Formatting Decimal Places with Unknown Number

Formatting decimal places with unknown number

A Float uses a binary (IEEE 754) representation and cannot represent
all decimal fractions precisely. For example,

let x : Float = 123.456

stores in x the bytes 42f6e979, which is approximately
123.45600128173828. So does x have 3 or 14 fractional digits?

You can use NSNumberFormatter if you specify a maximum number
of decimal digits that should be presented:

let fmt = NSNumberFormatter()
fmt.locale = NSLocale(localeIdentifier: "en_US_POSIX")
fmt.maximumFractionDigits = 3
fmt.minimumFractionDigits = 0

println(fmt.stringFromNumber(123)!)      // 123
println(fmt.stringFromNumber(123.4)!)    // 123.4
println(fmt.stringFromNumber(123.45)!)   // 123.45
println(fmt.stringFromNumber(123.456)!)  // 123.456
println(fmt.stringFromNumber(123.4567)!) // 123.457

Swift 3/4 update:

let fmt = NumberFormatter()
fmt.locale = Locale(identifier: "en_US_POSIX")
fmt.maximumFractionDigits = 3
fmt.minimumFractionDigits = 0

print(fmt.string(for: 123.456)!) // 123.456

MySQL format number with unknown number of decimal places

As suggested split the string drop the trailing zeros format the number before the decimal point and concat taking into account the possibility of no decimals being present at all for example

set @a = 1234.56;

select 
case when instr(@a,'.') > 0 then
  concat(
  format(substring_index(@a,'.',1),'###,###,###'),
  '.',
  trim(trailing '0' from substring_index(@a,'.',-1))
  ) 
else
format (@a,'###,###,###')
end formatted

How to nicely format floating numbers to string without unnecessary decimal 0's

If the idea is to print integers stored as doubles as if they are integers, and otherwise print the doubles with the minimum necessary precision:

public static String fmt(double d)
{
    if(d == (long) d)
        return String.format("%d",(long)d);
    else
        return String.format("%s",d);
}

Produces:

And does not rely on string manipulation.

Creating a decimal number with no decimal separator

Why would you use your own pattern? Java default implementation has pretty good patterns in most of locales. At least by looking Oracle's docs it looks that it should do what you need to:

Locale                  Formatted Numbers
German (Germany)        123.456,789
German (Switzerland)    123'456.789
English (United States) 123,456.789

So what you have to do ( besides dividing a number by 100 ) is set minimum fraction digits to "2":

public static String formatDecimal(double number) {
    NumberFormat german = NumberFormat.getNumberInstance(Locale.GERMAN);
    german.setMinimumFractionDigits(2);
    return german.format(number / 100);
}

Edited: prints numbers as expected:

0,00
0,01
0,02
0,09
0,10
0,11
0,99
1,00
1,01
9,99
10,00
10,01
99,99
100,00
100,01
999,99
1.000,00
1.000,01
9.999,99
10.000,00
10.000,01
99.999,99
100.000,00
100.000,01
999.999,99
1.000.000,00
1.000.000,01

Use DecimalFormat to get varying amount of decimal places

Create a method to generate a certain number of # to a string, like so:

public static String generateNumberSigns(int n) {

    String s = "";
    for (int i = 0; i < n; i++) {
        s += "#";
    }
    return s;
}

And then use that method to generate a string to pass to the DecimalFormat class:

double value = 1234.567890;
int numPlaces = 5;

String numberSigns = generateNumberSigns(numPlaces);
DecimalFormat fmt = new DecimalFormat ("0." + numberSigns);

System.out.println(fmt.format(value));

OR simply do it all at once without a method:

double value = 1234.567890;
int numPlaces = 5;

String numberSigns = "";
for (int i = 0; i < numPlaces; i++) {
    numberSigns += "#";
}

DecimalFormat fmt = new DecimalFormat ("0." + numberSigns);

System.out.println(fmt.format(value));

Formatting decimal places in a character column. as.numeric erase the values in the column

I assume you are somewhere which uses a comma for a decimal point, and perhaps a decimal point in place of a thousands separator.

As an example:

df <- c(',958229561278528615818098193915712388824', '2,05561009284393218251509777394193942492', '2,72096803821411321343605598060792704404', '2,00324997757400185789440370684992098409')

First, remove any decimal points, because they may be thousands separators. Then, replace the comma with a decimal point:

as.numeric(gsub(',', '.', gsub('\\.', '', df)))

Edit: however, if you intend to use more than the first few decimal places, you may run into problems with precision. Look into the package Rmpfr if you need arbitrary precision.

Displaying decimal values with a predefined format

You can use BigDecimal with the desired scale (e.g. 2) and rounding mode (e.g HALF_UP) like so:

import java.math.BigDecimal
import java.math.RoundingMode

fun main() {
    val roundingMode = RoundingMode.HALF_UP
    val doubles: List<Double> = listOf(
        -0.00123, -2.222154, -23.154, -2.13, -0.10001, -10.0012,
        -1.0023, 0.23, 0.56474, 1.000, 11.1111, 1.89566
    )
    doubles.map { BigDecimal(it).setScale(2, roundingMode) }.also { println(it)  }
    // [0.00, -2.22, -23.15, -2.13, -0.10, -10.00, -1.00, 0.23, 0.56, 1.00, 11.11, 1.90]
}

Same approach in a more functional flavour (partial function instead of constants):

val fancyRound: (scale: Int, roundingMode: RoundingMode) -> (Double) -> BigDecimal =
    { scale, roundingMode ->
        { d -> BigDecimal(d).setScale(scale, roundingMode) }
    }

fun main() {
    ...
    val myRound = fancyRound(2, RoundingMode.HALF_UP)
    doubles.map { myRound(it) }.also { println(it) }

}

Set n decimal places in BigDecimal and format the number Java

As mentioned by @JB Nizet, you need to tune NumberFormat acc. to your need.

Below is a working example:

int numberOfDecimalPlaces = 6;

BigDecimal bigDecimal = new BigDecimal(11212.122323);
bigDecimal.setScale(numberOfDecimalPlaces, RoundingMode.HALF_UP);

NumberFormat numberFormat = NumberFormat.getInstance();
numberFormat.setMinimumFractionDigits(numberOfDecimalPlaces);

System.out.println(numberFormat.format(bigDecimal));

Output:

11,212.122323

Formatting Decimal Places with Unknown Number