Why does `Repeated` not work in this case?

Question

I think I don't understand how to use Repeated.

I have a list for example

list={a,b,a,b,a,b,a,a,b}

I want to detect whether this list has repeated element a,a or b,b

I tried

Cases[list, Repeated[a,{2}]]

It doesn't work, why? How to do this?

Answer 1

Per request by the Pope:

Think about it - what are the cases that are tested?

It's the elements of the list.

Try, e.g., Cases[list, x_ /; (Print@x; True)] and see what you get (it will show that the test is done on an element-by-element basis).

Cases[Split@list, {Repeated[a | b, {2}]}] will spit out a list of matches, MemberQ[Split@list, {Repeated[a | b, {2}]}] will return True or False if any are there (you said "detect", hence I'm guessing that's what you meant.)

There are other ways, that should get you started.

Btw- putting a Print (or Sow, etc.) temporarily as part of a test/rule/etc. can be a quite handy debugging and learning tool - I use it every now and then when a complex rule and/or test is not doing what I think it should (and inevitably, it's my thinking that was off).

Answer 2

Why part is explained nicely in @ciao's answer. This post deals with the part:

I want to detect whether this list has repeated element a,a or b,b

You can get what you expected to get from Cases[...] using the new-in-v-10 function SequenceCases:

list={a,b,a,b,b,b,a,b,a,a,b};
SequenceCases[list,{Repeated[x_,{2}]}]

{{b, b}, {a, a}}

Or, define a Boolean function that returns True when the input list contains repeated elements:

containsRepeatedQ = SequenceCases[#,{Repeated[x_,{#2}]}] != {}&;

containsRepeatedQ[{a,b,a,b,a,b,a,b}, 2]

False

containsRepeatedQ[{a,b,a,b,a,b,a,a,b}, 2]

True