Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a user-defined function which takes three vectors as input. Suppose that is like this:

f[a_, b_, u_] := Total[a*Total[u]^b]

Usually I have vectors for a and b but a matrix format for u. For example,

a = {1.5, 2, 1}
b = {-1, 1, 0.5}
u = {{1, 2, 0}, {1, 0, 1.5}, {2, 1, 1}}

which should give the result

{8.23205, 7.18114, 10.375}

In other words, the function f should be applied to each row of u. I can do this using a Do loop, but I want to know how to do it using Map or other functional commands (but not Table).

share|improve this question
Have a look at what Total[a*Total[#]^b]&/@u will do. In fact, take a look at the result of Total[#]&/@u to see what this syntax does. – Jonathan Shock May 28 '13 at 23:09
up vote 5 down vote accepted

The equivalent to a Do loop using Map is

f[a, b, #] & /@ u

Which gives the desired results:

{8.23205, 7.18114, 10.375}

share|improve this answer
It's nice. I was trying MAP[f[a,b,#],u] but it didn't work. I always got confused by pure functions! – Amin May 29 '13 at 1:45
@Amin You were very close! It can be written Map[f[a, b, #] &, u] -- remember that it's not a pure function until you've marked where it ends, with an ampersand. The highlighting will help you with this, the # will turn from pink to green once you have an & in place. – C. E. May 29 '13 at 2:21

Anon's method is no doubt better but here is another method that may have some interest:

Block[{f, a, b}, Thread @ f[a, b, u]]
{8.23205, 7.18114, 10.375}

This makes f, a, b effectively inert, then uses Thread to distribute across the rows of u. This could also be formulated:

f @@@ Block[{a, b}, Thread@{a, b, u}]

You could also use Outer:

Outer[f, {a}, {b}, u, 1][[1, 1]]
{8.23205, 7.18114, 10.375}

Or a form of Table:

Table[f[a, b, x], {x, u}]
share|improve this answer
Thank you Mr.Wizard. Those gave me another insights to the problem. – Amin May 29 '13 at 1:47

Anon's answer is the way I would do it, but it's fun to figure out a solution that works for single vectors as well as a list of vectors:

f[a_, b_, u_] := Map[Composition[#^b &, Total], u, {-2}].a


f[a, b, u]
{8.23205, 7.18114, 10.375}
f[a, b, First@u]
share|improve this answer

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.