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    <p>try <a href="http://docs.python.org/library/itertools.html#itertools.groupby" rel="nofollow noreferrer"><code>itertools.groupby()</code></a> here:</p> <pre><code>&gt;&gt;&gt; from itertools import groupby,islice &gt;&gt;&gt; a = ['a', 'a', 'b', 'c', 'c', 'c', 'b'] &gt;&gt;&gt; [list(g) for k,g in groupby(a)] [['a', 'a'], ['b'], ['c', 'c', 'c'], ['b']] &gt;&gt;&gt; [k for k,g in groupby(a) if len(list(g))&gt;=2] ['a', 'c'] </code></pre> <p>using <code>islice()</code> :</p> <pre><code>&gt;&gt;&gt; [k for k,g in groupby(a) if len(list(islice(g,0,2)))==2] &gt;&gt;&gt; ['a', 'c'] </code></pre> <p>using <code>zip()</code> and <code>izip()</code>:</p> <pre><code>In [198]: set(x[0] for x in izip(a,a[1:]) if x[0]==x[1]) Out[198]: set(['a', 'c']) In [199]: set(x[0] for x in zip(a,a[1:]) if x[0]==x[1]) Out[199]: set(['a', 'c']) </code></pre> <p><code>timeit</code> results:</p> <pre><code>from itertools import * a='aaaabbbccccddddefgggghhhhhiiiiiijjjkkklllmnooooooppppppppqqqqqqsssstuuvv' def grp_isl(): [k for k,g in groupby(a) if len(list(islice(g,0,2)))==2] def grpby(): [k for k,g in groupby(a) if len(list(g))&gt;=2] def chn(): set(x[1] for x in chain(izip(*([iter(a)] * 2)), izip(*([iter(a[1:])] * 2))) if x[0] == x[1]) def dread(): set(a[i] for i in range(1, len(a)) if a[i] == a[i-1]) def xdread(): set(a[i] for i in xrange(1, len(a)) if a[i] == a[i-1]) def inrow(): inRow = [] last = None for x in a: if last == x and (len(inRow) == 0 or inRow[-1] != x): inRow.append(last) last = x def zipp(): set(x[0] for x in zip(a,a[1:]) if x[0]==x[1]) def izipp(): set(x[0] for x in izip(a,a[1:]) if x[0]==x[1]) if __name__=="__main__": import timeit print "islice",timeit.timeit("grp_isl()", setup="from __main__ import grp_isl") print "grpby",timeit.timeit("grpby()", setup="from __main__ import grpby") print "dread",timeit.timeit("dread()", setup="from __main__ import dread") print "xdread",timeit.timeit("xdread()", setup="from __main__ import xdread") print "chain",timeit.timeit("chn()", setup="from __main__ import chn") print "inrow",timeit.timeit("inrow()", setup="from __main__ import inrow") print "zip",timeit.timeit("zipp()", setup="from __main__ import zipp") print "izip",timeit.timeit("izipp()", setup="from __main__ import izipp") </code></pre> <p><strong>output:</strong></p> <pre><code>islice 39.9123107277 grpby 30.1204478987 dread 17.8041124706 xdread 15.3691785568 chain 17.4777339702 inrow 11.8577565327 zip 16.6348844045 izip 15.1468557105 </code></pre> <p><strong>Conclusion:</strong></p> <p><a href="https://stackoverflow.com/a/13020768/846892">Poke's solution</a> is the fastest solution in comparison to other alternatives.</p>
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    1. USAshwini Chaudhary
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    1. COI think you missed the "in a row" part. Nice simple solution otherwise, even if it is O(n^2).
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    2. COCounter still doesn't do "in a row".
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    3. CO@g.d.d.c what's "in a row"? I don't get it. :|
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