I want to calculate a simple number, and if the number is not an integer I want to round it up.
For instance, if after a calculation I get 1.2, I want to change it to 2. If the number is 3.7, I want to change it to 4 and so on.
You can use math.ceil to round a Double up and toInt to convert the Double to an Int.
def roundUp(d: Double) = math.ceil(d).toInt
roundUp(1.2) // Int = 2
roundUp(3.7) // Int = 4
roundUp(5) // Int = 5
The ceil function is also directly accessible on the Double:
3.7.ceil.toInt // 4
Having first imported math
import scala.math._ (the final dot & underscore are crucial for what comes next)
you can simply write
ceil(1.2)
floor(3.7)
plus a bunch of other useful math functions like
exp(1)
pow(2,2)
sqrt(pow(2,2)
Related
This question already has answers here:
How do you round a double in Dart to a given degree of precision AFTER the decimal point?
(28 answers)
Closed last year.
I want to set a double, let's call it Price, in Dart, so that it always gives me a double of 2 decimal places.
So 2.5 would return 2.50 and 2.50138263 would also return 2.50.
The simplest answer would be double's built-in toStringAsFixed.
In your case
double x = 2.5;
print('${x.toStringAsFixed(2)}');
x = 2.50138263;
print('${x.toStringAsFixed(2)}');
Would both return 2.50. Be aware that this truncates (e.g., 2.519 returns 2.51). It does not use the standard rounding (half-even) banker's algorithm.
I recommend using a NumberFormat from the intl package; The parsing and formatting rules are worth learning since they appear in other languages like Java.
double d = 2.519;
String s = NumberFormat.currency().format(d);
print(s);
returns USD2.52
s = NumberFormat('#.00').format(d);
returns 2.52
Since your are dealing with money, you should probably use NumberFormat.currency, which would add the currency symbol for the current locale.
Your question is more about how Dart handles the type double. Something like the following might work depending on your use-case:
void main() {
double num = 2.50138263;
num = double.parse(num.toStringAsFixed(2));
print(num);
}
More info about how Dart handles double can be found here.
I'd like to convert a double such as 1.1231053E7 to 11,231,053.0 in scala. Currently the way I am converting doubles is to do this f"$number" where number is a double value. Unfortunately this just gives me a string with 1.1231053E7.
I can convert it out of scientific notation using NumberFormat or DecimalFormat but these also force me to choose a predetermined precision. I want flexible precision. So...
val number1 = 1.2313215
val number2 = 100
val number4 = 3.333E2
... when converted should be...
1.2313215
100
333.3
Currently DecimalFormat makes me choose the precision during construction like so: new DecimalFormat(##.##). Each # after . signifies a decimal point.
If I use f"$number", it treats the decimal points correctly but, like I said before, it is unable to handle the scientific notation.
Just decide how many places after the . you need, write out the number hiding the zeros:
val fmt = new java.text.DecimalFormat("#,##0.##############")
for (x <- List[Double](1.2313215, 100, 3.333E2)) println(fmt.format(x))
prints:
1.2313215
100
333.3
Basically I would like to know if it is possible to get the exponent value in a number, ex:
number = 2.6e3
I want to get the value 3 of the exponent. I have been searching for quite a while now and have not found the answer to this. I am new to programming so I may not know exactly what to look for (which methods, etc).
Any help is much appreciated! Thanks!
Assuming I am interpreting your question correctly this is what you want to do:
B = A^X where A and B are known values. Solve for X.
1000 = 10^X (In this case, X = 3.)
The below code will work for any base. It requires either Foundation or UIKit. The function arguments "value" and "base" are B, A respectively. Try the code out in the Xcode Playground!
func getExponentForValueAndBase(value: Double, base: Double) -> Double {
return log(value)/log(base)
}
getExponentForValueAndBase(1000, base: 10) // = 3
Assuming this is your question: Given a number as an integer, find the interger value of the log base 10 of it.
import Foundation
func log(Int number) -> Int
{
return floor(log10(number))
}
Looking at various posts on this topic but still no luck. Is there a simple way to make division/conversion when dividing Double (or Float) with Int? Here is a simple example in playground returning and error "Double is not convertible to UInt8".
var score:Double = 3.00
var length:Int = 2 // it is taken from some an array lenght and does not return decimal or float
var result:Double = (score / length )
Cast the int to double with var result:Double=(score/Double(length))
What this will do is before computing the division it will create a new Double variable with int inside parentheses hence constructor like syntax.
You cannot combine or use different variable types together.
You need to convert them all to the same type, to be able to divide them together.
The easiest way I see to make that happen, would be to make the Int a Double.
You can do that quite simply do that by adding a ".0" on the end of the Integer you want to convert.
Also, FYI:
Floats are pretty rarely used, so unless you're using them for something specific, its also just more fluid to use more common variables.
Look at my simple math script
double s = 3/2;
NSLog (#"%.2f",s);
I get 1, it should be 1.50
How can I fix it?
double s = 3/2 means divide the int 3 by the int 2, and then cast the result (which is the int 1) to a double (giving you 1.0).
You want this:
double s = 3.0/2.0;
try double s = 3/2.0
both 3 and 2 are integers, so despite receiver being double, you still get an int result and it is then converted to double
3 and 2 are integers. So 3/2 is 1. You should try instead
double s = 3.0/2.0;
NSLog (#"%1.2f",s);
It will give you the right answer.