How to Get the Min/Max Possible Numeric

How can I get the min/max possible numeric?

help(numeric) sends you to help(double) which has

Double-precision values:

 All R platforms are required to work with values conforming tothe
 IEC 60559 (also known as IEEE 754) standard.  This basically works
 with a precision of 53 bits, and represents to that precision a
 range of absolute values from about 2e-308 to 2e+308.  It also has
 special values ‘NaN’ (many of them), plus and minus infinity and
 plus and minus zero (although R acts as if these are the same).
 There are also _denormal(ized)_ (or _subnormal_) numbers with
 absolute values above or below the range given above but
 represented to less precision.

 See ‘.Machine’ for precise information on these limits.  Note that
 ultimately how double precision numbers are handled is down to the
 CPU/FPU and compiler.

So you want to look at .Machine which on my 64-bit box has

$double.xmin
[1] 2.22507e-308

$double.xmax
[1] 1.79769e+308

Get min and max integer value containing X number of digits

Fox any x>1, the min value would be 10^x-1. E.g.:

• If x = 2, min=10^2-1 = 10¹ = 10
• If x = 3, min=10^3-1 = 10² = 100
• If x = 4, min=10^4-1 = 10³ = 1000

For x=1, since it's the only value that is not a one followed by a series of zeroes, you'll need special treatment.

public static int MinIntWithXDigits(this int x)
{
    if (x == 0) 
    {
        throw new ArgumentException(nameof(x), "An integer cannot contain zero digits.");
    }

    if (x == 1) {
        return 0;
    }

    try
    {
        return Convert.ToInt32(Math.Pow(10, x - 1));
    }
    catch 
    {
        throw new InvalidOperationException($"A number with {x} digits cannot be represented as an integer.");
    }
}

Side note:

If the result were restricted to positive integers instead of non-negative integers, the smallest integer with one digit would be 1, and the general formula would have applied for this case too.

What is the best way to get the minimum or maximum value from an Array of numbers?

The theoretical answers from everyone else are all neat, but let's be pragmatic. ActionScript provides the tools you need so that you don't even have to write a loop in this case!

First, note that Math.min() and Math.max() can take any number of arguments. Also, it's important to understand the apply() method available to Function objects. It allows you to pass arguments to the function using an Array. Let's take advantage of both:

var myArray:Array = [2,3,3,4,2,2,5,6,7,2];
var maxValue:Number = Math.max.apply(null, myArray);
var minValue:Number = Math.min.apply(null, myArray);

Here's the best part: the "loop" is actually run using native code (inside Flash Player), so it's faster than searching for the minimum or maximum value using a pure ActionScript loop.

Maximum and Minimum values for ints

Python 3

In Python 3, this question doesn't apply. The plain int type is unbound.

However, you might actually be looking for information about the current interpreter's word size, which will be the same as the machine's word size in most cases. That information is still available in Python 3 as sys.maxsize, which is the maximum value representable by a signed word. Equivalently, it's the size of the largest possible list or in-memory sequence.

Generally, the maximum value representable by an unsigned word will be sys.maxsize * 2 + 1, and the number of bits in a word will be math.log2(sys.maxsize * 2 + 2). See this answer for more information.

Python 2

In Python 2, the maximum value for plain int values is available as sys.maxint:

>>> sys.maxint
9223372036854775807

You can calculate the minimum value with -sys.maxint - 1 as shown here.

Python seamlessly switches from plain to long integers once you exceed this value. So most of the time, you won't need to know it.

Explanation on Integer.MAX_VALUE and Integer.MIN_VALUE to find min and max value in an array

but as for this method, I don't understand the purpose of Integer.MAX_VALUE and Integer.MIN_VALUE.

By starting out with smallest set to Integer.MAX_VALUE and largest set to Integer.MIN_VALUE, they don't have to worry later about the special case where smallest and largest don't have a value yet. If the data I'm looking through has a 10 as the first value, then numbers[i]<smallest will be true (because 10 is < Integer.MAX_VALUE) and we'll update smallest to be 10. Similarly, numbers[i]>largest will be true because 10 is > Integer.MIN_VALUE and we'll update largest. And so on.

Of course, when doing this, you must ensure that you have at least one value in the data you're looking at. Otherwise, you end up with apocryphal numbers in smallest and largest.

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Note the point Onome Sotu makes in the comments:

...if the first item in the array is larger than the rest, then the largest item will always be Integer.MIN_VALUE because of the else-if statement.

Which is true; here's a simpler example demonstrating the problem (live copy):

public class Example
{
    public static void main(String[] args) throws Exception {
        int[] values = {5, 1, 2};
        int smallest = Integer.MAX_VALUE;
        int largest  = Integer.MIN_VALUE;
        for (int value : values) {
            if (value < smallest) {
                smallest = value;
            } else if (value > largest) {
                largest = value;
            }
        }
        System.out.println(smallest + ", " + largest); // 1, 2 -- WRONG
    }
}

To fix it, either:

1. Don't use else, or

2. Start with smallest and largest equal to the first element, and then loop the remaining elements, keeping the else if.

Here's an example of that second one (live copy):

public class Example
{
    public static void main(String[] args) throws Exception {
        int[] values = {5, 1, 2};
        int smallest = values[0];
        int largest  = values[0];
        for (int n = 1; n < values.length; ++n) {
            int value = values[n];
            if (value < smallest) {
                smallest = value;
            } else if (value > largest) {
                largest = value;
            }
        }
        System.out.println(smallest + ", " + largest); // 1, 5
    }
}

The maximum value for an int type in Go

https://groups.google.com/group/golang-nuts/msg/71c307e4d73024ce?pli=1

The germane part:

Since integer types use two's complement arithmetic, you can infer the
min/max constant values for int and uint. For example,

const MaxUint = ^uint(0) 
const MinUint = 0 
const MaxInt = int(MaxUint >> 1) 
const MinInt = -MaxInt - 1

As per @CarelZA's comment:

uint8  : 0 to 255 
uint16 : 0 to 65535 
uint32 : 0 to 4294967295 
uint64 : 0 to 18446744073709551615 
int8   : -128 to 127 
int16  : -32768 to 32767 
int32  : -2147483648 to 2147483647 
int64  : -9223372036854775808 to 9223372036854775807

Get Minimum and maximum numbers from a list

Assuming you can observe any mutation to the list, you could store (cache) the min and max values and possibly update the cached values when a number is added / removed. At this point, the order of the list doens't matter.

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