How to go about formatting 1200 to 1.2k in Android studio

Question

I'd like to format following numbers into the numbers next to them with Android:

1000 to 1k 5821 to 5.8k 2000000 to 2m 7800000 to 7.8m

Possible duplicate of [How to go about formatting 1200 to 1.2k in java](http://stackoverflow.com/questions/4753251/how-to-go-about-formatting-1200-to-1-2k-in-java) — Hiren Jungi, Jan 25 '17 at 19:01
Try out this library https://github.com/tygalive/Number-Shortener [Disclaimer am the owner] — Richard Muvirimi, Apr 06 '21 at 09:17

score 25 · Answer 1 · answered Aug 14 '19 at 06:06

Function

public String prettyCount(Number number) {
    char[] suffix = {' ', 'k', 'M', 'B', 'T', 'P', 'E'};
    long numValue = number.longValue();
    int value = (int) Math.floor(Math.log10(numValue));
    int base = value / 3;
    if (value >= 3 && base < suffix.length) {
        return new DecimalFormat("#0.0").format(numValue / Math.pow(10, base * 3)) + suffix[base];
    } else {
        return new DecimalFormat("#,##0").format(numValue);
    }
}

Use

prettyCount(789); Output: 789
prettyCount(5821); Output: 5.8k
prettyCount(101808); Output: 101.8k
prettyCount(7898562); Output: 7.9M

failed for (9990), expected 9.9k found 10.0k – Ishaan Jun 23 '20 at 12:09 — Ishaan, Jun 23 '20 at 12:09
@Ishaan, it rounds up that's why. – Chuka_FC Mar 06 '22 at 02:45 — Chuka_FC, Mar 06 '22 at 02:45

score 16 · Answer 2 · answered Jan 30 '20 at 10:55

Try this method :

For Java

public static String formatNumber(long count) {
    if (count < 1000) return "" + count;
    int exp = (int) (Math.log(count) / Math.log(1000));
    return String.format("%.1f %c", count / Math.pow(1000, exp),"kMGTPE".charAt(exp-1));
}

For Kotlin(Android)

fun getFormatedNumber(count: Long): String {
if (count < 1000) return "" + count
val exp = (ln(count.toDouble()) / ln(1000.0)).toInt()
return String.format("%.1f %c", count / 1000.0.pow(exp.toDouble()), "kMGTPE"[exp - 1])
}

score 13 · Accepted Answer · edited Feb 07 '19 at 12:29

13

This should do the trick

String numberString = "";
if (Math.abs(number / 1000000) > 1) {
   numberString = (number / 1000000).toString() + "m";

} else if (Math.abs(number / 1000) > 1) {
   numberString = (number / 1000).toString() + "k";

} else {
   numberString = number.toString();

}

edited Feb 07 '19 at 12:29

Garvit Jain

1,862
2
19
27

answered Jan 25 '17 at 19:04

StackOverflower

5,463
13
58
89

5

Your logic will not show 1000 to 1k, because greater than equal to sign should be there. Also your logic will not show 5821 to 5.8k, it will show 5k instead of 5.8k. – Rahul Gupta Mar 17 '19 at 10:21

score 11 · Answer 4 · answered Oct 29 '18 at 10:37

11

Try this trick:

 private String numberCalculation(long number) {
    if (number < 1000) 
        return "" + number;
    int exp = (int) (Math.log(number) / Math.log(1000));
    return String.format("%.1f %c", number / Math.pow(1000, exp), "kMGTPE".charAt(exp-1));
}

answered Oct 29 '18 at 10:37

Tushar Lathiya

940
9
26

Doesn't work with negative numbers – Psijic Sep 05 '22 at 15:16
Modified: ```fun compactNumber(number: Int): String { if (number in 0 until 1000) return number.toString() val exp = (ln(abs(number).toDouble()) / ln(1000.0)).toInt() return String.format("%.1f%c", number / 1000.0.pow(exp), "KMGTPE"[exp - 1]) }``` – Psijic Sep 05 '22 at 15:27

Guillaume Husta · Answer 5 · 2021-02-15T16:39:31.343

As of Android 7.0 (API 24), it is possible to use CompactDecimalFormat.

For example :

    private String convertNumber(int number, Locale locale) {
        CompactDecimalFormat compactDecimalFormat =
                CompactDecimalFormat.getInstance(locale, CompactDecimalFormat.CompactStyle.SHORT);
        return compactDecimalFormat.format(number);
    }

This class is also part of ICU v49 (https://unicode-org.github.io/icu-docs/apidoc/dev/icu4j/com/ibm/icu/text/CompactDecimalFormat.html).

Furthermore, as of Android 11 (API 30), it is possible to use NumberFormatter.

This class is also part of ICU v60.

score 3 · Answer 6 · answered Jul 06 '21 at 08:21

// Kotlin Version

fun countviews(count:Long): String{
    val array = arrayOf(' ', 'k', 'M', 'B', 'T', 'P', 'E')
    val value = Math.floor(Math.log10(count.toDouble())).toInt()
    val  base = value / 3
        if (value >= 3 && base < array.size) {
            return DecimalFormat("#0.0").format(count/ Math.pow(10.0, (base * 3).toDouble())) + array[base]
        } else {
            return DecimalFormat("#,##0").format(count)
        }
    }

Wahib Ul Haq · Answer 7 · 2023-04-11T14:51:26.977

I needed to implement the same solution and took a hint from one of the answer in this thread. It suggested to use NumberFormatter for Android API 30+.

