Module talk:TableTools

removeDuplicate does not remove duplicate NaN

<source lang="lua"> function p.removeDuplicates(t) checkType('removeDuplicates', 1, t, 'table') local isNan = p.isNan local ret, exists = {}, {} for i, v in ipairs(t) do if isNan(v) then -- NaNs can't be table keys, and they are also unique, so we don't need to check existence. ret[#ret + 1] = v else if not exists[v] then ret[#ret + 1] = v exists[v] = true end end end return ret end </source> This should be: <source lang="lua"> function p.removeDuplicates(t) checkType('removeDuplicates', 1, t, 'table') local ret, isNan, exists, hasNan = {}, p.isNan, {}, nil for _, v in ipairs(t) do -- NaNs can't be table keys in exists[], and they are also equal to each other in Lua. if isNan(v) then -- But we want only one Nan in ret[], and there may be multiple Nan's in t[]. if not hasNan then hasNan = true ret[#ret + 1] = v end else if not exists[v] then exists[v] = true ret[#ret + 1] = v end end end return ret end </source> -- verdy_p (talk) 07:50, 2 February 2014 (UTC)

@Verdy p: This was by design, as comparing two NaNs always results in false. My reasoning was that since two NaNs can never be equal to each other - even if they were made by the exact same calculation - then they shouldn't be treated as duplicates by the algorithm. Although if there's some sort of precedent for doing things a different way, please let me know. I'm fairly new to the world of NaNs, after all. — Mr. Stradivarius ^{♪ talk ♪} 08:01, 2 February 2014 (UTC)