Evaluating ReplaceAll inside SetDelayed, first

Question

Consider the following (simplified for the purpose of example) definitions:

f[x_] := y /. y -> x
g[x_] := x

Where for example, the definition of f could be motivated e.g. by first getting some expression (y) from a previous calculation, and then deciding to make it into a function.

The problems begin when f is then passed on to a third function, that evaluates it many times (e.g. NIntegrate, Plot, etc), or maybe I then attempt to Compile it. Due to the replacement rule that is saved as part of the definition, everything will be much slower, as can be seen in the following trace:

In[23]:= Trace[f[1]]
Out[23]= {f[1],y/. y->1,{y->1,y->1},y/. y->1,1}

vs

Trace[g[1]]
{g[1],1}

I suspect though I am not certain and haven't profiled this, that the first operation would be significantly slower even when compared to evaluating a more complex mathematical expression, provided it consists of built in functions. I'm open to this being refuted, but my interest is only partly driven by efficiency, so I won't accept an answer that simply refutes my assumption without answering the actual question (I would still find it useful, however).

Now, a plausible way to avoid the previous conundrum is to wrap the RHS of SetDelayed by Evaluate This has its own problems however, namely evaluating unintended parts:

In[16]:= z=1
Out[16]= 1
...
In[42]:= k[x_]:=Evaluate[z+y/.y->x]
In[35]:= Trace[k[1]]
Out[35]= {k[1],1}
In[44]:= Definition[k]
Out[44]= k[x_]:=1+x

What are other methods making the ReplaceAll evaluate once, and what (possibly unintended) consequences do they have? I am mainly interested in something lightweight, that I might actually use in notebooks without my init.m file - not in defining a largs expression to manipulate the SetDelayed lines (though that would still be interesting from an intellectual point of view).

Ideally, this would work (it doesn't) (f[x_]:=y)/.y->x

Thanks in advance!

Answer 1

OK, let me extend my comment to an answer. As mentioned in your question:

Ideally, this would work (it doesn't) (f[x_]:=y)/.y->x

That attempt doesn't succeed because ReplaceAll (/.) is a function that doesn't have HoldAll etc. attribute and it evaluates its argument before the replacement. To resolve the problem we need to stop the auto evaluation in some way, for example:

z = 1;

(* Solution 1 *)
Unevaluated@(k[x_] := y + z) /. y -> x

(* Solution 2 *)
Hold[k2[x_] := y + z] /. y -> x // ReleaseHold

DownValues /@ {k, k2}
(* {{HoldPattern[k[x_]] :> x + z}, {HoldPattern[k2[x_]] :> x + z}} *)

Answer 2

Another idea is to use Set instead of SetDelayed, and use Unevaluated to prevent evaluation:

z = 1;
f[x_] = With[{y = x}, Unevaluated @ Unevaluated[y + z]];
DownValues[f]

{HoldPattern[f[x_]] :> x + z}