How to go through a number (float) and get every number separately with an int. Java

Question

Is there any good way to search through a floats first four numbers and return every number separately with int[]?

Example: the float 23,51 becomes the integer array, array[0]=2, array[1]=3, array[2]=5 and last array[3]=1

My code:

public void printNumber(float number){
    String string = String.valueOf(number);
    while(!numbers.isEmpty()){
        numbers.remove(0);
    }
    for(int i = 0; i < string.length(); i++) {
        int j = Character.digit(string.charAt(i), 10);
        this.number = new Number(j);
        numbers.add(this.number);
        System.out.println("digit: " + j);
    }
}

I should mention that Number is a class that only returns a different picture based on the number the constructor is given and ofcourse the number itself.

numbers is an ArrayList

"The first four numbers of a float" isn't well-defined for binary. — chrylis -cautiouslyoptimistic-, Aug 30 '16 at 19:15
convert to a string with a format specifier, then iterate through the string elements? — OldProgrammer, Aug 30 '16 at 19:17
I have tried that without any luck, may be wrong somewhere in my code so I'll post it above. — Viktor Lindblad, Aug 30 '16 at 19:18

score 5 · Accepted Answer · answered Aug 30 '16 at 19:26

Convert float to String using fixed-point format, then go through its characters one-by-one, and ignore the decimal point.

If the number could also be negative, you need to pay attention to the sign in the String output:

float v = 23.51F;
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
char[] d = df.format(v).toCharArray();
int count = 0;
for (int i = 0 ; i != d.length ; i++) {
    if (Character.isDigit(d[i])) {
        count++;
    }
}
int[] res = new int[count];
int pos = 0;
for (int i = 0 ; i != d.length ; i++) {
    if (Character.isDigit(d[i])) {
        res[pos++] = Character.digit(d[i], 10);
    }
}

Demo.

Important: Be aware that floats are inherently imprecise, so you may get a "stray" digit or two. For example, your example produces

[2 3 5 1 0 0 0 0 2 3]

with 2 and 3 at the end.

Tip: To avoid the mentioned inaccuracy of `float` introducing extraneous extra digits at the end, use [`BigDecimal`](https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html) objects to represent fractional numbers instead of `float` and `double` primitives. — Basil Bourque, Aug 30 '16 at 19:44
@BasilBourque Absolutely, `BigDecimal` would solve this problem. However, one needs to have `BigDecimal` to begin with. If you construct it from a `float` or `double`, the problem is back; if you start with a `String`, you can skip `BigDecimal` altogether. — Sergey Kalinichenko, Aug 30 '16 at 19:47

MaxZoom · Answer 2 · 2016-08-30T21:12:58.700

You can convert the float to String with 4 decimal places using String.format method, and then get each character to int array

float floatValue = 12.34567f;
String str = String.format("%.4f", floatValue);
// remove the minus, dot, or comma (used in some countries)
str = str.replaceAll("[-|.|,]", "");
int [] nums  = new int[str.length()];
for (int i=0; i<str.length(); i++) {
  nums [i] = str.charAt(i) - '0';
}

Here is DEMO

As for the code last line, decimal value of '0' char (which is 48) is subtracted from a decimal value of digit char, and the result is integer value of that digit (as specified in below table):

beatngu13 · Answer 3 · 2016-09-03T07:36:04.847

1

Java 8 flavored solution:

float number = -7.54f;
int[] digits = String.format("%.3f", number)
                     .chars()
                     .filter(Character::isDigit)
                     .limit(4L)
                     .map(Character::getNumericValue)
                     .toArray();

System.out.println(Arrays.toString(digits)); //=> [7, 5, 4, 0]

edited Sep 03 '16 at 07:36

answered Aug 31 '16 at 11:34

beatngu13

7,201
6
37
66

I'd be glad if the downvoter could explain what's wrong with the answer. – beatngu13 Sep 01 '16 at 23:07

score 0 · Answer 4 · answered Aug 30 '16 at 19:29

Use the static method Float.toString() to convert your float to a String.

Then go through each char and use Integer.parseInt() to get back to an int.

public void printNumber(float number){
    String string = Float.toString(number);

    for(int i = 0; i < string.length(); i++) {
        int j = Integer.parseInt(string.charAt(i));
        this.number = new Number(j);
        numbers.add(this.number);
        System.out.println("digit: " + j);
    }
}

score 0 · Answer 5 · answered Aug 30 '16 at 19:29

Another option

public static void main(String[] args) {

   float number = 23.51f;
   String strNumber = String.valueOf(number).replaceAll("\\D", "");
   int[] arrNumber = new int[strNumber.length()];

   for (int pos = 0; pos < strNumber.length(); ++pos) {
        arrNumber[pos] = Integer.valueOf(String.valueOf(strNumber.charAt(pos)));
        System.out.println(arrNumber[pos]);
    }
}

How to go through a number (float) and get every number separately with an int. Java

5 Answers5