       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>

```> 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[]]
SparseArray[<140>,{4,5,7}]

or

SparseArray[_->1,dims[]].(SparseArray[_->1,dims[]].(SparseArray[_->1,dims[]].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:
>> >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:= WAF = Total[W, {1, 2}]; // Timing
>>
>> >During evaluation of In:= 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= 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:= toy =
>>  SparseArray[{i_, j_, k_, l_} -> i + j - k + l, {4, 4, 4, 4}]
>>
>> Out= SparseArray[<252>,{4,4,4,4}]
>>
>> In:= Total[toy, {1, 2}]
>>
>> Out= {{80, 96, 112, 128}, {64, 80, 96, 112}, {48, 64, 80,
>>   96}, {32, 48, 64, 80}}
>>
>> In:= SparseArray[_ -> 1, 4].(SparseArray[_ -> 1, 4].toy)
>>
>> Out= SparseArray[<16>,{4,4}]
>>
>> In:= % // Normal
>>
>> Out= {{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

```

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