Scala - convert large or scientific numbers to Long - java

Need to be able to convert 00000924843571390729101 or 2.71000000000000000E+02 to Long
Expect 2.71000000000000000E+02 to become 271
funky results for this: 00000924843571390729101 became 924843571390729088
val signNumber = "00000924843571390729101"
val castnum = signNumber.toDouble.toLong.toString
First conversion below works for 2.71000000000000000E+02, 2nd one works for 00000924843571390729101
val castnum = signNumber.toDouble.toLong.toString
val castnum = signNumber.replaceAll("\\.[0-9]*$", "").toLong.toString
Do not want to keep any decimal places so not using java.math.BigDecimal
Input string may come in as 9028456928343.0000 in which case want 9028456928343 as the Long

The weird result in the first case is due to the fact that you are going through toDouble, limiting the precision due to how doubles are represented.
To reliably convert from these strings to Longs you can try out BigDecimal::longValueExact as follows:
import java.math.BigDecimal
new BigDecimal("00000924843571390729101").longValueExact()
// 924843571390729101: Long
new BigDecimal("9028456928343.0000").longValueExact()
// 9028456928343: Long
new BigDecimal("2.71000000000000000E+02").longValueExact()
// 271: Long
Since you also have strings that come with decimal digits, you have to use a form of number parsing that can recognize those, even if you don't want to keep that information.
This method will throw an ArithmeticInformation if you loose any information, meaning the number doesn't fit in a Long or it has a nonzero fractional part. If you want to be more lenient you can use BigDecimal::longValue
import java.math.BigDecimal
import scala.util.Try
​Try(new BigDecimal("9028456928343.0001").longValueExact())
// Failure(java.lang.ArithmeticException: Rounding necessary): scala.util.Try
Try(new BigDecimal("9028456928343.0001").longValue())
// Success(9028456928343): scala.util.Try
You can play around with this small snippet of code here on Scastie.

val signNumber = Try(numberCleaner(signNumber).replaceAll("\\.[0-9]*$", "").toLong.toString).getOrElse(numberCleaner(signNumber).toDouble.toLong.toString)

Related

getting wrong output while using numberFormat.parse("") method in java

I have below piece of code:
I am passing value "55.00000000000000" and getting output as 55.00000000000001.
But when i passed "45.00000000000000" and "65.00000000000000" i get output as 45.0 and 65.0.
Can someone please help me to get correct output as 55.0.
NumberFormat numberFormat = NumberFormat.getPercentInstance(Locale.US);
if (numberFormat instanceof DecimalFormat) {
DecimalFormat df = (DecimalFormat) numberFormat;
df.setNegativePrefix("(");
df.setNegativeSuffix("%)");
}
Number numericValue = numberFormat.parse("55.00000000000000%");
numericValue = new Double(numericValue.doubleValue() * 100);
System.out.println(numericValue);
The problem here is that numericValue is mathematically supposed to be 0.55. However, it will be a Double (because numberFormat.parse() can only return a Long or a Double). And a Double cannot hold the value 0.55 exactly. See this link for a complete explanation of why. The result is that as you do further computations with the inexact value, roundoff errors will occur, which is why the result being printed out is not quite the exact value. (A Double also cannot be exactly 0.45 or 0.65; it just happens that when multiplying by 100, the result rounds to the correct integer.)
When dealing with decimal values such as money or percentages, it's preferable to use BigDecimal. If the NumberFormat is a DecimalFormat, you can set things up so that parse returns a BigDecimal:
if (numberFormat instanceof DecimalFormat) {
DecimalFormat df = (DecimalFormat) numberFormat;
df.setNegativePrefix("(");
df.setNegativeSuffix("%)");
df.setParseBigDecimal(true); // ADD THIS LINE
}
Now, when you use numberFormat.parse(), the Number it returns will be a BigDecimal, which is able to hold the exact value 0.55. Now you have to avoid converting it to a double, which will introduce a roundoff error. Instead, you should say something like
Number numericValue = numberFormat.parse("55.00000000000000%");
if (numericValue instanceof BigDecimal) {
BigDecimal bdNumber = (BigDecimal) numericValue;
// use BigDecimal operations to multiply by 100, then print or format
// or whatever you want to do
} else {
// you're stuck doing things the old way, you might get some
// inaccuracy
numericValue = new Double(numericValue.doubleValue() * 100);
System.out.println(numericValue);
}
use this line of code
System.out.println(String.format("%.1f", numericValue));
Where format method use to format your data.

