Re: finding keys in a table
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- Subject: Re: finding keys in a table
- From: Hans Hagen <pragma@...>
- Date: 2008年2月21日 16:15:53 +0100
Alex Davies wrote:
Exactly as you list:
From: "Jelle Huisman" <jelle@jhnet.nl>
(1) lookup if the current character already exists as a key in the table.
(2a) If it does exist I increment the value for key currentchar with 1
(2b) else I add a new entry: currentchar=1
(3) Later I sort the table and print my freq.list.
if tab[char] ~= nil then -- (1) is the char in the table?
tab[char] = tab[char] + 1 -- (2a) then increment
else
tab[char] = 1 -- (2b) else add new entry
end
(It could also be written like this, for a bit more speed: )
local count = tab[char]
tab[char] = (count or 0) + 1 -- initializes count if necessary (an
explanation can be found at: http://www.lua.org/pil/3.3.html)
actually,
tab[char] = (count or 0) + 1
is slower than
if count then
tab[char] = count + 1
else
tab[char] = 1
end
(the if then variant takes 60% of the time of the or case; the
interpreter needs an extra LOADK)
in a similar fashion, parallel assignments can be faster
Hans
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Hans Hagen | PRAGMA ADE
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