I have a datastructure with a field of the float-type. A collection of these structures needs to be sorted by the value of the float. Is there a radix-sort implementation for this.
If there isn't, is there a fast way to access the exponent, the sign and the mantissa. Because if you sort the floats first on mantissa, exponent, and on exponent the last time. You sort floats in O(n).
Now, sort the elements based on digits at tens place. Finally, sort the elements based on the digits at hundreds place. Since radix sort is a non-comparative algorithm, it has advantages over comparative sorting algorithms.
Let the initial array be [121, 432, 564, 23, 1, 45, 788]. It is sorted according to radix sort as shown in the figure below. Please go through the counting sort before reading this article because counting sort is used as an intermediate sort in radix sort. Find the largest element in the array, i.e. max. Let X be the number of digits in max.
Radix Sort Complexity Since radix sort is a non-comparative algorithm, it has advantages over comparative sorting algorithms. For the radix sort that uses counting sort as an intermediate stable sort, the time complexity is O (d (n+k)). Here, d is the number cycle and O (n+k) is the time complexity of counting sort.
In C# this can be done by implementing the interface. methods of the integer and fractional parts. By specification, in fact, must return 0 if two numbers are equal. variables, we are able to perform comparisons with other floating-point types as well. This is because we have instructed the compiler that s. s are the same.
There's a nice explanation of how to perform radix sort on floats here: http://www.codercorner.com/RadixSortRevisited.htm
If all your values are positive, you can get away with using the binary representation; the link explains how to handle negative values.
By doing some fancy casting and swapping arrays instead of copying this version is 2x faster for 10M numbers as Philip Daubmeiers original with grouplength set to 8. It is 3x faster as Array.Sort for that arraysize.
static public void RadixSortFloat(this float[] array, int arrayLen = -1)
{
// Some use cases have an array that is longer as the filled part which we want to sort
if (arrayLen < 0) arrayLen = array.Length;
// Cast our original array as long
Span<float> asFloat = array;
Span<int> a = MemoryMarshal.Cast<float, int>(asFloat);
// Create a temp array
Span<int> t = new Span<int>(new int[arrayLen]);
// set the group length to 1, 2, 4, 8 or 16 and see which one is quicker
int groupLength = 8;
int bitLength = 32;
// counting and prefix arrays
// (dimension is 2^r, the number of possible values of a r-bit number)
var dim = 1 << groupLength;
int groups = bitLength / groupLength;
if (groups % 2 != 0) throw new Exception("groups must be even so data is in original array at end");
var count = new int[dim];
var pref = new int[dim];
int mask = (dim) - 1;
int negatives = 0, positives = 0;
// counting elements of the 1st group incuding negative/positive
for (int i = 0; i < arrayLen; i++)
{
if (a[i] < 0) negatives++;
count[(a[i] >> 0) & mask]++;
}
positives = arrayLen - negatives;
int c;
int shift;
for (c = 0, shift = 0; c < groups - 1; c++, shift += groupLength)
{
CalcPrefixes();
var nextShift = shift + groupLength;
//
for (var i = 0; i < arrayLen; i++)
{
var ai = a[i];
// Get the right index to sort the number in
int index = pref[( ai >> shift) & mask]++;
count[( ai>> nextShift) & mask]++;
t[index] = ai;
}
// swap the arrays and start again until the last group
var temp = a;
a = t;
t = temp;
}
// Last round
CalcPrefixes();
for (var i = 0; i < arrayLen; i++)
{
var ai = a[i];
// Get the right index to sort the number in
int index = pref[( ai >> shift) & mask]++;
// We're in the last (most significant) group, if the
// number is negative, order them inversely in front
// of the array, pushing positive ones back.
if ( ai < 0) index = positives - (index - negatives) - 1; else index += negatives;
//
t[index] = ai;
}
void CalcPrefixes()
{
pref[0] = 0;
for (int i = 1; i < dim; i++)
{
pref[i] = pref[i - 1] + count[i - 1];
count[i - 1] = 0;
}
}
}
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