5
$\begingroup$

If I have list1={x^a, x^b, x^c} and list2={1,2,3}, how do I substitute one into the other to get {1^a, 2^b, 3^c} ? ie. substitution of x

Edit: I need help with a more complicated example. Namely this:

If I have list1={e^(u[1]+a[1]), e^(u[2]+a[2]), e^(u[3]+a[3])} and list2={1,2,3}, how do I substitute one into the other to get {e^(1+a[1]), e^(2+a[2]), e^(3+a[3])} ? ie. substitution of u[i]

Improve this question
$\endgroup$
2
  • $\begingroup$ Use MapThread with pure functions: MapThread[#1 /. x -> #2 &, {list1, list2}]. $\endgroup$ Commented Oct 30, 2015 at 13:03
  • $\begingroup$ There are things to do after your question is answered. Its a good idea to stay vigilant for some time, better approaches may come later improving over previous replies. Experienced users may point alternatives, caveats or limitations. New users should test answers before voting and wait 24 hours before accepting the best one. Participation is essential for the site, please come back to do your part. $\endgroup$ Commented Nov 6, 2015 at 14:17

3 Answers 3

Reset to default
10
$\begingroup$ 
MapThread[#1 /. x -> #2 &, {list1, list2}]
Improve this answer
$\endgroup$
8
$\begingroup$

Response to the edit:

It's easy to adapt my 2nd and 3rd approach below to meet your requirement:

l = list1; l[[;; , 2, 2]] = list2; l i = 1; list1 /. u[_] :> list2[[i++]]

Original Answer:

Shortest so far:

list2^list1[[;; , 2]]

2nd shortest so far:

l = list1; l[[;; , 1]] = list2; l

If l isn't introduced to keep list1 unaltered, it'll be another shortest one.

3rd shortest so far:

i = 1; list1 /. x :> list2[[i++]]
Improve this answer
$\endgroup$
5
$\begingroup$ 
MapThread[Function[x, #][#2] &, {list1, list2}]
Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.