# Permuting symbols

Suppose I have a list like {a, b, a, c}. How can I apply a "symbolic permutation" of the symbols a, b, and c such as a->b, b->c, c->a to produce {b, c, b, a}?

Replace[] with multiple rules applies them all sequentially, and I get {a, b, a, c} back. Permute[] can only change the positions of things in a list, not their names. PermutationReplace[] requires a permutation to be specified with integers, not with symbols.

• You can try RepalceAll as in {a, b, a, c} /. {a -> b, b -> c, c -> a}. – b.gates.you.know.what Jan 10 '16 at 21:42
• Replace[{a, b, a, c}, {a -> b, b -> c, c -> a}, 1] or Replace[{a, b, a, c}, {a -> b, b -> c, c -> a}, {1}] would also work. You need to specify the level specification because you want to replace level 1 (inside a list), not level 0; Replace replaces level 0 only by default. – JungHwan Min Jan 10 '16 at 21:43
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• To clarify: My confusion was over how to use Replace[]. The problem wasn't that it applied all the rules in sequence; the problem was that I was applying the rules at the wrong level, and they weren't taking effect. When they are applied at the right level, only one takes effect on each element. – G. Smith Jan 12 '16 at 4:10

I realized that Map and Association can do the job:

Map[Association[a -> b, b -> c, c -> a][#] &, {a, b, a, c}]


I appreciate the quick answers from eldo and Kuba, but they were both too specific to my example. In general, I won't know the indexes of the elements I need to change, or the particular permutation I need to make.

For any arbitrary permutation, I can create an Association, and then Map each element of an arbitrary list.

• I now see that b.gatessucks and JHM provided general answers in comments to my original question. Thanks! – G. Smith Jan 11 '16 at 1:58
• Is there a way for me to approve an answer in a comment to the original question? – G. Smith Jan 11 '16 at 1:59
list = {a, b, a, c};

ReplacePart[list /. a -> b, {{2} -> c, {4} -> a}]


{b, c, b, a}

or simply

list[[{2, 4, 2, 1}]]


{b, c, b, a}