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I want to define

isGood[___] = False; isGood[#] = True & /@ list

where list is a list of several million integers. What's the fastest way of doing this?

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    $\begingroup$ I'd use Scan[] instead of Map[], for starters... is there no regular pattern to these "several million integers" that can possibly be exploited? $\endgroup$ Commented Oct 25, 2012 at 16:44
  • $\begingroup$ Those numbers have no pattern. $\endgroup$ Commented Oct 25, 2012 at 16:51

5 Answers 5

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Summary: undocumented HashTable is a bit faster (at least in version 9) both in storage and in retrieval than DownValues.

DownValues

list = RandomInteger[{-10^9, 10^9}, 10^6]; ret = RandomInteger[{-10^9, 10^9}, 10^6]; isGood[___] = False; Scan[(isGood[#] = True) &, list]; // AbsoluteTiming (* ==> {3.240005, Null} *) ClearAll[isGood]; isGood[___] = False; Do[isGood@i = True, {i, list}]; // AbsoluteTiming (* ==> {2.350003, Null} *)

On my computer, this takes less then 3 seconds for a million integers if Do is used instead of Scan. Isn't this fast enough?

Retrieval of the results is also quite quick, and is almost independent whether Table or Map is used:

isGood /@ ret; // AbsoluteTiming (* ==> {1.410002, Null} *) Table[isGood@i, {i, ret}]; // AbsoluteTiming (* ==> {1.450002, Null} *)

HashTable

Out of curiosity, I compared this to the undocumented HashTable (mentioned here) and got even better results. Note, that the hash table must be checked for existing value (as list might contain duplicates) otherwise HashTableAdd returns with error. Or it is even better to prefilter list by removing duplicates, but that is omitted here not to bias the comparison.

hash = System`Utilities`HashTable[]; Do[If[ Not@System`Utilities`HashTableContainsQ[hash, i], System`Utilities`HashTableAdd[hash, i, True] (* last argument can be omitted *) ], {i, list}]; // AbsoluteTiming (* ==> {2.010003, Null} *) System`Utilities`HashTableContainsQ[hash, #] & /@ ret; // AbsoluteTiming (* ==> {1.340002, Null} *) Table[System`Utilities`HashTableContainsQ[hash, i], {i, ret}]; // AbsoluteTiming (* ==> {1.050001, Null} *)

We see that both storage and retrieval are a bit faster.

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  • $\begingroup$ I hope you don't mind my edit: I did justice toHashTable. I wonder why it was so slow (~27 sec) in your original example. Table is obviously faster than Scan & With, but I also assume that in version 8 HashTable did not throw an error when encountering a duplicate hash, but silently failed with a slowdown. $\endgroup$ Commented Jan 11, 2014 at 14:30
  • $\begingroup$ I also expect version 10 associations to be at the performance level of HashTable, but that will need another post. $\endgroup$ Commented Jan 11, 2014 at 14:32
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Are you sure you want to use UpValues? You can use Dispatch which is pretty fast when generating the lookup table and is equally fast when accessing values:

n = 6; list = RandomInteger[{0, 10^(n + 1)}, {10^n}]; AbsoluteTiming[disp = Dispatch@Thread[list -> True];] 
{1.6220927, Null} 
Remove[isGood]; AbsoluteTiming[isGood[___] = False; Scan[(isGood[#] = True) &, list]] 
{3.5982058, Null}

Query values:

test = RandomInteger[{0, 10^(n + 1)}, {10^n}]; AbsoluteTiming[Count[test /. disp, True]] 
{1.9151096, 94844} 
AbsoluteTiming[Count[isGood /@ test, True]] 
{1.7601007, 94844}
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    $\begingroup$ In this case he wants to use DownValues not UpValues. $\endgroup$ Commented Oct 25, 2012 at 20:12
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This seems to be faster to define downvalues

list = RandomInteger[{-100000000, 100000000}, 1000000]; DownValues[isGood] = HoldPattern[isGood[#]] :> True & /@ list; // AbsoluteTiming isGood[___] = False;
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  • $\begingroup$ Again, Scan[] would be preferable to using Map[] in this case... $\endgroup$ Commented Oct 26, 2012 at 3:46
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    $\begingroup$ @J.M., I'm saving all downvalues at once $\endgroup$ Commented Oct 26, 2012 at 3:48
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My solution is ugly, but task-specific. It builds a bitmap out of machine-sized integers in imperative fashion and uses Compile. This works reasonably in memory usage for ranges that have at least couple percent of True values.

A million integers:

n = 6; list = RandomInteger[{0, 10^(n + 1)}, {10^n}];

Function itself:

<< Developer` isGood = With[ {bits = Floor[Log[2, $MaxMachineInteger + 1]], min = Min@list, max = Max@list}, With[ {bv = Compile[{{list, _Integer, 1}}, Module[ {bitvec = Table[0, {(max - min + bits - 1)~Quotient~bits}]}, Scan[(bitvec[[(# - min)~Quotient~bits + 1]] += 2^((# - min)~Mod~bits)) &, Union@list]; bitvec ], CompilationOptions -> {"ExpressionOptimization" -> True, "InlineExternalDefinitions" -> True} ]@list}, Compile[{{x, _Integer}}, min <= x <= max && bv[[(x - min)~Quotient~bits + 1]]~BitAnd~(2^((x - min)~Mod~bits)) != 0, CompilationOptions -> {"ExpressionOptimization" -> True, "InlineExternalDefinitions" -> True} ] ] ]; // AbsoluteTiming // First (* 0.347843 *)

Usage:

isGood /@ list // AbsoluteTiming // First (* 0.590347 *)

Originally I wanted to solve this problem using bitwise operations of arbitrarily large integers, but the issue with that is that functional programming with bigints has large return value overheads - even when just some individual bits are twiddled.

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    $\begingroup$ A few possible tweaks: use bits = BitLength[$MaxMachineInteger] to compute the maximum bit length, and you can use bitvec = ConstantArray[0, Quotient[max - min + bits - 1, bits]] for initialization. $\endgroup$ Commented Oct 27, 2012 at 1:13
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What you aim at can be done quicker now with Association. I compare only to the Do method, which is also reasonably fast:

list = RandomInteger[{-10^9, 10^9}, 10^6]; ret = RandomInteger[{-10^9, 10^9}, 10^6]; ClearAll[isGood]; isGood[___] = False; Do[isGood@i = True, {i, list}]; // AbsoluteTiming isAlsoGood = With[{data = Union[list]}, AssociationThread[data -> ConstantArray[True, Length[data]]] ]; // AbsoluteTiming (* 1.43326 *) (* 0.673688 *)

And looking up values is even four times faster:

aa = Map[isGood, ret]; // AbsoluteTiming // First bb = Lookup[isAlsoGood, ret, False]; // AbsoluteTiming // First aa == bb (* 0.959888 *) (* 0.2606 *) (* True *)

Even deallocation is considerably faster:

ClearAll[isGood]; // AbsoluteTiming // First ClearAll[isAlsoGood]; // AbsoluteTiming // First (* 0.656773 *) (* 0.279496 *)
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  • $\begingroup$ Note that Boolean arrays cannot be packed. Only integer, real and complex arrays can be packed. $\endgroup$ Commented Oct 22, 2017 at 17:04
  • $\begingroup$ @CarlWoll Yes I know. This ToPackedArrayQ was what remained from may experiments with Boolean. I removed it. $\endgroup$ Commented Oct 22, 2017 at 17:07

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