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Replace unevaluated symbols

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  Replace unevaluated symbols Antonio De Juárez <adejuarez@gmail.com> - 2011-05-13 10:29 +0000
    Re: Replace unevaluated symbols David Reiss <dbreiss@gmail.com> - 2011-05-14 07:10 +0000
    Re: Replace unevaluated symbols Albert Retey <awnl@gmx-topmail.de> - 2011-05-14 07:15 +0000
      Re: Replace unevaluated symbols Antonio De Juárez <adejuarez@gmail.com> - 2011-05-15 11:05 +0000

#2343 — Replace unevaluated symbols

Given an expression like

{x+y,2*x-y}

I would like to replace x and y with some values even if these
variables have some value assigned. For example, the code

expr = Hold[x+y,2x-y]
expr/.x->3

produces

Hold[3 + y, 2*3 - y]

but the code

x=2.5;
expr = Hold[x+y,2x-y]
expr/.x->3

produces the wrong result

Hold[x + y, 2 x - y]

I know this can be done using Block; however, I don't know the
variables to be replaced beforehand. I would like to have a function
with attribute HoldAll like

fun[expr,var1,value1,var2,value2,...]

that replaces the variables var1, var2, ... by the corresponding
values value1, value2,..., even if any of the variables var1, var2,...
has a preassigned value.

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#2352

The reason why this happens can be gleaned by Tracing an example:

z = 0;

THen

TracePrint[
 Hold[z] /. z -> 7
 ]

gives

 Hold[z]/. z->7

  ReplaceAll

  Hold[z]

   Hold

  z->7

   Rule

   z

   0

   7

  0->7

  0->7

   Rule

   0

   7

 Hold[z]/. 0->7

 Hold[z]




>From this you can see that the replacement rule z->7 is evaluated to 0-
>7 and hence the result is correct.  This is all a consequence of
Mathematica's standard evaluation process.

So indeed an approach using Block or Module for example allows you to
by pass this:

Block[{z},
 TracePrint[
  Hold[z] /. z -> 7
  ]

 ]

gives


 Hold[z]/. z->7

  ReplaceAll

  Hold[z]

   Hold

  z->7

   Rule

   z

   7

  z->7

   Rule

   z

   7

 Hold[z]/. z->7

 Hold[7]

  Hold


In essence, if you wish to bypass  Mathematica's standard evaluation
order, you will need to localize the variable as you suggest.

Best,
David


On May 13, 6:29 am, Antonio De Ju=E1rez <adejua...@gmail.com> wrote:
> Given an expression like
>
> {x+y,2*x-y}
>
> I would like to replace x and y with some values even if these
> variables have some value assigned. For example, the code
>
> expr = Hold[x+y,2x-y]
> expr/.x->3
>
> produces
>
> Hold[3 + y, 2*3 - y]
>
> but the code
>
> x=2.5;
> expr = Hold[x+y,2x-y]
> expr/.x->3
>
> produces the wrong result
>
> Hold[x + y, 2 x - y]
>
> I know this can be done using Block; however, I don't know the
> variables to be replaced beforehand. I would like to have a function
> with attribute HoldAll like
>
> fun[expr,var1,value1,var2,value2,...]
>
> that replaces the variables var1, var2, ... by the corresponding
> values value1, value2,..., even if any of the variables var1, var2,...
> has a preassigned value.

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#2377

Am 13.05.2011 12:29, schrieb Antonio De Ju=E1rez:
> Given an expression like
>
> {x+y,2*x-y}
>
> I would like to replace x and y with some values even if these
> variables have some value assigned. For example, the code
>
> expr = Hold[x+y,2x-y]
> expr/.x->3
>
> produces
>
> Hold[3 + y, 2*3 - y]
>
> but the code
>
> x=2.5;
> expr = Hold[x+y,2x-y]
> expr/.x->3
>
> produces the wrong result
>
> Hold[x + y, 2 x - y]
>
> I know this can be done using Block; however, I don't know the
> variables to be replaced beforehand. I would like to have a function
> with attribute HoldAll like
>
> fun[expr,var1,value1,var2,value2,...]
>
> that replaces the variables var1, var2, ... by the corresponding
> values value1, value2,..., even if any of the variables var1, var2,...
> has a preassigned value.

so what you want to do is to hold the pattern in your replacement rule
-- and what a surprise: there is HoldPattern :-)

expr /. HoldPattern[x] -> 3

hth,

albert

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#2398

Thanks Albert. I think HoldPattern may solve the problem.

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