So I am making a fitness app for android and now I am asking the user to input a number e.g. 72.5
I would take this number and take percentages of it and apply functions to this etc.
I need to make sure that the percentage I take of that number is rounded to 2.5. This is because in a UK gym you only have the following plates: 1.25x2=2.5 2.5x2=5 5+2.5=7.5 , 10, 15, 20, 25
What I mean is that it would be numbers like these: 40, 42.5, 45, 47.5, 50
How can I round a Number N to the nearest 2.5? I understand that math.Round() rounds to nearest whole but what about to a custom number like this?
Do it as follows:
public class Main {
public static void main(String args[]) {
// Tests
System.out.println(roundToNearest2Point5(12));
System.out.println(roundToNearest2Point5(14));
System.out.println(roundToNearest2Point5(13));
System.out.println(roundToNearest2Point5(11));
}
static double roundToNearest2Point5(double n) {
return Math.round(n * 0.4) / 0.4;
}
}
Output:
12.5
15.0
12.5
10.0
Explanation:
It will be easier to understand with the following example:
double n = 20 / 3.0;
System.out.println(n);
System.out.println(Math.round(n));
System.out.println(Math.round(n * 100.0));
System.out.println(Math.round(n * 100.0) / 100.0);
Output:
6.666666666666667
7
667
6.67
As you can see here, rounding 20 / 3.0 returns 7 (which is the floor value after adding 0.5 to 20 / 3.0. Check this to understand the implementation). However, if you wanted to round it up to the nearest 1/100th place (i.e. up to 2 decimal places), an easier way (but not so precise. Check this for more precise way) would be to round n * 100.0 (which would make it 667) and then divide it by 100.0 which would give 6.67 (i.e. up to 2 decimal places). Note that 1 / (1 / 100.0) = 100.0
Similarly, to round the number to the nearest 2.5th place, you will need to do the same thing with 1 / 2.5 = 0.4 i.e. Math.round(n * 0.4) / 0.4.
To round a number to the nearest 100th place, you will need to do the same thing with 1 / 100 = 0.01 i.e. Math.round(n * 0.1) / 0.1.
To round a number to the nearest 0.5th place, you will need to do the same thing with 1 / 0.5 = 2.0 i.e. Math.round(n * 2.0) / 2.0.
I hope, it is clear.
Late reply but you can also do 2.5*(Math.round(number/2.5))
Same way if you wanted to round in lbs to the nearest 5th its 5*(Math.round(number/5))
Related
I have used this one
Math.round(input * 100.0) / 100.0;
but it didn't work with my requirement. I want to round into two decimal points according to nearest decimal value.
firstly I want to check fourth decimal value
if it is 5 or above I want to add 1 into 3rd decimal value
then want to check 3rd one
if it is 5 or above I want to add 1 into 2nd one
ex: if I have 22.3246
refer above example number. 4th decimal no is 6.
we can add 1 into 3rd decimal value.
result : 22.325
now 3rd decimal no is 5. we can add 1 into 2nd decimal value
result : 22.33
I want to get result 22.33
If you want to respect the 4th decimal digit, you can do it in this way:
double val = Math.round(22.3246 * 1000.0) / 1000.0;
double result = Math.round(val * 100.0) / 100.0;
System.out.println(result); // print: 22.33
You can use BigDecimal class to accomplish this task, in combination with a scale and RoundingMode.HALF_UP.
Sample code:
System.out.println(new BigDecimal("22.3246").divide(BigDecimal.ONE,3,RoundingMode.HALF_UP)
.divide(BigDecimal.ONE,2,RoundingMode.HALF_UP));
Output:
22.33
According to your requirement, the greatest number to be rounded down to 22.32 is 22.3244444444.... From 22.32444...4445 on, iteratively rounding digit per digit will lead to 22.33.
So, you can add a bias of 0.005 - 0.0044444444 (standard rounding threshold minus your threshold), being 5/9000 and then round to the next 1/100 the standard way:
double BIAS = 5.0 / 9000.0;
double result = Math.round((input + BIAS) * 100.0) / 100.0;
This question already has an answer here:
Java rounding to nearest 0.05
(1 answer)
Closed 8 years ago.
I'm trying to find a way on how to round to the nearest 0.05 in java. Let's say that I have the following numbers:
0.33
0.02
0.874
0.876
This should become:
0.35
0.00
0.85
0.90
I tried many things and I can only get it to round to n places behind the comma by using BigDecimal, but I can't seem to find a way for this one.
Can someone help me?
EDIT: Thank you for all your help, I am amazed at how easy this could be done. And how do I get the double converted into a string properly? I can't use Double.toString(double d) because for example the string will be "0.9" instead of "0.90"?
