Here it is given a possible method in order to count data with weights. That solution works. However, it is ~140 times slower than the Bincounts
function. See, e.g. the following code:
Ndata = 10^6;
SeedRandom[321];
data = RandomReal[{0, 10}, Ndata];
weights = RandomReal[{.9, 1.1}, Ndata];
bins = Table[i, {i, 0, 10}];
AbsoluteTiming[Last[HistogramList[WeightedData[data, weights], {bins}]]]
AbsoluteTiming[BinCounts[data, {bins}]]
I would like to know if it is possible to make a faster function.
A faster method could be to apply the function proposed here to the output of BinLists
:
myFn = Merge[KeyIntersection[PositionIndex /@ {##}], Identity] &;
blist = BinLists[data, {bins}]
myFn[blist, Partition[data, 1]]
(*<|{0.986147} -> {{1}, {3}}, {1.49106} -> {{2}, {6}}, {3.23491} -> {{4}, {7}}, {7.15785} -> {{8}, {5}}, {8.9058} -> {{9}, {9}}|>*)
Then I could use the latter output (the positions) to sum the weights. However, myFn
does not identify the positions if there are more elements inside a bin.
Does someone know how to improve upon this attempt? Other solutions are most welcome. Perhaps one could see how does the Python function numpy.bincount
work?
Answer
I would use a solution based on Pick
:
wBinCounts[data_, weights_, bins_] := Module[{min, max, b, d},
{min, max} = MinMax[data];
If[min<0 && max>0,
d = data - min + 1;
b = bins - min + 1,
d = data;
b = bins
];
With[{ranges = Partition[b, 2, 1]},
Total @ Pick[weights, Unitize@Clip[d, {##}, {0, 0}], 1]& @@@ ranges
]
]
For your data:
Ndata=10^6;
SeedRandom[321];
data=RandomReal[{0,10},Ndata];
weights=RandomReal[{.9,1.1},Ndata];
bins=Table[i,{i,0,10}];
AbsoluteTiming[Last[HistogramList[WeightedData[data,weights],{bins}]]]
AbsoluteTiming[BinCounts[data, {bins}]]
AbsoluteTiming[wBinCounts[data,weights,bins]]
{19.7562, {99667.3, 100028., 99756.9, 100022., 100024., 100361., 100513., 99787.1, 99992.4, 99881.1}}
{0.132353, {99688, 100024, 99735, 100031, 99998, 100371, 100488, 99778, 100001, 99886}}
{0.096202, {99667.3, 100028., 99756.9, 100022., 100024., 100361., 100513., 99787.1, 99992.4, 99881.1}}
Update
Clip[x, {min, max}]
uses $min \leq x \leq max$, and so values of $x$ at the end points will be included in both intervals. A slightly slower version of wBinCounts
that avoids this issue (using $min \leq x < max$):
wBinCountsLeft[data_, weights_, bins_] := With[{ranges=Partition[bins,2,1]},
Total @ Pick[
weights,
BitXor[UnitStep[Subtract[data, #1]], UnitStep[Subtract[data, #2]]],
1
]& @@@ ranges
]
And a version using $min Example: {0, 2, 1} {1, 2, 0}wBinCountsRight[data_, weights_, bins_] := With[{ranges=Partition[bins,2,1]},
Total @ Pick[
weights,
BitXor[UnitStep[Subtract[#1,data]], UnitStep[Subtract[#2,data]]],
1
]& @@@ ranges
]wBinCountsLeft[{1,1.5,2}, {1,1,1}, {0,1,2,3}]
wBinCountsRight[{1,1.5,2}, {1,1,1}, {0,1,2,3}]
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