Extracting RHS from Rule

Question

In an example from another question, the rhs of a list of rules can be simply extracted:

In[1]:= rules1 = Solve[x + y == 3 && x - y == 6, {x, y}][[1]]

              9         3
Out[1]= {x -> -, y -> -(-)}
              2         2

In[2]:= rules1 /. Rule -> (#2 & )

          9    3
 Out[2]= {-, -(-)}
          2    2

However, when I try the same replacement with a list of rules where the RHS is a pure function, it does not result in a list of the RHS.

In[3]:= rules2 = {MySum -> Total[#1] & , MyProd -> Times[#1] & }

Out[3]= {MySum -> Total[#1] & , MyProd -> Times[#1] & }

In[4]:= rules2 /. Rule -> (#2 & )

Out[4]= {(#2 & )[MySum, Total[#1]] & , (#2 & )[MyProd, Times[#1]] & }

How do I correct this?

score 4 · Answer 1 · edited Apr 13 '17 at 12:55

4

Jason shows the solution. Your input is not in the correct form as in fact the RHS is not a "pure function" but rather the entire rule is. Using the methods described here will show you this:

Mathematica graphics

Another way to look at this is that Rule has a greater binding power than Function:

Precedence /@ {Rule, Function}

{120., 90.}

Also see: Parentheses in pure functions: # & vs. ( # &)

edited Apr 13 '17 at 12:55

Community

1

answered Nov 12 '12 at 21:55

Mr.Wizard

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score 3 · Answer 2 · answered Nov 12 '12 at 21:43

3

It seems to make a difference to wrap pure functions inside a set of parentheses, including the ampersand.

rules2 = {MySum -> (Total[#1] &), MyProd -> (Times[#1] &)};
rules2 /. Rule -> (#2 &)

gives the result you are looking for.

answered Nov 12 '12 at 21:43

Jason B.

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score 1 · Answer 3 · answered Nov 12 '12 at 21:07

1

This works:

rules2 = {MySum -> Total[#1] &, MyProd -> Times[#1] &}
rules2 /. Rule[_, p_] :> p
(* {Total[#1] &, Times[#1] &} *)

answered Nov 12 '12 at 21:07

ssch

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Thanks. It makes sense why your solution works. I still can't see why the other version fails. Your solution seems more general and should work in all cases. What is it about using a the original patter of just 'Rule' that allowed it to work at all on rules1? – ebergerson Nov 12 '12 at 21:30
1

It should be noted that while this solution works, it does so by taking advantage of the structure of rules2 (see FullForm or TreeForm) – tkott Nov 13 '12 at 20:11

score -1 · Answer 4 · answered Nov 13 '12 at 16:21

-1

This also works:

 rules2 = {MySum -> Total[#1] &, MyProd -> Times[#1] &}
 Last /@ rules2
 (* {Total[#1] &, Times[#1] &} *)

answered Nov 13 '12 at 16:21

murray

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The output I get does not match what you show. I get: {MySum -> 1, MyProd -> #1}. This is no mystery as the Function has a body of only one argument so Last (or First) is simply returning that body, evaluated. Sorry, but -1. – Mr.Wizard Nov 13 '12 at 17:45
As a point of reference you can, in this limited case, get the desired results with rules2[[All, {1}, 2]] for the reason described here but it is far better to enter the expression correctly in the first place. – Mr.Wizard Nov 13 '12 at 17:50

Extracting RHS from Rule

4 Answers4