ReplacePart -> eliminate

• To: mathgroup at smc.vnet.net
• Subject: [mg131656] ReplacePart -> eliminate
• From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>
• Date: Mon, 16 Sep 2013 05:27:02 -0400 (EDT)
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```I'd like to eliminate the Indeterminate  (or other non-numerical terms,
as Infinite, etc...) items. Es.

x = {-1.25664, 0.628319, -0.628319, 1.25664, Indeterminate, -1.25664,
0.628319, -0.628319, 1.25664, Indeterminate};

or

x=Table[1/(3 - n), {n, 0, 5}])
{1/3, 1/2, 1, ComplexInfinity, -1,-1/2}

How can I do it with a simple command?
Many thanks, Roberto

-----Messaggio originale-----
Da: Bob Hanlon [mailto:hanlonr357 at gmail.com]
Inviato: domenica 15 settembre 2013 13:06
A: mathgroup at smc.vnet.net
Oggetto: Re: ReplacePart

{5,10} is apparently interpreted as a position within an array rather
than a list of positions. Not sure why the algorithm isn't able to
resolve the ambiguity given that the input is a simple list. Or why
there is no error stating that position {5,10} does not exist.

x = {-1.25664, 0.628319, -0.628319, 1.25664, Indeterminate, -1.25664,
0.628319, -0.628319, 1.25664, Indeterminate};

x2 = x /. Indeterminate -> 0

{-1.25664, 0.628319, -0.628319, 1.25664, 0, -1.25664, 0.628319,
-0.628319, \ 1.25664, 0}

x2 ==
ReplacePart[x, {5 -> 0, 10 -> 0}] ==
ReplacePart[x, Thread[{5, 10} -> 0]] ==  ReplacePart[x, {{5} -> 0, {10}
-> 0}] ==  ReplacePart[x, {{5}, {10}} -> 0] ==  ReplacePart[x, List=
/@
{5, 10} -> 0] ==  ReplacePart[x, Position[x, Indeterminate] -> 0]

True

Bob Hanlon

On Sat, Sep 14, 2013 at 6:02 AM, man21 <man21 at free.fr> wrote:

> Hello,
>
> As a result of a calculation, I end up with a list of numerical values

> which contains  some "Indeterminate".
>
> x ={-1.25664, 0.628319, -0.628319, 1.25664, Indeterminate, -1.25664,
> 0.628319, -0.628319, 1.25664, Indeterminate}
>
> I try to replace the "Indeterminate" by "0", using :
>
> ReplacePart[x, {5, 10} -> 0.]
>
> but this dosen't work. Any idea why, and how to do it ?
>
> Thanks,
>
> Michel
>
>
>

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```

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