Re: lpeg pattern repetition

[Nick Gammon <nick@...>]

On 02/06/2008, at 2:57 PM, Roberto Ierusalimschy wrote:
> Actually, consuming no input is slightly less efficient, because the
> machine must consume the input (otherwise how to match it?) and then
> backtrack. But in this particular case we have '-line', so it does
> not consume input anyway.

I was working out some timings while you wrote that, and my results  
agree with your analysis. :)



Further to my question about whether "consuming" the pattern takes  
time or not, here are some test results:

local match = "(.*) says, (.*)"
local target = "Nick Gammon says, But does it get goat's blood out?"
local count = 10000000
local who, what

start = os.time ()
for i = 1, count do
 who, what = string.match (target, match)
end -- for
period = os.time () - start

print ("Time taken = ", period, " secs")  --> 15 seconds

require "lpeg"

local line = lpeg.P " says, "
local patt = lpeg.C((1 - line)^1) * line * lpeg.C (lpeg.P(1)^1)

start = os.time ()
for i = 1, count do
 who, what = lpeg.match (patt, target)
end -- for
period = os.time () - start

print ("Time taken = ", period, " secs")  --> 12 seconds

local patt = lpeg.C((1 - #line)^1) * line * lpeg.C (lpeg.P(1)^1)

start = os.time ()
for i = 1, count do
 who, what = lpeg.match (patt, target)
end -- for
period = os.time () - start

print ("Time taken = ", period, " secs")  --> 16 seconds

-------
Summary
-------


All tests produced the same (correct) results.

string.match took 15 seconds
lpeg.match took 12 seconds using (1 - line)
lpeg.match took 16 seconds using (1 - #line)

So it seems that leaving the # out is somewhat faster.

- Nick