On Tue, 15 Feb 2005 09:48:23 -0500, Wiggins d'Anconia <[EMAIL PROTECTED]> wrote: > Please bottom post, and reply-all so that everyone can help and be helped. > > Nope. That's a here are the tools that you should be able to determine > on your own which is faster. ".... teach a man to fish ...." > > http://danconia.org > > Ken Gillett wrote: > > That's a no then? > > > > On 14 Feb 2005, at 15:00, Wiggins d'Anconia wrote: > > > >> Ken Gillett wrote: > >> > >>> I have a script that creates a hash, up to several thousand > >>> key=>value pairs. Each value is a string that is created by adding to > >>> it repeatedly, maybe hundreds of times, each addition probably about > >>> 10 bytes. > >>> I can do this in (at least) 2 ways. One is to repeatedly concatenate > >>> ( .= ) the additional string onto a scalar variable and then, once it > >>> has been fully created, to add this variable to the hash with its > >>> appropriate key. The other is to directly add onto the hash value > >>> itself, no other variable involved. > >>> My question is:- > >>> Which is faster in operation? > >>> I don't know enough about the internal workings of perl's memory > >>> structures (actually I know nothing about that:-) to be able to > >>> hazard a guess at this. Maybe it makes no measurable difference, but > >>> maybe one is definitely the better modus operandi. > >>> Can anyone answer this? > >> > >> > >> perldoc Benchmark > >> > >> Check the list archives for examples of usage on other problems.
To elaborate on the "no" a little: there are a host of things that affect execution. Diifferent processes use system resources in differnt ways. Some take more memory, some take more processor time. Some scale lineraly, some have a high initial overhead. Different architectures and processors are optimised for different tasks. The list goes on. Often, you'll find that the fastest solution for a small data set isn't the fastest for a larger one, and vice versa. No one on this list is you, sitting at your computer, with your data. So none of us can tell you what's best. Here are some things to think about. Generally, assigning to a temporary variable is slower than modifying a vaule directly (because it incurs the overhead of creating the variable). On the other hand, hashes take up a lot of memory, and modifying large hases can be time consuming. So it would be reasonbale to expect that for a "small" data set, modifying the hash directly would be faster, but that on a "large"--in terms of bytes--data set assigning to a temp variable might possibly be faster, because it only modifies the hash once. But only Becnchmark will tell you what large and small mean for your system in this situation, and where the tradeoff happens for you. Remember, too, that your results will change depending on what the rest of your program is doing, and the available system resources. Each platform also has its own quirks. Look into the archives for some of the recent threads on Benchmarking, and you'll see why the answer to your original question has to be no. HTH, --jay -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
