Python and STL efficiency

[andrei.zavidei at gmail.com]

---------------------------- C++ --------------------------
#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <algorithm>
#include <windows.h>

using namespace std;

int main()
{
	DWORD ticks = ::GetTickCount();

	const string s1("What do you know");
	const string s2("So long...");
	const string s3("chicken crosses road");
	const string s4("fool");

	typedef vector<const string*> str_vector_t;
	typedef str_vector_t::iterator vec_iter;
	typedef set<const string*> str_set_t;
	typedef str_set_t::const_iterator set_iter;


	const int size = 1000000;
	str_vector_t vec;
	vec.reserve(size*4);

	for(int i = 0; i < size; ++i){
		vec.push_back(&s1);
		vec.push_back(&s2);
		vec.push_back(&s3);
		vec.push_back(&s4);
	}

	vec_iter new_end = unique(vec.begin(), vec.end());
	str_set_t set_(vec.begin(), new_end);

	for(set_iter it = set_.begin(); it != set_.end(); ++it){
		cout<<*it<<endl;
	}
	cout<<::GetTickCount()-ticks<<endl;

	return 0;
}

In MS VC+ 2005 in release configuration it gets the work done in 187
ms.

---------------- Python + Psyco----------------
def f():
		a = []
		for i in range(1000000):
				a.append('What do you know')
				a.append('so long...')
				a.append('chicken crosses road')
				a.append('fool')
		b = set(a)
		for s in b:
				print s

import time
from time import clock

import psyco
psyco.full()
f_start = clock()
f()
f_end   = clock()

print "Elapsed: %f seconds" % (f_end - f_start)

In Python in manages to do the same in 1.8 secs (psyco boosts it to
0.7; see below)
so long...
What do you know
fool
chicken crosses road
Elapsed: 0.772899 seconds


Well, that's how it is in my book.

Regards,
Andrei