Java: Format number in millions

Question

Is there a way to use DecimalFormat (or some other standard formatter) to format numbers like this:

1,000,000 => 1.00M

1,234,567 => 1.23M

1,234,567,890 => 1234.57M

Basically dividing some number by 1 million, keeping 2 decimal places, and slapping an 'M' on the end. I've thought about creating a new subclass of NumberFormat but it looks trickier than I imagined.

I'm writing an API that has a format method that looks like this:

public String format(double value, Unit unit); // Unit is an enum

Internally, I'm mapping Unit objects to NumberFormatters. The implementation is something like this:

public String format(double value, Unit unit) { NumberFormatter formatter = formatters.get(unit); return formatter.format(value); }

Note that because of this, I can't expect the client to divide by 1 million, and I can't just use String.format() without wrapping it in a NumberFormatter.

Do you want to handle M(ega) only, or also (G)iga, (T)era, etc? — Zach Scrivena
– Zach Scrivena, Commented Feb 9, 2009 at 19:22
It actually represents a security's volume, so it's M(illions) and potentially B(illions) but I'll be happy with just the M. — Tim Frey
– Tim Frey, Commented Feb 9, 2009 at 19:31
This related question has been getting much attention recently (due to a bounty). — Amos M. Carpenter
– Amos M. Carpenter, Commented Jun 11, 2015 at 1:05

jjnguy · Accepted Answer · 2009-02-09 19:21:03Z

20

String.format("%.2fM", theNumber/ 1000000.0);

For more information see the String.format javadocs.

answered Feb 9, 2009 at 19:21

jjnguy

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7 Comments

Zach Scrivena Over a year ago

Just in case: use 1000000.0 instead.

wds Over a year ago

Minor nitpick, looks like his code is using longs or ints so you can dispense with the floating point arithmetic:

jjnguy Over a year ago

If I leave off the floating point math, I would be left with a whole number, no decimal places at all. We need at least 2.

Tim Frey Over a year ago

This works but I still need to wrap it with a NumberFormat object because of an additional constraint that I forgot to mention. Basically there is a MAP[Type => NumberFormatter] that this needs to play nicely with.

jjnguy Over a year ago

Well, I have never used a NumberFormatter...so I don't know the easiest way to wrap it.

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oxbow_lakes · Accepted Answer · 2009-02-09 22:32:59Z

Note that if you have a BigDecimal, you can use the movePointLeft method:

new DecimalFormat("#.00").format(value.movePointLeft(6));

Muhammad Bilal ahmad · Accepted Answer · 2019-08-30 06:50:51Z

For someone looking out there to convert a given digit in human readable form.

public static String getHumanReadablePriceFromNumber(long number){ if(number >= 1000000000){ return String.format("%.2fB", number/ 1000000000.0); } if(number >= 1000000){ return String.format("%.2fM", number/ 1000000.0); } if(number >= 100000){ return String.format("%.2fL", number/ 100000.0); } if(number >=1000){ return String.format("%.2fK", number/ 1000.0); } return String.valueOf(number); }

Tim Frey · Accepted Answer · 2009-02-09 19:18:31Z

Here's a subclass of NumberFormat that I whipped up. It looks like it does the job but I'm not entirely sure it's the best way:

private static final NumberFormat MILLIONS = new NumberFormat() { private NumberFormat LOCAL_REAL = new DecimalFormat("#,##0.00M"); public StringBuffer format(double number, StringBuffer toAppendTo, FieldPosition pos) { double millions = number / 1000000D; if(millions > 0.1) LOCAL_REAL.format(millions, toAppendTo, pos); return toAppendTo; } public StringBuffer format(long number, StringBuffer toAppendTo, FieldPosition pos) { return format((double) number, toAppendTo, pos); } public Number parse(String source, ParsePosition parsePosition) { throw new UnsupportedOperationException("Not implemented..."); } };

I like the solution from Outlaw, especially as it can also create k/M/G "human readable" formatting without the need for the user of this api to do calculations. He always gets the shortest possible number. 999999 999.99k 999.99M 999.99G 999.99T

toolkit · Accepted Answer · 2009-02-09 19:27:07Z

3

Why not simply?

DecimalFormat df = new DecimalFormat("0.00M"); System.out.println(df.format(n / 1000000));

answered Feb 9, 2009 at 19:27

toolkit

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1 Comment

Tim Frey Over a year ago

The division needs to be encapsulated inside the formatter. Guess I'll update the question to more clearly explain the problem I'm having.

Evstropov V · Accepted Answer · 2020-11-04 23:00:35Z

In Kotlin language, you can make extention function:

fun Long.formatToShortNumber(): String { return when { this >= 1000000000 -> String.format("%.2fB", this / 1000000000.0) this >= 1000000 -> String.format("%.2fM", this / 1000000.0) this >= 1000 -> String.format("%.2fK", this / 1000.0) else -> this.toString() } }

sblundy · Accepted Answer · 2009-02-09 19:15:14Z

0

Take a look at ChoiseFormat.

A more simplistic way would be to use a wrapper that auto divided by 1m for you.

edited Feb 9, 2009 at 19:15

answered Feb 9, 2009 at 19:11

sblundy

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1 Comment

Tim Frey Over a year ago

Took a look a the docs but I'm really not sure how that's going to help me here. Seems like ChoiceFormat basically contains a bunch of formats and somehow matches the input with one of these sub-formats. I think I want all input to be handled the same.

Heath Borders · Accepted Answer · 2019-03-18 14:50:45Z

For now, you should use ICU's CompactDecimalFormat, which will localize the formatting result for non-english locales. Other locales might not use a "Millions" suffix.

This functionality will be standard Java in JDK 12 with CompactNumberFormat.

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