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Re: ReplaceAll doesn't replace


Ken:

The reason nothing is happening to x in  (#^2 & /@ x)  is that you're using
Map, which will map the function #^2& over the arguments of x, of which there
are none.    You probably want (#^2 &  @ x); or if you really want to use Map:
Map[#1^2 & , x, {0}], where the {0} specifies that the function is applied to
level 0, not level 1.

Ken Levasseur
Math. Sci.
UMass Lowell


Ken Morgan wrote:

> I understand how the following will replace x with the list
>
> #^2 & /@ (x /. x -> {a, b, c})
>
> to generate
>
> {a^2, b^2, c^2}
>
> But, why isn't x replaced at the beginning of the evaluation in the
> following
>
> (#^2 & /@ x) /. x -> {a, b, c}
>
> since it generates
>
> {a, b, c}
>
> What I really want to know is: What is it about the Function function that
> doesn't allow ReplaceAll to "replace all" at the beginning of an evaluation?
>
> Thanks,
> Ken Morgan
> kmorga51 at calvin.edu



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