This is what is working for me. Additional benefit of using NumberFromatter API is that it adds the appropriate suffix as per Locale. Notation.compactShort() is the one adding engineering notations (K, M, B, T)

/**
 * Converts Number to appropriate rounded off format with max 1 decimal place
 * e.g. 1345 --> 1.3K, 15555900 --> 15.6M
*/
private fun formatCount(number: Long) = NumberFormatter.with()
    .notation(Notation.compactShort())
    .decimal(NumberFormatter.DecimalSeparatorDisplay.ALWAYS)
    .precision(Precision.fixedFraction(1))
    .locale(Resources.getSystem().configuration.locales.get(0))
    .format(number)
    .toString()

score 1 · Answer 8 · edited Dec 01 '20 at 09:09

This a Kotlin "extension on Double" that worked for me.
This also doesn't round the number:

fun Double.currencyCountWithSuffix(): String {
val suffixChars = "KMGTPE"
val formatter = DecimalFormat("###.#")
formatter.roundingMode = RoundingMode.DOWN

return if (this < 1000.0) formatter.format(this)
else {
    val exp = (ln(this) / ln(1000.0)).toInt()
    formatter.format(this / 1000.0.pow(exp.toDouble())) + suffixChars[exp - 1]
  }
}

score 0 · Answer 9 · edited Jan 20 '21 at 02:27

public static String prettyCount(Double number) {
    DecimalFormat df = new DecimalFormat("#.#");
    String numberString = "";

    if (Math.abs(number / 1000000) >= 1) {
        numberString = df.format(number / 1000000.0) + "m";

    } else if (Math.abs(number / 1000.0) >= 1) {
        numberString = df.format(number / 1000.0) + "k";

    } else {
        numberString = number.toString();

    }
    return numberString;
}

score 0 · Answer 10 · answered May 28 '23 at 18:19

Summary of all answers

private val Long.shortNotation
    get(): String {
        return if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
            val locale = AppCompatDelegate.getApplicationLocales()[0]
                ?: Resources.getSystem().configuration.locales.get(0)
            if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.R) NumberFormatter.with()
                .notation(Notation.compactShort())
                .locale(locale)
                .format(this)
                .toString()
            else CompactDecimalFormat
                .getInstance(locale, CompactDecimalFormat.CompactStyle.SHORT)
                .format(this)
        } else {
            val base = 1_000f
            if (this < base) return "$this"
            val ratio = ln(this.toDouble()) / ln(base)
            val exp = ratio.toInt()
            val format = if (ratio % 1 * 3 < 1) "%.1f%c" else "%.0f%c"
            @Suppress("SpellCheckingInspection")
            String.format(format, this / base.pow(exp), "kMGTPE"[exp - 1])
        }
    }

score 0 · Answer 11 · answered Jun 20 '23 at 11:37

Simple use this function for kotlin

fun formatNumber(number: Long): String {
val suffix = charArrayOf(' ', 'k', 'M', 'B', 'T', 'P', 'E')
val value = floor(log10(number.toDouble())).toInt()
val base = value / 3
val result = if (value >= 3 && base < suffix.size) {
    DecimalFormat("#.#").format(
        number / 10.0.pow((base * 3).toDouble())
    ) + suffix[base]
} else {
    DecimalFormat("#,##0").format(number)
}
return result.convertToEnglishDigits()

}

HAZEEM JOONUS · Answer 12 · 2021-02-26T01:08:40.063

This is a modified version of Sujith Manjavana answer.

here i have used DecimalFormat to control the integers after the decimal point.

long kil = 1000;
long meg = kil * 1000;
long gig = meg * 1000;
long ter = gig * 1000;

double bytes = 56789; // your integer. if you are using a String then -->  double bytes = Double.parseDouble(56789);
DecimalFormat form = new DecimalFormat("0.0"); // you can also use 0.00 or 0.000 


if ((bytes >= 0) && (bytes < kil)) {
String oneDecimal = form.format(bytes);
Toast.makeText(MainActivity.this, oneDecimal, Toast.LENGTH_LONG).show();

} else if ((bytes >= kil) && (bytes < meg)) {
    String oneDecimal = form.format(bytes / kil);
    Toast.makeText(MainActivity.this, oneDecimal+" K", Toast.LENGTH_LONG).show();
    
     } else if ((bytes >= meg) && (bytes < gig)) {
         String oneDecimal = form.format(bytes / meg);
         Toast.makeText(MainActivity.this, oneDecimal+" M", Toast.LENGTH_LONG).show();
         
         } else if ((bytes >= gig) && (bytes < ter)) {
             String oneDecimal = form.format(bytes / gig);
             Toast.makeText(MainActivity.this, oneDecimal+" B", Toast.LENGTH_LONG).show();
            
            } else if (bytes >= ter) {
                String oneDecimal = form.format(bytes / ter);
                Toast.makeText(MainActivity.this, oneDecimal+" T", Toast.LENGTH_LONG).show();
            } else {
                String oneDecimal = form.format(bytes);
                Toast.makeText(MainActivity.this, oneDecimal, Toast.LENGTH_LONG).show();
            }

How to go about formatting 1200 to 1.2k in Android studio

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