0.05 == 1/20, right? Therefore, what you need is just the nearest number with dividing by 1/20, so, you may multiply this number by 20, get the nearest number with dividing by 1, then get the initial things.
TL;DR: you just may just multiply it by 20, round and divide by 20 again:
public double customRound(double num) {
return Math.round(num * 20) / 20.0;
}
A simple way would be:
double d = 0.33;
double roundedTimes20 = Math.round(d * 20);
double rounded = roundedTimes20 / 20; //0.35
but note that the resulting double is not necessarily the exact representation of the rounded number (usual floating point caveat) and that the method assumes that your original double times 20 can fit in a long.
Try a function:
public static double round05(double num) {
return Math.round(num * 20) / 20.0;
}
You can use String.format to format value to String
String s = String.format("%.2f", 0.9);
I'd like to round manually without the round()-Method.
So I can tell my program that's my number, on this point i want you to round.
Let me give you some examples:
Input number: 144
Input rounding: 2
Output rounded number: 140
Input number: 123456
Input rounding: 3
Output rounded number: 123500
And as a litte addon maybe to round behind the comma:
Input number: 123.456
Input rounding: -1
Output rounded number: 123.460
I don't know how to start programming that...
Has anyone a clue how I can get started with that problem?
Thanks for helping me :)
I'd like to learn better programming, so i don't want to use the round and make my own one, so i can understand it a better way :)
A simple way to do it is:
Divide the number by a power of ten
Round it by any desired method
Multiply the result by the same power of ten in step 1
Let me show you an example:
You want to round the number 1234.567 to two decimal positions (the desired result is 1234.57).
x = 1234.567;
p = 2;
x = x * pow(10, p); // x = 123456.7
x = floor(x + 0.5); // x = floor(123456.7 + 0.5) = floor(123457.2) = 123457
x = x / pow(10,p); // x = 1234.57
return x;
Of course you can compact all these steps in one. I made it step-by-step to show you how it works. In a compact java form it would be something like:
public double roundItTheHardWay(double x, int p) {
return ((double) Math.floor(x * pow(10,p) + 0.5)) / pow(10,p);
}
As for the integer positions, you can easily check that this also works (with p < 0).
Hope this helps
if you need some advice how to start,
step by step write down calculations what you need to do to get from 144,2 --> 140
replace your math with java commands, that should be easy, but if you have problem, just look here and here
public static int round (int input, int places) {
int factor = (int)java.lang.Math.pow(10, places);
return (input / factor) * factor;
}
Basically, what this does is dividing the input by your factor, then multiplying again. When dividing integers in languages like Java, the remainder of the division is dropped from the results.
edit: the code was faulty, fixed it. Also, the java.lang.Math.pow is so that you get 10 to the n-th power, where n is the value of places. In the OP's example, the number of places to consider is upped by one.
Re-edit: as pointed out in the comments, the above will give you the floor, that is, the result of rounding down. If you don't want to always round down, you must also keep the modulus in another variable. Like this:
int mod = input % factor;
If you want to always get the ceiling, that is, rounding up, check whether mod is zero. If it is, leave it at that. Otherwise, add factor to the result.
int ceil = input + (mod == 0 ? 0 : factor);
If you want to round to nearest, then get the floor if mod is smaller than factor / 2, or the ceiling otherwise.
Divide (positive)/Multiply (negative) by the "input rounding" times 10 - 1 (144 / (10 * (2 - 1)). This will give you the same in this instance. Get the remainder of the last digit (4). Determine if it is greater than or equal to 5 (less than). Make it equal to 0 or add 10, depending on the previous answer. Multiply/Divide it back by the "input rounding" times 10 - 1. This should give you your value.
If this is for homework. The purpose is to teach you to think for yourself. I may have given you the answer, but you still need to write the code by yourself.
Next time, you should write your own code and ask what is wrong
For integers, one way would be to use a combination of the mod operator, which is the percent symbol %, and the divide operator. In your first example, you would compute 144 % 10, resulting in 4. And compute 144 / 10, which gives 14 (as an integer). You can compare the result of the mod operation to half of the denominator, to find out if you should round the 14 up to 15 or not (in this case not), and then multiply back by the denominator to get your answer.
