Suppose I have two very large lists {a1, a2, …} and {b1, b2, …} where all ai and bj are large sparse arrays. For the sake of memory efficiency I store each list as one comprehen
I will propose a solution which is rather complex but allows one to only use about twice as much memory during the computation as is needed to store the final result as a SparseArray
. The price to pay for this will be a much slower execution.
Here is the code. First, a slightly modified (to address higher-dimensional sparse arrays) sparse array construction - deconstruction API, taken from this answer:
ClearAll[spart, getIC, getJR, getSparseData, getDefaultElement,
makeSparseArray];
HoldPattern[spart[SparseArray[s___], p_]] := {s}[[p]];
getIC[s_SparseArray] := spart[s, 4][[2, 1]];
getJR[s_SparseArray] := spart[s, 4][[2, 2]];
getSparseData[s_SparseArray] := spart[s, 4][[3]];
getDefaultElement[s_SparseArray] := spart[s, 3];
makeSparseArray[dims_List, jc_List, ir_List, data_List, defElem_: 0] :=
SparseArray @@ {Automatic, dims, defElem, {1, {jc, ir}, data}};
The following functions produce iterators. Iterators are a good way to encapsulate the iteration process.
ClearAll[makeTwoListIterator];
makeTwoListIterator[fname_Symbol, a_List, b_List] :=
With[{indices = Flatten[Outer[List, a, b, 1], 1]},
With[{len = Length[indices]},
Module[{i = 0},
ClearAll[fname];
fname[] := With[{ind = ++i}, indices[[ind]] /; ind <= len];
fname[] := Null;
fname[n_] :=
With[{ind = i + 1}, i += n;
indices[[ind ;; Min[len, ind + n - 1]]] /; ind <= len];
fname[n_] := Null;
]]];
Note that I could have implemented the above function more memory - efficiently and not use Outer
in it, but for our purposes this won't be the major concern.
Here is a more specialized version, which produces interators for pairs of 2-dimensional indices.
ClearAll[make2DIndexInterator];
make2DIndexInterator[fname_Symbol, i : {iStart_, iEnd_}, j : {jStart_, jEnd_}] :=
makeTwoListIterator[fname, Range @@ i, Range @@ j];
make2DIndexInterator[fname_Symbol, ilen_Integer, jlen_Integer] :=
make2DIndexInterator[fname, {1, ilen}, {1, jlen}];
Here is how this works:
In[14]:=
makeTwoListIterator[next,{a,b,c},{d,e}];
next[]
next[]
next[]
Out[15]= {a,d}
Out[16]= {a,e}
Out[17]= {b,d}
We can also use this to get batch results:
In[18]:=
makeTwoListIterator[next,{a,b,c},{d,e}];
next[2]
next[2]
Out[19]= {{a,d},{a,e}}
Out[20]= {{b,d},{b,e}}
, and we will be using this second form.
This function will build a SparseArray
object iteratively, by getting chunks of data (also in SparseArray
form) and gluing them together. It is basically code used in this answer, packaged into a function. It accepts the code piece used to produce the next chunk of data, wrapped in Hold
(I could alternatively make it HoldAll
)
Clear[accumulateSparseArray];
accumulateSparseArray[Hold[getDataChunkCode_]] :=
Module[{start, ic, jr, sparseData, dims, dataChunk},
start = getDataChunkCode;
ic = getIC[start];
jr = getJR[start];
sparseData = getSparseData[start];
dims = Dimensions[start];
While[True, dataChunk = getDataChunkCode;
If[dataChunk === {}, Break[]];
ic = Join[ic, Rest@getIC[dataChunk] + Last@ic];
jr = Join[jr, getJR[dataChunk]];
sparseData = Join[sparseData, getSparseData[dataChunk]];
dims[[1]] += First[Dimensions[dataChunk]];
];
makeSparseArray[dims, ic, jr, sparseData]];
This function is the main one, putting it all together:
ClearAll[sparseArrayOuter];
sparseArrayOuter[f_, a_SparseArray, b_SparseArray, chunkSize_: 100] :=
Module[{next, wrapperF, getDataChunkCode},
make2DIndexInterator[next, Length@a, Length@b];
wrapperF[x_List, y_List] := SparseArray[f @@@ Transpose[{x, y}]];
getDataChunkCode :=
With[{inds = next[chunkSize]},
If[inds === Null, Return[{}]];
wrapperF[a[[#]] & /@ inds[[All, 1]], b[[#]] & /@ inds[[All, -1]]]
];
accumulateSparseArray[Hold[getDataChunkCode]]
];
Here, we first produce the iterator which will give us on demand portions of index pair list, used to extract the elements (also SparseArrays
). Note that we will generally extract more than one pair of elements from two large input SparseArray
-s at a time, to speed up the code. How many pairs we process at once is governed by the optional chunkSize
parameter, which defaults to 100
. We then construct the code to process these elements and put the result back into SparseArray
, where we use an auxiliary function wrapperF
. The use of iterators wasn't absolutely necessary (could use Reap
-Sow
instead, as with other answers), but allowed me to decouple the logic of iteration from the logic of generic accumulation of sparse arrays.
First we prepare large sparse arrays and test our functionality:
In[49]:=
arr = {SparseArray[{{1,1,1,1}->1,{2,2,2,2}->1}],SparseArray[{{1,1,1,2}->1,{2,2,2,1}->1}],
SparseArray[{{1,1,2,1}->1,{2,2,1,2}->1}],SparseArray[{{1,1,2,2}->-1,{2,2,1,1}->1}],
SparseArray[{{1,2,1,1}->1,{2,1,2,2}->1}],SparseArray[{{1,2,1,2}->1,{2,1,2,1}->1}]};
In[50]:= list=SparseArray[arr]
Out[50]= SparseArray[<12>,{6,2,2,2,2}]
In[51]:= larger = sparseArrayOuter[Dot,list,list]
Out[51]= SparseArray[<72>,{36,2,2,2,2,2,2}]
In[52]:= (large= sparseArrayOuter[Dot,larger,larger])//Timing
Out[52]= {0.047,SparseArray[<2592>,{1296,2,2,2,2,2,2,2,2,2,2}]}
In[53]:= SparseArray[Flatten[Outer[Dot,larger,larger,1],1]]==large
Out[53]= True
In[54]:= MaxMemoryUsed[]
Out[54]= 21347336
Now we do the power tests
In[55]:= (huge= sparseArrayOuter[Dot,large,large,2000])//Timing
Out[55]= {114.344,SparseArray[<3359232>,{1679616,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2}]}
In[56]:= MaxMemoryUsed[]
Out[56]= 536941120
In[57]:= ByteCount[huge]
Out[57]= 262021120
In[58]:= (huge1 = Flatten[Outer[Dot,large,large,1],1]);//Timing
Out[58]= {8.687,Null}
In[59]:= MaxMemoryUsed[]
Out[59]= 2527281392
For this particular example, the suggested method is 5 times more memory-efficient than the direct use of Outer
, but about 15 times slower. I had to tweak the chunksize
parameter (default is 100
, but for the above I used 2000
, to get the optimal speed / memory use combination). My method only used as a peak value twice as much memory as needed to store the final result. The degree of memory-savings as compared to Outer
- based method will depend on the sparse arrays in question.