For example I need number with minimum 3 digit
"512" --> 512
"24" --> 24.0
"5" --> 5.00
One option is write small function. Using answers here for my case it will be something like this
function f(value, w)
local p = math.ceil(math.log10(value))
local prec = value <= 1 and w - 1 or p > w and 0 or w - p
return string.format('%.' .. prec .. 'f', value)
end
print(f(12, 3))
But may be it is possible just using string.format() or any other simple way?
Ok, it seems this case beyond the string.format power. Thanks to #Schollii, this is my current variant
function f(value, w)
local p = math.ceil(math.log10(value))
local prec = value <= 1 and w - 1 or p > w and 0 or w - p
return string.format('%.' .. prec .. 'f', value)
end
print(f(12, 3))
There is no format code specifically for this since string.format uses printf minus a few codes (like * which would hace simplified the solution I give below). So you have to implement yourself, for example:
function f(num, w)
-- get number of digits before decimal
local intWidth = math.ceil(math.log10(num))
-- if intWidth > w then ... end -- may need this
local fmt='%'..w..'.' .. (w-intWidth) .. 'f'
return string.format(fmt, num)
end
print(f(12, 4))
print(f(12, 3))
print(f(12, 2))
print(f(512, 3))
print(f(24, 3))
print(f(5, 3))
You should probably handle case where integer part doesn't fit in field width given (return ceil or floor?).
You can't. Maximum you can reach - specify floating point precision or digit number, but you can't force output to be like your example. Lua uses C like printf with few limitations reference. Look here for full specifiers list link. Remember unsupported ones.
Writing a function would be the best and only solution, especially as your task looks strange, as it doesn't count decimal dot.
Related
I'm having problem returning spesific amount of decimal numbers from this function, i would like it to get that info from "dec" argument, but i'm stuck with this right now.
Edit: Made it work with the edited version bellow but isn't there a better way?
local function remove_decimal(t, dec)
if type(dec) == "number" then
for key, num in pairs(type(t) == "table" and t or {}) do
if type(num) == "number" then
local num_to_string = tostring(num)
local mod, d = math.modf(num)
-- find only decimal numbers
local num_dec = num_to_string:sub(#tostring(mod) + (mod == 0 and num < 0 and 3 or 2))
if dec <= #num_dec then
-- return amount of deciamls in the num by dec
local r = d < 0 and "-0." or "0."
local r2 = r .. num_dec:sub(1, dec)
t[key] = mod + tonumber(r2)
end
end
end
end
return t
end
By passing the function bellow i want a result like this:
result[1] > 0.12
result[2] > -0.12
result[3] > 123.45
result[4] > -1.23
local result = remove_decimal({0.123, -0.123, 123.456, -1.234}, 2)
print(result[1])
print(result[2])
print(result[3])
print(result[4])
I tried this but it seems to only work with one integer numbers and if number is 12.34 instead of 1.34 e.g, the decimal place will be removed and become 12.3. Using other methods
local d = dec + (num < 0 and 2 or 1)
local r = tonumber(num_to_string:sub(1, -#num_to_string - d)) or 0
A good approach is to find the position of the decimal point (the dot, .) and then extract a substring starting from the first character to the dot's position plus how many digits you want:
local function truncate(number, dec)
local strnum = tostring(number)
local i, j = string.find(strnum, '%.')
if not i then
return number
end
local strtrn = string.sub(strnum, 1, i+dec)
return tonumber(strtrn)
end
Call it like this:
print(truncate(123.456, 2))
print(truncate(1234567, 2))
123.45
1234567
To bulk-truncate a set of numbers:
local function truncate_all(t, dec)
for key, value in pairs(t) do
t[key] = truncate(t[key], dec)
end
return t
end
Usage:
local result = truncate_all({0.123, -0.123, 123.456, -1.234}, 2)
for key, value in pairs(result) do
print(key, value)
end
1 0.12
2 -0.12
3 123.45
4 -1.23
One could use the function string.format which is similar to the printf functions from C language. If one use the format "%.2f" the resulting string will contain 2 decimals, if one use "%.3f" the resulting string will be contain 3 decimals, etc. The idea is to dynamically create the format "%.XXXf" corresponding to the number of decimal needed by the function. Then call the function string.format with the newly created format string to generate the string "123.XXX". The last step would be to convert back the string to a number with the function tonumber.
Note that if one want the special character % to be preserved when string.format is called, you need to write %%.
function KeepDecimals (Number, DecimalCount)
local FloatFormat = string.format("%%.%df", DecimalCount)
local String = string.format(FloatFormat, Number)
return tonumber(String)
end
The behavior seems close to what the OP is looking for:
for Count = 1, 5 do
print(KeepDecimals(1.123456789, Count))
end
This code should print the following:
1.1
1.12
1.123
1.1235
1.12346
Regarding the initial code, it's quite straight-forward to integrate the provided solution. Note that I renamed the function to keep_decimal because in my understanding, the function will keep the requested number of decimals, and discard the rest.
function keep_decimal (Table, Count)
local NewTable = {}
local NewIndex = 1
for Index = 1, #Table do
NewTable[NewIndex] = KeepDecimal(Table[Index], Count)
NewIndex = NewIndex + 1
end
return NewTable
end
Obviously, the code could be tested easily, simply by copy and pasting into a Lua interpreter.
