jeff@ndcheg.cheg.nd.edu (Jeffrey C. Kantor) (08/02/88)
After spending the weekend doing Mathematica, I thought I would post a
simple question. Suppose you want to replace every element of a list
with a corresponding element of another list, i.e.,
expr /. {a,b,c,d}->{1,2,3,4} (NOTE: NOT A VALID STATEMENT)
where -> works on an element-by-element basis. What is the most efficient
way of doing this? I hacked together a rather messy bit of code that is
not very efficient. I was looking for a one-liner.
This came up when trying to get the RungeKutta Notebook to work for a system
of ODE's rather than a single ODE.
--
Jeff Kantor US Mail: Department of Chemical Engineering
University of Notre Dame
uucp: iuvax!ndmath!ndcheg!jeff Notre Dame, IN 46556 USA
internet: jeff@ndcheg.cheg.nd.eduzardoz@Apple.COM (Phil Wayne) (08/04/88)
In article <605@ndcheg.cheg.nd.edu> jeff@ndcheg.cheg.nd.edu (Jeffrey C. Kantor) writes: >After spending the weekend doing Mathematica, I thought I would post a >simple question. Suppose you want to replace every element of a list >with a corresponding element of another list, i.e., > >expr /. {a,b,c,d}->{1,2,3,4} (NOTE: NOT A VALID STATEMENT) > >where -> works on an element-by-element basis. What is the most efficient >way of doing this? I hacked together a rather messy bit of code that is >not very efficient. I was looking for a one-liner. > >This came up when trying to get the RungeKutta Notebook to work for a system >of ODE's rather than a single ODE. > I hate to point this out but... If both lists are the same length (which I assume they are) then if we have x = { some list of elements... } y = { some other list of elements of the same length as x... } then the *EASY* *CHEAP* and *QUICK* way to do the replacement is... (Ohhhh, god, how can I be this devious...) x = y Tah....Dahhhh --- zardoz