Converting large scientific number to long

I've spend a long time now, trying to convert the number 1.2846202978398e+19 in java, without any luck. Currently what I'm trying to do (long)Double.parseDouble(hashes), however this gives 9223372036854775807, which is obviously incorrect. The actual number should look something like this 12855103593745000000.
Using int val = new BigDecimal(stringValue).intValue(); returns -134589568 as it's unable to hold the result. Switching the code to long val = new BigDecimal(hashes).longValue(); gives me -5600541095311551616 which is also incorrect.
I'm assuming this is happening due to the size of a double compared to a long.
Any ideas?
Did you try to use String.format :
String result = String.format("%.0f", Double.parseDouble("1.2846202978398e+19"));
System.out.println(result);
Output
12846202978398000000
Edit
Why you don't work with BigDecimal to do the arithmetic operations, for example :
String str = "1.2846202978398e+19";
BigDecimal d = new BigDecimal(str).multiply(BigDecimal.TEN);
// ^^^^^^^^------example of arithmetic operations
System.out.println(String.format("%.0f", d));
System.out.println(String.format("%.0f", Double.parseDouble(str)));
Output
128462029783980000000
12846202978398000000
Your value exceeds the maximum size of long. You can not use long in this situation.
Try
BigDecimal value = new BigDecimal("1.2846202978398e+19");
After that, you can call
value.toBigInteger()
or
value.toBigIntegerExact()
if needed.
What about:
System.out.println(new BigDecimal("1.2846202978398e+19").toBigInteger());

How to add a very small number and a very large number

I am pretty new to Java. I am learning numerical computation at the moment. How does one add and multiply a very small number and a very large number, say something of order $10^{-20}$ and something of order $10^{20}$ to arbitrary precision.
Take a look at the BigDecimal class. From the Javadoc:
Immutable, arbitrary-precision signed decimal numbers.
and:
The BigDecimal class gives its user complete control over rounding behavior.
For your example:
import java.math.BigDecimal;
public class Main {
public static void main(String[] args) {
BigDecimal big = new BigDecimal("10e20");
BigDecimal small = new BigDecimal("10e-20");
BigDecimal ans = big.add(small);
System.err.println("Answer: " + ans);
}
}
Running gives the following:
$ java Main
Answer: 1000000000000000000000.00000000000000000010
Try the following (didn't count the zeros). You may find other methods to construct 10^20/10^-20 more suitable.
System.out.println( new BigDecimal("0.0000000000000000000000000000001").add( new BigDecimal
("100000000000000000000000000000000")));

Java, math (adding, dividing, multiplying) all together

This is my code and it whines about calc 2 and the result.
BigDecimal costNum1 = new BigDecimal(number3.getText().toString());
BigDecimal costNum2 = new BigDecimal(number1.getText().toString());
BigDecimal costNum3 = new BigDecimal(number2.getText().toString());
BigDecimal calc1 = costNum1.multiply(costNum2);
BigDecimal calc2 = calc1.divide("100");
BigDecimal calc3 = calc2.multiply(costNum3);
result.setText(calc3).toString());
Safe to say I'm quite new in this, I'm almost there but I can't make up what is wrong. It's for my first Android App.
BigDecimal#divide accepts another BigDecimal, not a String.
Try
calc2.divide(new BigDecimal("100"));
Also, you have one too many parentheses in your last line.
Try
result.setText(calc3.toString());
You should always count the number of left parens, and see if it matches the number of right parens. If you use an IDE like eclipse, it should point these problems out to you automatically.

