wolfram mathematica - Using Evaluate with a Pure Function and SetDelayed -

i want evaluate f below passing list function:

f = {z[1] z[2], z[2]^2}; = % /. {z[1]-> #1,z[2]-> #2}; f[z_] := evaluate[a] & @@ z ;  

so if try f[{1,2}] {2, 4} expected. looking closer ?f returns definition

f[z_] := (evaluate[a] &) @@ z 

which depends on value of a, if set a=3 , evaluate f[{1,2}], 3. know adding last & makes evaluate[a] hold, elegant work around? need force evaluation of evaluate[a], improve efficiency, a in fact quite complicated.

can please out, , take consideration f has contain array[z,2] given unknown calculation. writing

f[z_] := {z[[1]]z[[2]],z[[2]]^2} 

would not enough, need generated automatically our f.

many contribution.

please consider asking future questions @ dedicated stackexchange site mathematica.
questions less become tumbleweeds , may viewed many experts.

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you can inject value of a body of both function , setdelayed using with:

with[{body = a},  f[z_] := body & @@ z ] 

check definition:

definition[f] 

f[z$_] := ({#1 #2, #2^2} &) @@ z$ 

you'll notice z has become z$ due automatic renaming within nested scoping constructs behavior same.

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in comments said:

and again bothers me if values of z[i] changed, workaround fail.

while should not problem after f[z_] defined above, if wish protect replacement done a use formal symbols instead of z. these entered e.g. esc$zesc formal z. formal symbols have attribute protected , exist avoid such conflicts this.

this looks better in notebook here:

f = {\[formalz][1] \[formalz][2], \[formalz][2]^2}; = f /. {\[formalz][1] -> #1, \[formalz][2] -> #2}; 

another approach replacements inside hold expression, , protect rules evaluation using unevaluated:

clearall[f, z, a, f, z]  z[2] = "fail!";  f = hold[{z[1] z[2], z[2]^2}]; = f /. unevaluated[{z[1] -> #1, z[2] -> #2}] // releasehold;  with[{body = a},  f[z_] := body & @@ z ]  definition[f] 

f[z$_] := ({#1 #2, #2^2} &) @@ z$