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MathGroup Archive 2006

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Re: replacement rule and sparsearray question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71352] Re: replacement rule and sparsearray question
  • From: Bill Rowe <readnewsciv at sbcglobal.net>
  • Date: Wed, 15 Nov 2006 06:45:03 -0500 (EST)

On 11/14/06 at 5:06 AM, Arkadiusz.Majka at gmail.com wrote:

>How can I apply a replacement rule to sparse array, e.g

>SparseArray[{1,2,3}]/.{2->222}

>and obtain the same result as for {1,2,3}/.{2->222} (what is
>{1,222,3}.

>I DON'T want to call Normal before.

In this particular example where you know the position of the 
element to be replaced, use ReplacePart, i.e.,

In[45]:=
s=SparseArray[Range@5];
d=ReplacePart[s,222,2];
Normal@d

Out[47]=
{1,222,3,4,5}

if you don't know the position of the element to be replaced, it 
can be obtained as follows:

In[48]:=
pos = DeleteCases[ArrayRules[s] /.
     HoldPattern[{a_} -> 2] :> a, _Rule]

Out[48]=
{2}

Now ReplacePart can be used as above, i.e.

In[49]:=
d=ReplacePart[s,222,pos];
Normal@d

Out[50]=
{1,222,3,4,5}

but if the element to be replaced occurs multiple times, this 
last would need to be done as

In[51]:=
d=ReplacePart[s,222,Partition[pos,1]];
Normal@d

Out[52]=
{1,222,3,4,5}

It is also possible to construct a pattern that operates 
directly on the SparseArray object. But I think that approach is 
a bit more complex than what I've outlined here.
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