Re: Re: Taking sums across indices of a SparseArray

*To*: mathgroup at smc.vnet.net*Subject*: [mg96066] Re: [mg96053] Re: Taking sums across indices of a SparseArray*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Wed, 4 Feb 2009 05:18:45 -0500 (EST)*References*: <200901301042.FAA06408@smc.vnet.net> <gm0q2n$rle$1@smc.vnet.net>*Reply-to*: drmajorbob at longhorns.com

> What if I want to sum across just the 3rd index of an n-dimensional > array? Can this still be made to work? Depending on what you mean by that: toy=SparseArray[{i_,j_,k_,l_}->i+j-k+l,{4,5,6,7}]; dims=Dimensions@toy {4,5,6,7} Transpose[toy,{1,2,4,3}].SparseArray[_->1,dims[[3]]] SparseArray[<140>,{4,5,7}] or SparseArray[_->1,dims[[4]]].(SparseArray[_->1,dims[[2]]].(SparseArray[_->1,dims[[1]]].Transpose[toy,{1,2,4,3}])) SparseArray[<6>,{6}] Bobby On Tue, 03 Feb 2009 05:32:52 -0600, D. Grady <D.C.Grady at gmail.com> wrote: > On Jan 31, 12:11 am, Carl Woll <ca... at wolfram.com> wrote: >> D. Grady wrote: >> >Suppose we've got a four-dimensional array: >> >> >t = Array[Subscript[w, ##] &, {3, 3, 3, 3}] >> >> >If we want to take the sum across one index of this array (which will >> >reduce its dimension), we can use the Total function: >> >> >Dimensions@Total[t, {2}] >> >{3, 3, 3} >> >> >In the problem I'm working on, I've got an array and I need to sum >> >across the first two dimensions. Using this toy array, I can see that >> >Total[t,{1,2}] gives me exactly the object that I want. The problem >> >is that I'm working with a four-dimensional sparse array, and Total >> >will apparently always try to convert its first argument to a normal >> >array. This fails because the array is too big to fit in memory: >> >> >In[26]:= WAF = Total[W, {1, 2}]; // Timing >> >> >During evaluation of In[26]:= SparseArray::ntb: Cannot convert the >> >sparse array SparseArray[Automatic,{489,489,489,489},0,{<<1>>}] to an >> >ordinary array because the 57178852641 elements required exceeds the >> >current size limit. >> >> >> >Out[26]= SystemException[SparseArrayNormalLimit,Normal[SparseArray >> >[<1400152>,{489,489,489,489}]]] >> >> >I can roll my own function to do this computation just by sorting >> >through the ArrayRules: >> >> >Timing[ >> > WAF = >> > SparseArray@( >> > (#[[1, 1, 3 ;; 4]] -> Total[#[[All, 2]]] &) /@ >> > (SplitBy[#, Drop[First@#, 2] &] &)@ >> > (SortBy[#, Drop[First@#, 2] &] &)@ >> > Most@ >> > ArrayRules@ >> > W)] >> >> >{22.1335,SparseArray[<21122>,{489,489}]} >> >> >The point is that actually doing the computation isn't particularly >> >memory or time intensive, but I can't find a simple way to do this >> >directly using built-in functions like Total. Does anyone know if >> >there is a way? If there isn't, why not? Thanks a lot! >> >> >-Daniel >> >> Another workaround is to use Dot. Here is a toy array: >> >> In[31]:= toy = >> SparseArray[{i_, j_, k_, l_} -> i + j - k + l, {4, 4, 4, 4}] >> >> Out[31]= SparseArray[<252>,{4,4,4,4}] >> >> In[32]:= Total[toy, {1, 2}] >> >> Out[32]= {{80, 96, 112, 128}, {64, 80, 96, 112}, {48, 64, 80, >> 96}, {32, 48, 64, 80}} >> >> In[33]:= SparseArray[_ -> 1, 4].(SparseArray[_ -> 1, 4].toy) >> >> Out[33]= SparseArray[<16>,{4,4}] >> >> In[34]:= % // Normal >> >> Out[34]= {{80, 96, 112, 128}, {64, 80, 96, 112}, {48, 64, 80, >> 96}, {32, 48, 64, 80}} >> >> Carl Woll >> Wolfram Research > > What if I want to sum across just the 3rd index of an n-dimensional > array? Can this still be made to work? > > -Daniel > -- DrMajorBob at longhorns.com