line at the end many times I get suitable random results.
This is not my intention as a proper Markov pseudo-random text generator should be able to run the same program with the same inputs multiple times and output different pseudo-random text every time. I have tried resetting the seed using math.randomseed(os.time()) and this makes it so the random number distribution is no longer uniform. My goal is to be able to re-run the above program and receive a randomly selected number every time.
You need to run math.randomseed() once before using math.random(), like this:
math.randomseed(os.time())
From your comment that you saw the first number is still the same. This is caused by the implementation of random generator in some platforms.
The solution is to pop some random numbers before using them for real:
math.randomseed(os.time())
math.random(); math.random(); math.random()
Note that the standard C library random() is usually not so uniformly random, a better solution is to use a better random generator if your platform provides one.
Reference: Lua Math Library
Standard C random numbers generator used in Lua isn't guananteed to be good for simulation. The words "Markov chain" suggest that you may need a better one. Here's a generator widely used for Monte-Carlo calculations:
local A1, A2 = 727595, 798405 -- 5^17=D20*A1+A2
local D20, D40 = 1048576, 1099511627776 -- 2^20, 2^40
local X1, X2 = 0, 1
function rand()
local U = X2*A2
local V = (X1*A2 + X2*A1) % D20
V = (V*D20 + U) % D40
X1 = math.floor(V/D20)
X2 = V - X1*D20
return V/D40
end
It generates a number between 0 and 1, so r = math.floor(rand()*10) + 1 would go into your example.
(That's multiplicative random number generator with period 2^38, multiplier 5^17 and modulo 2^40, original Pascal code by http://osmf.sscc.ru/~smp/)
math.randomseed(os.clock()*100000000000)
for i=1,3 do
math.random(10000, 65000)
end
Always results in new random numbers. Changing the seed value will ensure randomness. Don't follow os.time() because it is the epoch time and changes after one second but os.clock() won't have the same value at any close instance.
There's the Luaossl library solution: (https://github.com/wahern/luaossl)
local rand = require "openssl.rand"
local randominteger
if rand.ready() then -- rand has been properly seeded
-- Returns a cryptographically strong uniform random integer in the interval [0, n−1].
randominteger = rand.uniform(99) + 1 -- randomizes an integer from range 1 to 100
end
http://25thandclement.com/~william/projects/luaossl.pdf
I have an SPSS variable containing lines like:
|2|3|4|5|6|7|8|10|11|12|13|14|15|16|18|20|21|22|23|24|25|26|27|28|29|
Every line starts with pipe, and ends with one. I need to refactor it into boolean variables as the following:
var var1 var2 var3 var4 var5
|2|4|5| 0 1 0 1 1
I have tried to do it with a loop like:
loop # = 1 to 72.
compute var# = SUBSTR(var,2#,1).
end loop.
exe.
My code won't work with 2 or more digits long numbers and also it won't place the values into their respective variables, so I've tried nest the char.substr(var,char.rindex(var,'|') + 1) into another loop with no luck because it still won't allow me to recognize the variable number.
How can I do it?
This looks like a nice job for the DO REPEAT command. However the type conversion is somewhat tricky:
DO REPEAT var#i=var1 TO var72
/i=1 TO 72.
COMPUTE var#i = CHAR.INDEX(var,CONCAT("|",LTRIM(STRING(i,F2.0)),"|"))>0).
END REPEAT.
Explanation: Let's go from the inside to the outside:
STRING(value,F2.0) converts the numeric values into a string of two digits (with a leading white space where the number consist of just one digit), e.g. 2 -> " 2".
LTRIM() removes the leading whitespaces, e.g. " 2" -> "2".
CONCAT() concatenates strings. In the above code it adds the "|" before and after the number, e.g. "2" -> "|2|"
CHAR.INDEX(stringvar,searchstring) returns the position at which the searchstring was found. It returns 0 if the searchstring wasn't found.
CHAR.INDEX(stringvar,searchstring)>0 returns a boolean value indicating if the searchstring was found or not.
It's easier to do the manipulations in Python than native SPSS syntax.
You can use SPSSINC TRANS extension for this purpose.
/* Example data*/.
data list free / TextStr (a99).
begin data.
"|2|3|4|5|6|7|8|10|11|12|13|14|15|16|18|20|21|22|23|24|25|26|27|28|29|"
end data.
/* defining function to achieve task */.
begin program.
def runTask(x):
numbers=map(int,filter(None,[i.strip() for i in x.lstrip('|').split("|")]))
answer=[1 if i in numbers else 0 for i in xrange(1,max(numbers)+1)]
return answer
end program.
/* Run job*/.
spssinc trans result = V1 to V30 type=0 /formula "runTask(TextStr)".
exe.
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.
Easy question here, probably, but searching did not find a similar question.
The # operator finds the length of a string, among other things, great. But with Lua being dynamically typed, thus no conversion operators, how does one type a number as a string in order to determine its length?
For example suppose I want to print the factorials from 1 to 9 in a formatted table.
i,F = 1,1
while i<10 do
print(i.."! == "..string.rep("0",10-#F)..F)
i=i+1
F=F*i
end
error: attempt to get length of global 'F' (a number value)
why not use tostring(F) to convert F to a string?
Alternatively,
length = math.floor(math.log10(number)+1)
Careful though, this will only work where n > 0!
There are probably a dozen ways to do this. The easy way is to use tostring as Dan mentions. You could also concatenate an empty string, e.g. F_str=""..F to get F_str as a string representation. But since you are trying to output a formatted string, use the string.format method to do all the hard work for you:
i,F = 1,1
while i<10 do
print(string.format("%01d! == %010d", i, F))
i=i+1
F=F*i
end
Isn't while tostring(F).len < 10 do useful?
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );