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I'm trying to get rid of unnecessary symbols after decimal seperator of my double value. I'm doing it this way:

DecimalFormat format = new DecimalFormat("#.#####");
value = Double.valueOf(format.format(41251.50000000012343));

But when I run this code, it throws:

java.lang.NumberFormatException: For input string: "41251,5"
    at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:1224)
    at java.lang.Double.valueOf(Double.java:447)
    at ...

As I see, Double.valueOf() works great with strings like "11.1", but it chokes on strings like "11,1". How do I work around this? Is there a more elegant way then something like

Double.valueOf(format.format(41251.50000000012343).replaceAll(",", "."));

Is there a way to override the default decimal separator value of DecimalFormat class? Any other thoughts?

folone
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10 Answers10

86

By

get rid of unnecessary symbols after decimal seperator of my double value

do you actually mean you want to round to e.g. the 5th decimal? Then just use

value = Math.round(value*1e5)/1e5;

(of course you can also Math.floor(value*1e5)/1e5 if you really want the other digits cut off)

edit

Be very careful when using this method (or any rounding of floating points). It fails for something as simple as 265.335. The intermediate result of 265.335 * 100 (precision of 2 digits) is 26533.499999999996. This means it gets rounded down to 265.33. There simply are inherent problems when converting from floating point numbers to real decimal numbers. See EJP's answer here at https://stackoverflow.com/a/12684082/144578 - How to round a number to n decimal places in Java

Tobias Kienzler
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  • Won't this cause a problem when value*1e5 exceeds the maximum double value, effectivly reducing the maximum value `value` can have to `Double.MAX_VALUE/1e5`? – Jimmy R.T. Jan 26 '21 at 12:49
  • @JimmyR.T. Indeed it will. However, to quote from [Wikipedia on Doubles](https://en.wikipedia.org/wiki/Double-precision_floating-point_format): _The 11 bit width of the exponent allows the representation of numbers between 10⁻³⁰⁸ and 10³⁰⁸, with full 15–17 decimal digits precision._ So you cannot store a number with 308-5=303 digits to round to anyway. – Tobias Kienzler Jan 27 '21 at 06:48
  • I'm aware that it won't be necessary in that case, but I think a solution should just keep the number the same if nothing is to be cut off instead of generating an overflow; or that restriction should be mentioned. Just a minor remark for completeness – Jimmy R.T. Jan 27 '21 at 12:55
  • @JimmyR.T. Absolutely, I never thought about it. It's a corner case IMHO, but of course one could work around it. Or just refer to [another answer here](https://stackoverflow.com/a/153785/321973) ;) – Tobias Kienzler Jan 27 '21 at 13:35
24

The problem is that your decimal format converts your value to a localized string. I'm guessing that your default decimal separator for your locale is with a ','. This often happens with French locales or other parts of the world.

Basically what you need to do is create your formatted date with the '.' separator so Double.valueOf can read it. As indicated by the comments, you can use the same format to parse the value as well instead of using Double.valueOf.

DecimalFormatSymbols symbols = DecimalFormatSymbols.getInstance();
symbols.setDecimalSeparator('.');
DecimalFormat format = new DecimalFormat("#.#####", symbols);
value = format.parse(format.format(41251.50000000012343));
the Tin Man
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Chris Dail
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    or `DecimalFormat format = new DecimalFormat("#"+symbols.getDecimalSeparator+"#####", symbols;` without changing the decimal separator – Tobias Kienzler Apr 22 '10 at 13:30
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    If you already have the DecimalFormat, why not use it for parsing as well? Much more robust than trying to build a DecimalFormat whose formatted output can be parsed by Double.valueOf() – Michael Borgwardt Apr 22 '10 at 13:37
  • @Michael Borgwardt Good point. I've updated the answer to include that – Chris Dail Apr 22 '10 at 13:53
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    I'm tempted to downvote this answer - not because it doesn't solve the problem, but because it supports this insane attempt at rounding - what the [OP admitted to be the goal](http://stackoverflow.com/questions/2691018/decimalformat-and-double-valueof/2691162#comment2712294_2691162). But technically it's the answer the OP asked for :-/ – Tobias Kienzler Aug 22 '12 at 21:53
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    @VipulPurohit For this error, yes, but for the problem behind it: NO! One should not use string operations on floats to perform MATHS – Tobias Kienzler Aug 16 '13 at 09:10
6

The fact that your formatting string uses . as the decimal separator while the exception complains of , points to a Locale issue; i.e. DecimalFormat is using a different Locale to format the number than Double.valueOf expects.

In general, you should construct a NumberFormat based on a specific Locale.

Locale myLocale = ...;
NumberFormat f = NumberFormat.getInstance(myLocale);

From the JavaDocs of DecimalFormat:

To obtain a NumberFormat for a specific locale, including the default locale, call one of NumberFormat's factory methods, such as getInstance(). In general, do not call the DecimalFormat constructors directly, since the NumberFormat factory methods may return subclasses other than DecimalFormat.

However as BalusC points out, attempting to format a double as a String and then parse the String back to the double is a pretty bad code smell. I would suspect that you are dealing with issues where you expect a fixed-precision decimal number (such as a monetary amount) but are running into issues because double is a floating point number, which means that many values (such as 0.1) cannot be expressed precisely as a double/float. If this is the case, the correct way to handle a fixed-precision decimal number is to use a BigDecimal.

Community
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matt b
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5

Use Locale.getDefault() to get your system's decimal separator which you can also set. You can't have two different separators at the same time since the other is then usually used as the separator for thousands: 2.001.000,23 <=> 2,001,000.23

Vincent Tjeng
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Tobias Kienzler
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4

The real solution is: don't use floating-point numbers for anything that needs to be counted with precision:

  • If you are dealing with currency, don't use a double number of dollars, use an integer number of cents.
  • If you are dealing with hours of time and need to count quarter-hours and 10-minute intervals, use an integer number of minutes.

A floating point number is almost always an approximation of some real value. They are suitable for measurements and calculation of physical quantities (top a degree of precision) and for statistical artifacts.

Fooling about with rounding floating point to a number of digits is a code smell: it's wasteful and you can never really be sure that your code will work properly in all cases.

Mark Rotteveel
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PaulMurrayCbr
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3

For the ',' instead of the '.' , you'd have to change the locale.

For the number of decimals, use setMaximumFractionDigits(int newValue).

For the rest, see the javadoc.

the Tin Man
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Maxime ARNSTAMM
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3

looks like your local use a comma "," as a decimal separation.To get the "." as a decimal separator, you will have to declare:

DecimalFormat dFormat =new DecimalFormat("#.#", new DecimalFormatSymbols(Locale.ENGLISH));
Vaibhav Barad
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2

You can't change the internal representation of double/Double that way.

If you want to change the (human) representation, just keep it String. Thus, leave that Double#valueOf() away and use the String outcome of DecimalFormat#format() in your presentation. If you ever want to do calculations with it, you can always convert back to a real Double using DecimalFormat and Double#valueOf().

By the way, as per your complain I'm trying to get rid of unnecessary symbols after decimal seperator of my double value, are you aware of the internals of floating point numbers? It smells a bit like that you're using unformatted doubles in the presentation layer and that you didn't realize that with the average UI you can just present them using DecimalFormat without the need to convert back to Double.

BalusC
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  • The problem is when I convert it to `String`, it then won't convert back to `Double`, using `Double.valueOf()`, because `DecimalFormat` places `,`, where `valueOf()` is expecting `.`. So is there a way to override this in `DecimalFormat` class (so that I don't have to place a call to `replaceAll(",", ".")` there)? – folone Apr 22 '10 at 13:14
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    Use `DecimalFormat` to reformat it into the format as expected by `Double`. – BalusC Apr 22 '10 at 13:16
  • Yeah, but `DecimalFormat` places it's default decimal separator (which is `,`). How do I make it place `.` instead? – folone Apr 22 '10 at 13:18
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    The separator is locale dependent. Ensure that you get the right `DecimalFormat` instance for the desired locale. – BalusC Apr 22 '10 at 13:20
  • @folone See my answer, e.g. `Locale.setDefault(Locale.ENGLISH)` – Tobias Kienzler Apr 22 '10 at 13:20
  • Goldberg's paper is *way* too high-level to serve as a basic introduction to floating point to be thrown at newbies in the manner or RTFM. For this reason I've written http://floating-point-gui.de/ – Michael Borgwardt Apr 22 '10 at 13:24
  • @Michael: now that's a nice one. I updated the link accordingly. – BalusC Apr 22 '10 at 13:32
  • I don't actually use doubles in gui or currency presentations (here I use them for Ole Date representation). But thanks, that's a great link. – folone Apr 22 '10 at 13:40
2

Somewhat related to this, but not an answer to the question: try switching to BigDecimal instead of doubles and floats. I was having a lot of issue with comparisons on those types and now I'm good to go with BigDecimal.

Joao Coelho
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0

My code function :

 private static double arrondi(double number){
         DecimalFormatSymbols symbols = DecimalFormatSymbols.getInstance();
         symbols.setDecimalSeparator('.');
         DecimalFormat format = new DecimalFormat("#.#####", symbols);
         return Double.valueOf(format.format(number));
     }
Neoray
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