CODEWITHSUNDEEP
mathc#-3.0roundingtruncate
Spilarix
Spilarix

Reputation: 1468

How to simply truncate a number for display

I would like to know the simplest way to transform a number to a truncated and rounded form.
I know many way to do it "manually" (truncate manually, put a comma then round the number after) but I think there are easier ways to do it (probably with Maths methods).

Example :
1 234 567 should become 1,2 M
1 567 890 should become 1.6 M

Upvotes: 0

Views: 79

Answers (2)

user1196549
user1196549

Reputation:

The integer part of the base-10 logarithm gives you the exponent as appearing in the scientific notation.

If you want s significant digits, normalize by dividing by ten to the exponent, rescale by ten to s-1 and round.

e= floor(log10(x)); // => e = 6
x= Round(x * pow(10, s - 1 - e)); // => x = 12, x = 16

Upvotes: 1

David Gourde
David Gourde

Reputation: 3914

Something like that:

static string FormatNumber(uint n)
{
    if (n < 1000)
        return n.ToString();
    if (n < 10000)
        return String.Format("{0:#,.##}K", n - 5);
    if (n < 100000)
        return String.Format("{0:#,.#}K", n - 50);
    if (n < 1000000)
        return String.Format("{0:#,.}K", n - 500);
    if (n < 10000000)
       return String.Format("{0:#,,.##}M", n - 5000);
    if (n < 100000000)
       return String.Format("{0:#,,.#}M", n - 50000);
    if (n < 1000000000)
       return String.Format("{0:#,,.}M", n - 500000);
    return String.Format("{0:#,,,.##}B", n - 5000000);
    }

Will give you this output:

1249            1.24K
12499           12.4K
124999          124K
1249999         1.24M
12499999        12.4M
124999999       124M
1249999999      1.24B

I don't think there are built-in libraries to do this.

Upvotes: 1

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