Module talk:TableTools: Difference between revisions
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local ret, isNan, exists, hasNan = {}, p.isNan, {}, nil | local ret, isNan, exists, hasNan = {}, p.isNan, {}, nil | ||
for _, v in ipairs(t) do | 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 | if isNan(v) then | ||
-- But we want only one Nan in ret[], and there may be multiple Nan's in t[]. | -- But we want only one Nan in ret[], and there may be multiple Nan's in t[]. | ||
if not hasNan then | if not hasNan then | ||
Revision as of 07:54, 2 February 2014
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)