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Re: Sum of terms --> list
*To*: mathgroup at smc.vnet.net
*Subject*: [mg110687] Re: Sum of terms --> list
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Fri, 2 Jul 2010 02:56:41 -0400 (EDT)
On 7/1/10 at 8:27 AM, nma at 12000.org (Nasser M. Abbasi) wrote:
>it seems in the second case, this happens becuase the
>expression was evaluated first? By holding evaluation, one can
>get the same result as in the first case:
>List @@ (a + b*I)
>{a, I*b}
>List @@ (3 + 4*I)
>3 + 4*I
>ReleaseHold@(List @@ (3 + Hold@4*I))
>{3,4 I}
Not quite the right explanation. The key is 3 + 4 I isn't an
expression in the same sense as 3.4 isn't an expression. That is
it has no parts. It makes no more sense to have List@@(3+4 I)
yield {3, 4 I} then it does to have List@@{3.4} yield either
{3.4} or {3, 4}.
By adding Hold, you create a new expression which has parts. Now
you can use Apply to change this to a list and rid yourself of
the Hold part using ReleaseHold.
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