There's a faster way to do this that requires only a single scan of the list.
Consider picking a line at random from a text file when you don't know how many lines are in the file, and the file is too large to fit in memory. The obvious solution is to read the file once to count the lines, pick a random number in the range of 0 to Count-1, and then read the file again up to the chosen line number. That works, but requires you to read the file twice.
A faster solution is to read the first line and save it as the selected line. You replace the selected line with the next line with probability 1/2. When you read the third line, you replace with probability 1/3, etc. When you've read the entire file, you have selected a line at random, and every line had equal probability of being selected. The code looks something like this:
string selectedLine = null;
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
double prob = 1.0/numLines;
if (rnd.Next() >= prob)
selectedLine = line;
}
Now, what if you want to select 2 lines? You select the first two. Then, as each line is read the probability that it will replace one of the two lines is 2/n, where n is the number of lines already read. If you determine that you need to replace a line, you randomly select the line to be replaced. You can follow that same basic idea to select any number of lines at random. For example:
string[] selectedLines = new int[M];
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
if (numLines <= M)
{
selectedLines[numLines-1] = line;
}
else
{
double prob = (double)M/numLines;
if (rnd.Next() >= prob)
{
int ix = rnd.Next(M);
selectedLines[ix] = line;
}
}
}
You can apply that to your BitArray
quite easily:
int[] selected = new int[quantity];
int num = 0; // number of True items seen
Random rnd = new Random();
for (int i = 0; i < items.Length; ++i)
{
if (items[i])
{
++num;
if (num <= quantity)
{
selected[num-1] = i;
}
else
{
double prob = (double)quantity/num;
if (rnd.Next() > prob)
{
int ix = rnd.Next(quantity);
selected[ix] = i;
}
}
}
}
You'll need some special case code at the end to handle the case where there aren't quantity
set bits in the array, but you'll need that with any solution.
This makes a single pass over the BitArray
, and the only extra memory it uses is for the list of selected indexes. I'd be surprised if it wasn't significantly faster than the LINQ version.
Note that I used the probability calculation to illustrate the math. You can change the inner loop code in the first example to:
if (rnd.Next(numLines+1) == numLines)
{
selectedLine = line;
}
++numLines;
You can make a similar change to the other examples. That does the same thing as the probability calculation, and should execute a little faster because it eliminates a floating point divide for each item.