Result = keep_decimal({0.123, -0.123, 123.456, -1.234}, 2)
for Index = 1, #Result do
print(Result[Index])
end
This should print the following:
0.12
-0.12
123.46
-1.23
Edit due to the clarification of the need of truncate:
function Truncate (Number, Digits)
local Divider = Digits * 10
local TruncatedValue = math.floor(Number * Divider) / Divider
return TruncatedValue
end
On my computer, the code is working as expected:
> Truncate(123.456, 2)
123.45
I have one program downloaded from internet and need to get each digit printed out from a three digit number. For example:
Input: 123
Expected Output:
1
2
3
I have 598
Need to Get:
5
9
8
I try using this formula but the problem is when number is with decimal function failed:
FIRST_DIGIT = (number mod 1000) / 100
SECOND_DIGIT = (number mod 100) / 10
THIRD_DIGIT = (number mod 10)
Where number is the above example so here is calulation:
FIRST_DIGIT = (598 mod 1000) / 100 = 5,98 <== FAILED...i need to get 5 but my program shows 0 because i have decimal point
SECOND_DIGIT = (598 mod 100) / 10 = 9,8 <== FAILED...i need to get 9 but my program shows 0 because i have decimal point
THIRD_DIGIT = (598 mod 10) = 8 <== CORRECT...i get from program output number 8 and this digit is correct.
So my question is is there sample or more efficient code that get each digit from number without decimal point? I don't want to use round to round nearest number because sometime it fill failed if number is larger that .5.
Thanks
The simplest solution is to use integer division (\) instead of floating point division (/).
If you replace each one of your examples with the backslash (\) instead of forward slash (/) they will return integer values.
FIRST_DIGIT = (598 mod 1000) \ 100 = 5
SECOND_DIGIT = (598 mod 100) \ 10 = 9
THIRD_DIGIT = (598 mod 10) = 8
You don't have to do any fancy integer calculations as long as you pull it apart from a string:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
PRINT MID$(X$, Z, 1)
NEXT
Then, for example, you could act upon each string element:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
NEXT
Additionally, you could tear apart the string character by character:
INPUT X
X$ = STR$(X)
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " ", ".", "+", "-", "E", "D"
' special char
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
I'm no expert in Basic but looks like you have to convert floating point number to Integer. A quick google search told me that you have to use Int(floating_point_number) to convert float to integer.
So
Int((number mod 100)/ 10)
should probably the one you are looking for.
And, finally, all string elements could be parsed:
INPUT X
X$ = STR$(X)
PRINT X$
FOR Z = 1 TO LEN(X$)
SELECT CASE MID$(X$, Z, 1)
CASE " "
' nul
CASE "E", "D"
Exponent = -1
CASE "."
Decimal = -1
CASE "+"
UnaryPlus = -1
CASE "-"
UnaryNegative = -1
CASE ELSE
Q = VAL(MID$(X$, Z, 1))
N = N + 1
PRINT "Digit"; N; " equals"; Q
END SELECT
NEXT
IF Exponent THEN PRINT "There was an exponent."
IF Decimal THEN PRINT "There was a decimal."
IF UnaryPlus THEN PRINT "There was a plus sign."
IF UnaryNegative THEN PRINT "There was a negative sign."
I'm trying to translate a code from C to Lua and I'm facing a problem.
How can I translate a Bitwise AND in Lua?
The source C code contains:
if ((command&0x80)==0)
...
How can this be done in Lua?
I am using Lua 5.1.4-8
Implementation of bitwise operations in Lua 5.1 for non-negative 32-bit integers
OR, XOR, AND = 1, 3, 4
function bitoper(a, b, oper)
local r, m, s = 0, 2^31
repeat
s,a,b = a+b+m, a%m, b%m
r,m = r + m*oper%(s-a-b), m/2
until m < 1
return r
end
print(bitoper(6,3,OR)) --> 7
print(bitoper(6,3,XOR)) --> 5
print(bitoper(6,3,AND)) --> 2
Here is a basic, isolated bitwise-and implementation in pure Lua 5.1:
function bitand(a, b)
local result = 0
local bitval = 1
while a > 0 and b > 0 do
if a % 2 == 1 and b % 2 == 1 then -- test the rightmost bits
result = result + bitval -- set the current bit
end
bitval = bitval * 2 -- shift left
a = math.floor(a/2) -- shift right
b = math.floor(b/2)
end
return result
end
usage:
print(bitand(tonumber("1101", 2), tonumber("1001", 2))) -- prints 9 (1001)
Here's an example of how i bitwise-and a value with a constant 0x8000:
result = (value % 65536) - (value % 32768) -- bitwise and 0x8000
In case you use Adobe Lightroom Lua, Lightroom SDK contains LrMath.bitAnd() method for "bitwise AND" operation:
-- x = a AND b
local a = 11
local b = 6
local x = import 'LrMath'.bitAnd(a, b)
-- x is 2
And there are also LrMath.bitOr(a, b) and LrMath.bitXor(a, b) methods for "bitwise OR" and "biwise XOR" operations.
