Re: finding keys in a table

[Hans Hagen <pragma@...>]

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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