In psuedo code, assuming n is the number to round, p is the power of 10 representing the position of the significant digits:
denom = power(10, p)
remainder = n % denom
dividend = n / denom
if (remainder < denom/2)
return dividend * denom
else
return (dividend + 1) * denom
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Round a double to 2 significant figures after decimal point
I am trying to work with converting a decimal degree (23.1248) into a minutes style degree(23 7'29.3").
this is what I have so far:
double a=23.1248;
int deg=(int)a;//gives me the degree
float b=(float) (a-deg);
int min=(int) (b*60);//gives me the minutes
double sec= (double) ((c*60)-min);//gives me my seconds
everything works fine, but I would like to round the seconds up to either the nearest tenth or hundrenth. I have looked at decimal formatting, but would prefer not to cast it to a string. I have also looked at bigdecimal but do not think that would be helpful,
Try using Math.round(double) on the number after scaling it up, then scaling it back down.
double x = 1.234;
double y = Math.round(x * 100.0) / 100.0; // => 1.23
You can also use BigDecimal if you want to get really heavyweight:
BigDecimal a = new BigDecimal("1.234");
BigDecimal b = a.setScale(2, RoundingMode.DOWN); // => BigDecimal("1.23")
First off, there are library functions to do this, so why not just use those? See Math.round(). No need to reinvent the wheel. If you wanted to, though, you could try what follows. To round a double to the hundredth's place:
x = 0.01 * floor(x * 100.0)
To round a double to the tenth's place:
x = 0.1 * floor(x * 10.0)
To round a double to the 10^k place:
x = 10^k * floor(x / 10^k)
The implementation in any language - including Java - should be straightforward. A problem with this is that it doesn't really round, but truncates, to your position. To fix this, you can simply add 0.5 * 10^k to your number before rounding. If you just want to round up, use the versions above, and add 10^k before or after the computation.
I have implemented a Kahan floating point summation algorithm in Java. I want to compare it against the built-in floating point addition in Java and infinite precision addition in Mathematica. However the data set I have is not good for testing, because the numbers are close to each other. (Condition number ~= 1)
Running Kahan on my data set gives all most the same result as the built-in +.
Could anyone suggest how to generate a large amount of data that can potentially cause serious rounding off error?
However the data set I have is not good for testing, because the numbers are close to each other.
It sounds like you already know what the problem is. Get to it =)
There are a few things that you will want:
Numbers of wildly different magnitudes, so that most of the precision of the smaller number is lost with naive summation.
Numbers with different signs and nearly equal (or equal) magnitudes, such that catastrophic cancellation occurs.
Numbers that have some low-order bits set, to increase the effects of rounding.
To get you started, you could try some simple three-term sums, which should show the effect clearly:
1.0 + 1.0e-20 - 1.0
Evaluated with simple summation, this will give 0.0; clearly incorrect. You might also look at sums of the form:
a0 + a1 + a2 + ... + an - b
Where b is the sum a0 + ... + an evaluated naively.
You want a heap of high precision numbers? Try this:
double[] nums = new double[SIZE];
for (int i = 0; i < SIZE; i++)
nums[i] = Math.rand();
Are we talking about number pairs or sequences?
If pairs, start with 1 for both numbers, then in every iteration divide one by 3, multiply the other by 3. It's easy to calculate the theoretical sums of those pairs and you'll get a whole host of rounding errors. (Some from the division and some from the addition. If you don't want division errors, then use 2 instead of 3.)
By experiment, I found following pattern:
public static void main(String[] args) {
System.out.println(1.0 / 3 - 0.01 / 3);
System.out.println(1.0 / 7 - 0.01 / 7);
System.out.println(1.0 / 9 - 0.001 / 9);
}
I've subtracted close negative powers of prime numbers (which should not have exact representation in binary form). However, there are cases then such expression evaluates correctly, for example
System.out.println(1.0 / 9 - 0.01 / 9);
You can automate this approach by iterating power of subtrahend and stopping when multiplication by appropriate value doesn't yield integer number, for example:
System.out.println((1.0 / 9 - 0.001 / 9) * 9000);
if (1000 - (1.0 / 9 - 0.001 / 9) * 9000 > 1.0)
System.out.println("Found it!");
Scalacheck might be something for you. Here is a short sample:
cat DoubleSpecification.scala
import org.scalacheck._
object DoubleSpecification extends Properties ("Doubles") {
/*
(a/1000 + b/1000) = (a+b) / 1000
(a/x + b/x ) = (a+b) / x
*/
property ("distributive") = Prop.forAll { (a: Int, b: Int, c: Int) =>
(c == 0 || a*1.0/c + b*1.0/c == (a+b) * 1.0 / c) }
}
object Runner {
def main (args: Array[String]) {
DoubleSpecification.check
println ("...done")
}
}
To run it, you need scala, and the schalacheck-jar. I used version 2.8 (I don't have to say, that your c-path will vary):
scalac -cp /opt/scala/lib/scalacheck.jar:. DoubleSpecification.scala
scala -cp /opt/scala/lib/scalacheck.jar:. DoubleSpecification
! Doubles.distributive: Falsified after 6 passed tests.
> ARG_0: 28 (orig arg: 1030341)
> ARG_1: 9 (orig arg: 2147483647)
> ARG_2: 5
Scalacheck takes some random values (orig args) and tries to simplify these, if the test fails, in order to find simple examples.