Using BigDecimal to work with currencies

I was trying to make my own class for currencies using longs, but apparently I should use BigDecimal instead. Could someone help me get started? What would be the best way to use BigDecimals for dollar currencies, like making it at least but no more than 2 decimal places for the cents, etc. The API for BigDecimal is huge, and I don't know which methods to use. Also, BigDecimal has better precision, but isn't that all lost if it passes through a double? if I do new BigDecimal(24.99), how will it be different than using a double? Or should I use the constructor that uses a String instead?
Here are a few hints:
Use BigDecimal for computations if you need the precision that it offers (Money values often need this).
Use the NumberFormat class for display. This class will take care of localization issues for amounts in different currencies. However, it will take in only primitives; therefore, if you can accept the small change in accuracy due to transformation to a double, you could use this class.
When using the NumberFormat class, use the scale() method on the BigDecimal instance to set the precision and the rounding method.
PS: In case you were wondering, BigDecimal is always better than double, when you have to represent money values in Java.
PPS:
Creating BigDecimal instances
This is fairly simple since BigDecimal provides constructors to take in primitive values, and String objects. You could use those, preferably the one taking the String object. For example,
BigDecimal modelVal = new BigDecimal("24.455");
BigDecimal displayVal = modelVal.setScale(2, RoundingMode.HALF_EVEN);
Displaying BigDecimal instances
You could use the setMinimumFractionDigits and setMaximumFractionDigits method calls to restrict the amount of data being displayed.
NumberFormat usdCostFormat = NumberFormat.getCurrencyInstance(Locale.US);
usdCostFormat.setMinimumFractionDigits( 1 );
usdCostFormat.setMaximumFractionDigits( 2 );
System.out.println( usdCostFormat.format(displayVal.doubleValue()) );
I would recommend a little research on Money Pattern. Martin Fowler in his book Analysis pattern has covered this in more detail.
public class Money {
private static final Currency USD = Currency.getInstance("USD");
private static final RoundingMode DEFAULT_ROUNDING = RoundingMode.HALF_EVEN;
private final BigDecimal amount;
private final Currency currency;
public static Money dollars(BigDecimal amount) {
return new Money(amount, USD);
}
Money(BigDecimal amount, Currency currency) {
this(amount, currency, DEFAULT_ROUNDING);
}
Money(BigDecimal amount, Currency currency, RoundingMode rounding) {
this.currency = currency;
this.amount = amount.setScale(currency.getDefaultFractionDigits(), rounding);
}
public BigDecimal getAmount() {
return amount;
}
public Currency getCurrency() {
return currency;
}
#Override
public String toString() {
return getCurrency().getSymbol() + " " + getAmount();
}
public String toString(Locale locale) {
return getCurrency().getSymbol(locale) + " " + getAmount();
}
}
Coming to the usage:
You would represent all monies using Money object as opposed to BigDecimal. Representing money as big decimal will mean that you will have the to format the money every where you display it. Just imagine if the display standard changes. You will have to make the edits all over the place. Instead using the Money pattern you centralize the formatting of money to a single location.
Money price = Money.dollars(38.28);
System.out.println(price);
Or, wait for JSR-354. Java Money and Currency API coming soon!
1) If you are limited to the double precision, one reason to use BigDecimals is to realize operations with the BigDecimals created from the doubles.
2) The BigDecimal consists of an arbitrary precision integer unscaled value and a non-negative 32-bit integer scale, while the double wraps a value of the primitive type double in an object. An object of type Double contains a single field whose type is double
3) It should make no difference
You should have no difficulties with the $ and precision. One way to do it is using System.out.printf
Use BigDecimal.setScale(2, BigDecimal.ROUND_HALF_UP) when you want to round up to the 2 decimal points for cents. Be aware of rounding off error when you do calculations though. You need to be consistent when you will be doing the rounding of money value. Either do the rounding right at the end just once after all calculations are done, or apply rounding to each value before doing any calculations. Which one to use would depend on your business requirement, but generally, I think doing rounding right at the end seems to make a better sense to me.
