Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I need a table with the elements made of pure functions and list elements. This is a simplified example:

I need a list as:


and, my failed try is : Table[a[[i]]*Sin[#]&,{i,3}]

Why is the failure and how can I improve it?

share|improve this question
What's a supposed to be? Do you need something like the result of Function[c, c Sin[#] &] /@ Range[3] or Table[With[{cs = c}, cs Sin[#] &], {c, Range[3]}]? –  J. M. Jul 1 '12 at 17:13
This may be relevant. –  Leonid Shifrin Jul 1 '12 at 17:23
@R.M, that puts the constant outside the Function[] as opposed to the inside, no? –  J. M. Jul 1 '12 at 17:23
@WReach, nice to see you around. Undelete your post!! –  Rojo Jul 1 '12 at 17:27
My favorites for this problem would still be either Range[3] /. i_Integer :> (a[[i]] Sin[#] &) or Array[Function[x, a[[x]] Sin[#] &], {3}]. –  Leonid Shifrin Jul 1 '12 at 17:38
show 8 more comments

3 Answers 3

up vote 18 down vote accepted

Function has the attribute HoldAll, so the reference to i in the Table expression will not be expanded. However, you can use With to inject the value into the held expressions:

Table[With[{i = i}, a[[i]]*Sin[#] &], {i, 3}]

{a[[1]] Sin[#1] &, a[[2]] Sin[#1] &, a[[3]] Sin[#1] &}
share|improve this answer
If I do a = Range[3]; Table[With[{i = i}, a[[i]] Sin[#] &], {i, 3}], then the a[[i]] remain frozen as Part[] expressions as opposed to whatever the actual values of the a[[i]] are, but maybe this is what the OP wants... –  J. M. Jul 1 '12 at 17:33
@J.M. It gets substituted when the function is evaluated: Through[Table[With[{i = i}, a[[i]] Sin[#] &], {i, 3}][x]] –  rm -rf Jul 1 '12 at 17:37
@R.M, yes, that's true, but it's still a bit jarring for me to see the list of Function[]s still carrying Part[] objects around... –  J. M. Jul 1 '12 at 17:40
@JM, the thing is that the OP showed an example where a is undefined. If you try to evaluate it and it's undefined you get an error –  Rojo Jul 1 '12 at 17:44
@Rojo: hence my "what's a supposed to be?" question in my first comment. ;) –  J. M. Jul 1 '12 at 17:51
show 2 more comments

This works, but only because j is undefined:

Table[(a[[j]]*Sin[#] &) /. j -> i, {i, 3}]

(if we do j = 5; Table[(a[[j]]*Sin[#] &) /. j -> i, {i, 3}] then it fails; one could localize this with Module to get it to work anyway).

Or, if you hate brevity and compactness:

cF = Function[{j}, a[[j]]*Sin[#] &];
 {j, 1, 3}

Personally I'd use either this last form or WReach's/Rojo's way.

share|improve this answer
@LeonidShifrin thanks. Yes, that it would need to be localized is what I meant (it's accidental that j is undefined). Bad choice of words, I suppose (and, oops, I hadn't seen your comment... why not an answer?) –  acl Jul 1 '12 at 18:13
I've already answered a variant of this question twice (in the links I give in the comments to the question). Trying not to be greedy :-) –  Leonid Shifrin Jul 1 '12 at 18:15
add comment

. . . & is a held expression. (Function has attribute HoldAll.)

Injector pattern to the rescue:

Range@3 /. i_Integer :> (a[[i]] Sin[#] &)

Replace[Range@3, i_ :> (a[[i]] Sin[#] &), 1]

Table[j /. i_ :> (a[[i]] Sin[#] &), {j, 3}]

In this case you could do the replacement the other direction but you will need to hold i to protect it from a global value:

Table[a[[i]] Sin[#] & /. HoldPattern[i] -> j, {j, 3}]

Or use Block:

  Table[a[[i]] Sin[#] & /. i -> j, {j, 3}]
share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.