This answer is specifically for Lua 5.1.X
you can use
if( (bit.band(command,0x80)) == 0) then
...
in Lua 5.3.X and onwards it's very straight forward...
print(5 & 6)
hope that helped 😉
I want to implement a 3 point crossover for genetic programming but I don't know how to do it and where to start.
My input is:
a = {(first pair), (second pair), ... etc.}
For example a = {(12345,67890), (09876,54321)} (those are numbers, not strings)
Output: Something like this:
Example: a_1 = {(12895), (67340)} also numbers.
Thanks for reply and sorry for my bad English.
Here is my quick implementation of k-point crossover for integers using mostly integer arithmetic. Starting with this, you can extend it to crossover your chromosomes of many pairs of integers using a loop.
math.randomseed(111)
-- math.randomseed(os.time())
a = 12345
b = 67890
len = 5 -- number of digits
function split(mum, dad, len, base)
local split = math.pow(base, math.random(len))
local son = math.floor(dad / split) * split + mum % split
local daughter = math.floor(mum / split) * split + dad % split
return son, daughter
end
function kpoint(mum, dad, len, base, k)
for i=1, k do
mum, dad = split(mum, dad, len, base)
end
return mum, dad
end
s, d = kpoint(a, b, len, 10, 3) -- 3 point crossover in base 10
print(s) -- 67395
print(d) -- 12840
-- binary, (crossover binary representation)
s, d = kpoint(tonumber("10001", 2), tonumber("10110", 2), 5, 2, 3)
print(s) -- 23 which is (10111) in base 2
print(d) -- 16 which is (10000) in base 2
-- binary, (crossover base 10, but interpret as binary)
s, d = kpoint(1101, 1010, 4, 10, 3)
print(s) -- 1001
print(d) -- 1110
I am trying to understand how the bound variables are indexed in z3.
Here in a snippet in z3py and the corresponding output. ( http://rise4fun.com/Z3Py/plVw1 )
x, y = Ints('x y')
f1 = ForAll(x, And(x == 0, Exists(y, x == y)))
f2 = ForAll(x, Exists(y, And(x == 0, x == y)))
print f1.body()
print f2.body()
Output:
ν0 = 0 ∧ (∃y : ν1 = y)
y : ν1 = 0 ∧ ν1 = y
In f1, why is the same bound variable x has different index.(0 and 1). If I modify the f1 and bring out the Exists, then x has the same index(0).
Reason I want to understand the indexing mechanism:
I have a FOL formula represented in a DSL in scala that I want to send to z3. Now ScalaZ3 has a mkBound api for creating bound variables that takes index and sort as arguments. I am not sure what value should I pass to the index argument. So, I would like to know the following:
If I have two formulas phi1 and phi2 with maximum bound variable indexes n1 and n2, what would be the index of x in ForAll(x, And(phi1, phi2))
Also, is there a way to show all the variables in an indexed form? f1.body() just shows me x in indexed form and not y. (I think the reason is that y is still bound in f1.body())
Z3 encodes bound variables using de Bruijn indices.
The following wikipedia article describes de Bruijn indices in detail:
http://en.wikipedia.org/wiki/De_Bruijn_index
Remark: in the article above the indices start at 1, in Z3, they start at 0.
Regarding your second question, you can change the Z3 pretty printer.
The Z3 distribution contains the source code of the Python API. The pretty printer is implemented in the file python\z3printer.py.
You just need to replace the method:
def pp_var(self, a, d, xs):
idx = z3.get_var_index(a)
sz = len(xs)
if idx >= sz:
return seq1('Var', (to_format(idx),))
else:
return to_format(xs[sz - idx - 1])
with
def pp_var(self, a, d, xs):
idx = z3.get_var_index(a)
return seq1('Var', (to_format(idx),))
If you want to redefine the HTML pretty printer, you should also replace.
def pp_var(self, a, d, xs):
idx = z3.get_var_index(a)
sz = len(xs)
if idx >= sz:
# 957 is the greek letter nu
return to_format('ν<sub>%s</sub>' % idx, 1)
else:
return to_format(xs[sz - idx - 1])
with
def pp_var(self, a, d, xs):
idx = z3.get_var_index(a)
return to_format('ν<sub>%s</sub>' % idx, 1)