Use a String when you construct BigDecimal for money value. If you use double, it will have a trailing floating point values at the end. This is due to computer architecture regarding how double/float values are represented in binary format.
Primitive numeric types are useful for storing single values in memory. But when dealing with calculation using double and float types, there is a problems with the rounding.It happens because memory representation doesn't map exactly to the value. For example, a double value is supposed to take 64 bits but Java doesn't use all 64 bits.It only stores what it thinks the important parts of the number. So you can arrive to the wrong values when you adding values together of the float or double type.
Please see a short clip https://youtu.be/EXxUSz9x7BM
I would be radical. No BigDecimal.
Here is a great article
https://lemnik.wordpress.com/2011/03/25/bigdecimal-and-your-money/
Ideas from here.
import java.math.BigDecimal;
public class Main {
public static void main(String[] args) {
testConstructors();
testEqualsAndCompare();
testArithmetic();
}
private static void testEqualsAndCompare() {
final BigDecimal zero = new BigDecimal("0.0");
final BigDecimal zerozero = new BigDecimal("0.00");
boolean zerosAreEqual = zero.equals(zerozero);
boolean zerosAreEqual2 = zerozero.equals(zero);
System.out.println("zerosAreEqual: " + zerosAreEqual + " " + zerosAreEqual2);
int zerosCompare = zero.compareTo(zerozero);
int zerosCompare2 = zerozero.compareTo(zero);
System.out.println("zerosCompare: " + zerosCompare + " " + zerosCompare2);
}
private static void testArithmetic() {
try {
BigDecimal value = new BigDecimal(1);
value = value.divide(new BigDecimal(3));
System.out.println(value);
} catch (ArithmeticException e) {
System.out.println("Failed to devide. " + e.getMessage());
}
}
private static void testConstructors() {
double doubleValue = 35.7;
BigDecimal fromDouble = new BigDecimal(doubleValue);
BigDecimal fromString = new BigDecimal("35.7");
boolean decimalsEqual = fromDouble.equals(fromString);
boolean decimalsEqual2 = fromString.equals(fromDouble);
System.out.println("From double: " + fromDouble);
System.out.println("decimalsEqual: " + decimalsEqual + " " + decimalsEqual2);
}
}
It prints
From double: 35.7000000000000028421709430404007434844970703125
decimalsEqual: false false
zerosAreEqual: false false
zerosCompare: 0 0
Failed to devide. Non-terminating decimal expansion; no exact representable decimal result.
How about storing BigDecimal into a database? Hell, it also stores as a double value??? At least, if I use mongoDb without any advanced configuration it will store BigDecimal.TEN as 1E1.
Possible solutions?
I came with one - use String to store BigDecimal in Java as a String into the database. You have validation, for example #NotNull, #Min(10), etc... Then you can use a trigger on update or save to check if current string is a number you need. There are no triggers for mongo though.
Is there a built-in way for Mongodb trigger function calls?
There is one drawback I am having fun around - BigDecimal as String in Swagger defenition
I need to generate swagger, so our front-end team understands that I pass them a number presented as a String. DateTime for example presented as a String.
There is another cool solution I read in the article above...
Use long to store precise numbers.
A standard long value can store the current value of the Unites States national debt (as cents, not dollars) 6477 times without any overflow. Whats more: it’s an integer type, not a floating point. This makes it easier and accurate to work with, and a guaranteed behavior.
Update
https://stackoverflow.com/a/27978223/4587961
Maybe in the future MongoDb will add support for BigDecimal.
https://jira.mongodb.org/browse/SERVER-1393
3.3.8 seems to have this done.
It is an example of the second approach. Use scaling.
http://www.technology-ebay.de/the-teams/mobile-de/blog/mapping-bigdecimals-with-morphia-for-mongodb.html
There is an extensive example of how to do this on javapractices.com. See in particular the Money class, which is meant to make monetary calculations simpler than using BigDecimal directly.
The design of this Money class is intended to make expressions more natural. For example:
if ( amount.lt(hundred) ) {
cost = amount.times(price);
}
The WEB4J tool has a similar class, called Decimal, which is a bit more polished than the Money class.
NumberFormat.getNumberInstance(java.util.Locale.US).format(num);

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