Is there a way to call a function and have that function use an argument defined in an enclosing function?


set = Table@Range@50;

posE[set_] := Select[set, EvenQ]

tform1[mult_,set_] := (#*mult) & /@ set[[posE]];

t1set = tform1[10, set]

which leads to the following message/error,

Part::pspec: Part specification posE is neither an integer nor a list of integers. >>

I would like posE to get its argument set_ from the second argument in tform1 (also named set_).

Is this possible? is there a better way to do this?

(Note that I do not want to combine the two functions into a single function).

  • $\begingroup$ how about: Cases[set, x_?EvenQ :> 10 x]? (without knowing what you actually want to achieve) $\endgroup$ May 2 '13 at 7:59

You have defined a function posE[set] but when you call it, it has no argument. You must call it with an argument. So change the function definition to

 tform1[mult_,set_] := (#*mult) & /@ set[[posE[set]]];

If what you are really after is a list that goes from 20 to 500 by 20s (which is what this code seems to do) then there is much better way to do it.

Range[20, 500, 20]
  • $\begingroup$ No, I was just using these functions as an example. Thanks anyway. $\endgroup$
    – geordie
    May 2 '13 at 7:57

is it too early to add an answer to my own question?

I just found a way to do it using With:

tform1[mult_, set_, pos_] := 
  With[{posf = pos[set]}, (#*mult) & /@ set[[posf]]];

t1set = tform1[10, set, posE]

This works very similar to @bill s' answer.

  • 1
    $\begingroup$ You could eliminate the With and just use tform1[mult_, set_, pos_] := (#*mult) & /@ set[[pos[set]]] as per Bill s answer. $\endgroup$ May 2 '13 at 8:28

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