我正在研究一些在Ruby中以不同格式解析时间值的文本转换例程。这个例程越来越复杂,我目前正在测试更好的方法来解决这个问题。为什么ruby scanf太慢了?
我目前正在测试一种使用方法scanf
。为什么?我总是认为比正则表达式快,但是Ruby中发生了什么?这太慢了!
我在做什么错?
注:我使用的红宝石1.9.2-P290 [x86_64的(MRI)
第一台红宝石测试:
require "scanf"
require 'benchmark'
def duration_in_seconds_regex(duration)
if duration =~ /^\d{2,}\:\d{2}:\d{2}$/
h, m, s = duration.split(":").map{ |n| n.to_i }
h * 3600 + m * 60 + s
end
end
def duration_in_seconds_scanf(duration)
a = duration.scanf("%d:%d:%d")
a[0] * 3600 + a[1] * 60 + a[2]
end
n = 500000
Benchmark.bm do |x|
x.report { for i in 1..n; duration_in_seconds_scanf("00:10:30"); end }
end
Benchmark.bm do |x|
x.report { for i in 1..n; duration_in_seconds_regex("00:10:30"); end }
end
这是我使用scanf
第一和第二的正则表达式:
user system total real
95.020000 0.280000 95.300000 (96.364077)
user system total real
2.820000 0.000000 2.820000 ( 2.835170)
第二测试使用C:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <sys/types.h>
#include <string.h>
#include <regex.h>
char *regexp(char *string, char *patrn, int *begin, int *end) {
int i, w = 0, len;
char *word = NULL;
regex_t rgT;
regmatch_t match;
regcomp(&rgT, patrn, REG_EXTENDED);
if ((regexec(&rgT, string, 1, &match, 0)) == 0) {
*begin = (int) match.rm_so;
*end = (int) match.rm_eo;
len = *end - *begin;
word = malloc(len + 1);
for (i = *begin; i<*end; i++) {
word[w] = string[i];
w++;
}
word[w] = 0;
}
regfree(&rgT);
return word;
}
int main(int argc, char** argv) {
char * str = "00:01:30";
int h, m, s;
int i, b, e;
float start_time, end_time, time_elapsed;
regex_t regex;
regmatch_t * pmatch;
char msgbuf[100];
char *pch;
char *str2;
char delims[] = ":";
char *result = NULL;
start_time = (float) clock()/CLOCKS_PER_SEC;
for (i = 0; i < 500000; i++) {
if (sscanf(str, "%d:%d:%d", &h, &m, &s) == 3) {
s = h * 3600L + m * 60L + s;
}
}
end_time = (float) clock()/CLOCKS_PER_SEC;
time_elapsed = end_time - start_time;
printf("sscanf_time (500k iterations): %.4f", time_elapsed);
start_time = (float) clock()/CLOCKS_PER_SEC;
for (i = 0; i < 500000; i++) {
char * match = regexp(str, "[0-9]{2,}:[0-9]{2}:[0-9]{2}", &b, &e);
if (strcmp(match, str) == 0) {
str2 = (char*) malloc(sizeof (str));
strcpy(str2, str);
h = strtok(str2, delims);
m = strtok(NULL, delims);
s = strtok(NULL, delims);
s = h * 3600L + m * 60L + s;
}
}
end_time = (float) clock()/CLOCKS_PER_SEC;
time_elapsed = end_time - start_time;
printf("\n\nregex_time (500k iterations): %.4f", time_elapsed);
return (EXIT_SUCCESS);
}
分
C代码的结果是明显加快,和正则表达式的结果是慢scanf
结果不出所料:
sscanf_time (500k iterations): 0.1774
regex_time (500k iterations): 3.9692
很明显,在C运行时间比较快,所以请不要评论说,Ruby是解释和像这样的东西请。
这是相关的gist。
难道你不是在C中每次迭代重新编译表达式吗?我不认为Ruby这样做。如果只编译一次表达式,我会很有兴趣看到C结果。另外,你为什么甚至使用拆分?您正在匹配字符串,因此您可以直接捕获值,而无需对字符串进行进一步操作。 – Qtax 2012-03-01 20:35:55
是的,我正在重新编译,它可能比这更快,但我有时需要更改exp。 – AndreDurao 2012-03-01 20:41:40
然后,只需在更改时重新编译它即可。但我只想看看这些数字。 ;-) – Qtax 2012-03-01 20:45:31