One of n lines in a file: Difference between revisions
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} |
} |
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inline int |
inline int one_of_n(int n) |
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{ |
{ |
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int i, r = 0; |
int i, r = 0; |
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int i, r[10] = {0}; |
int i, r[10] = {0}; |
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for (i = 0; i < 1000000; i++, r[ |
for (i = 0; i < 1000000; i++, r[one_of_n(10)]++); |
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for (i = 0; i < 10; i++) |
for (i = 0; i < 10; i++) |
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printf("%d%c", r[i], i == 9 ? '\n':' '); |
printf("%d%c", r[i], i == 9 ? '\n':' '); |
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return 0; |
return 0; |
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}</lang>output<lang>100561 99814 99816 99721 99244 99772 100790 100072 99997 100213</lang> |
}</lang>output<lang>100561 99814 99816 99721 99244 99772 100790 100072 99997 100213</lang> |
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=={{header|Perl 6}}== |
=={{header|Perl 6}}== |
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{{trans|Python}} |
{{trans|Python}} |
Revision as of 04:21, 8 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>
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]