One of n lines in a file: Difference between revisions
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=={{header|Icon}} and {{header|Unicon}}== |
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<lang Icon>procedure main() # one of n |
<lang Icon>procedure main() # one of n |
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one_of_n_test(10,1000000) |
one_of_n_test(10,1000000) |
Revision as of 00:23, 9 September 2011
A method of choosing a line randomly from a file:
- Without reading the file more than once
- When substantial parts of the file cannot be held in memory
- Without knowing how many lines are in the file
Is to:
- keep the first line of the file as a possible choice, then
- Read the second line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/2.
- Read the third line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/3.
- ...
- Read the Nth line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/N
- Return the computed possible choice when no further lines exist in the file.
- Task
- Create a function/method/routine called
one_of_n
that givenn
, the number of actual lines in a file, follows the algotrithm above to return an integer - the line number of the line chosen from the file.
The number returned can vary, randomly, in each run. - Use
one_of_n
in a simulation to find what woud be the chosen line of a 10 line file simulated 1,000,000 times. - Print and show how many times each of the 10 lines is chosen as a rough measure of how well the algorithm works.
Note: You may choose a smaller number of repetitions if necessary, but mention this up-front.
C
<lang c>#include <stdio.h>
- include <stdlib.h>
inline int irand(int n) { int r, randmax = RAND_MAX/n * n; while ((r = rand()) >= randmax); return r / (randmax / n); }
inline int one_of_n(int n) { int i, r = 0; for (i = 1; i < n; i++) if (!irand(i + 1)) r = i; return r; }
int main(void) { int i, r[10] = {0};
for (i = 0; i < 1000000; i++, r[one_of_n(10)]++); for (i = 0; i < 10; i++) printf("%d%c", r[i], i == 9 ? '\n':' ');
return 0; }</lang>output<lang>100561 99814 99816 99721 99244 99772 100790 100072 99997 100213</lang>
Icon and Unicon
<lang Icon>procedure main() # one of n
one_of_n_test(10,1000000)
end
procedure one_of_n(n)
every i := 1 to n do choice := (?0 < 1. / i, i) return \choice | fail
end
procedure one_of_n_test(n,trials)
bins := table(0) every i := 1 to trials do bins[one_of_n(n)] +:= 1 every writes(bins[i := 1 to n]," ") return bins
end</lang>
Sample output:
99470 99806 99757 99921 100213 100001 99778 100385 100081 100588
Perl 6
<lang perl6>sub one_of_n($n) {
my $choice; $choice = $_ if rand * $_ < 1 for 1 .. $n; $choice - 1;
}
sub one_of_n_test($n = 10, $trials = 1000000) {
my @bins; @bins[one_of_n($n)]++ for ^$trials; @bins;
}
say one_of_n_test();</lang> Output:
100288 100047 99660 99773 100256 99633 100161 100483 99789 99910
Python
<lang python>from random import random as rnd
def one_of_n(n):
# Zero based line numbers choice = 0 for i in range(1, n): if rnd() < 1. / (i + 1.): choice = i return choice
def one_of_n_test(n=10, trials=1000000):
bins = [0] * n if n: for i in range(trials): bins[one_of_n(n)] += 1 return bins
print(one_of_n_test())</lang>
- Sample output
[99833, 100303, 99902, 100132, 99608, 100117, 99531, 100017, 99795, 100762]