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."
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
Quick explanation, I have recently started using codewars to further improve my programming skills and my first challenge was to make a roman numeral decoder, I went through many versions because I wasnt satisfied with what I had, So I am asking if there is an easier way of handling all the patterns that roman numerals have, for example I is 1 but if I is next to another number it takes it away for example V = 5 but IV = 4.
here is my CODE:
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local number = 0
local i = 1
while i < #roman + 1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
if roman:sub(i,i) == "I" and roman:sub(i + 1,i + 1) ~= "I" and roman:sub(i + 1,i + 1) ~= "" then -- Checks for the I pattern when I exists and next isnt I
number = number + (Dict[roman:sub(i +1,i + 1)] - Dict[roman:sub(i,i)]) -- Taking one away from the next number
i = i + 2 -- Increase the counter
else
number = number + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
i = i + 1
end
end
return number
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII
at the moment I am trying to get 1049 (MXLIX) to work but I am getting 1069, obviously I am not following a rule and I feel like its more wrong then it should be because usually if its not correct its 1 or 2 numbers wrong.
The algorithm is slightly different: you need to consider subtraction when the previous character has less weight than the next one.
function Roman_Numerals_Decoder (roman)
local Dict = {I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000}
local num = 0
local i = 1
for i=1, #roman-1 do
local letter = roman:sub(i,i) -- Gets the current character in the string roman
local letter_p = roman:sub(i+1,i+1)
if (Dict[letter] < Dict[letter_p]) then
num = num - Dict[letter] -- Taking one away from the next number
print("-",Dict[letter],num)
else
num = num + Dict[letter] -- Adds the numbers together if no pattern is found, currently checking only I
print("+",Dict[letter],num)
end
end
num = num + Dict[roman:sub(-1)];
print("+",Dict[roman:sub(-1)], num)
return num
end
print(Roman_Numerals_Decoder("MXLIX")) -- 1049 = MXLIX , 2008 = MMVIII
The var c return 3 but 10/7=1.4285, the rest is 0.4285, operator % has a bug?
void main() {
var a = 10;
var b = 7;
var c;
c = a % b;
print(c);
}
From the documentation of the % operator on num in Dart:
Euclidean modulo operator.
Returns the remainder of the Euclidean division. The Euclidean division of two integers a and b yields two integers q and r such that a == b * q + r and 0 <= r < b.abs().
The Euclidean division is only defined for integers, but can be easily extended to work with doubles. In that case r may have a non-integer value, but it still verifies 0 <= r < |b|.
The sign of the returned value r is always positive.
See remainder for the remainder of the truncating division.
https://api.dart.dev/stable/2.8.4/dart-core/num/operator_modulo.html
The '%' operator returns the remainder left after dividing two numbers. It does not return the decimal part. For example:
10 / 7
1
______
7 ) 10
- 7
______
3
So it returns 3 which is what remains after dividing 10 by 7 without any decimals.
10 / 7 = 1 3/7
What you want to do can be accomplished like this:
var floatNumber = 12.5523;
var x = floatNumber - floatNumber.truncate();
I'm trying to solve the FizzBuzz game.
I need to check if a number is divisible by 3. So, when we take a number and divide it by 3, we need that operation to have NO REST, or 0 REST.
The solution given to me is this:
def fizzbuzz(number)
if number % 3 == 0
return "Fizz"
end
Why does they propose to use the "%"symbol? Why not the "/"symbol? Or both work as the same?
How should I check if the division has NO REST?
division operator / - gives the quotient of the division whatever the remainder of the division is. So you cannot determine if a number is perfectly divisible (remainder = 0) or not perfectly divisible (with non-zero remainder) using a division operator (/).
10 / 3
#=> 3
modulo operator % - gives the remainder of the division. If perfectly divisible, the output is 0, if not-perfectly divisible the output is non-zero value.
10 % 3
#=> 1
In your case number % 3 == 0 is true only if number is divisible by 3 with 0 remainder (i.e if number passed into the method frizzbuzz is a multiple of 3 like -12, -3, 3, 6, 9, etc )
Ruby has 4 division operators.
divmod returns the division and the remainder
number = 15
number.divmod(7) # => 2, 1
modulo alias % returns the remainder only
number = 15
number % 7 # => 1
number.modulo(7) # => 1
div alias / returns the integer division if both operands are integers, and floating-point division if either operand is a float.
number = 15
number / 7 # => 2
number.div(7) # => 2
1.0 * number / 7 # => 2.142857142857143
fdiv always returns a full precision floating-point division
number = 15
number.fdiv(7) # => 2.142857142857143
% is the Modulus - Divides left hand operand by right hand operand and returns remainder. \ is just the Divider. No Rest means that x % y == 0.
lets take an example to understand better:
number1 = 12
number2 = 13
lets see if number1 and number2 is devisible by 4 ok?
number1 / 4 = 3 and the rest = 0
number2 / 4 = 3 and the rest = 1
so the "/" operation let us know the result of devision operation
and the "%" operation let us know the rest of devision operation
so if we take our examples the number1 is devisible by 3 because
number1 % 3 = 0 ( the rest )
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.