Repeating elements in list Python

Question

Let's say I have a list of strings:

a = ['a', 'a', 'b', 'c', 'c', 'c', 'd']

I want to make a list of items that appear at least twice in a row:

result = ['a', 'c']

I know I have to use a for loop, but I can't figure out how to target the items repeated in a row. How can I do so?

EDIT: What if the same item repeats twice in a? Then the set function would be ineffective

a = ['a', 'b', 'a', 'a', 'c', 'a', 'a', 'a', 'd', 'd']
result = ['a', 'a', 'd']

Answer 1

try itertools.groupby() here:

>>> from itertools import groupby,islice
>>> a = ['a', 'a', 'b', 'c', 'c', 'c', 'b']

>>> [list(g) for k,g in groupby(a)]
[['a', 'a'], ['b'], ['c', 'c', 'c'], ['b']] 

>>> [k for k,g in groupby(a) if len(list(g))>=2]
['a', 'c']

using islice() :

>>> [k for k,g in groupby(a) if len(list(islice(g,0,2)))==2]
>>> ['a', 'c']

using zip() and izip():

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'])

timeit results:

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))>=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")

output:

islice 39.9123107277
grpby 30.1204478987
dread 17.8041124706
xdread 15.3691785568
chain 17.4777339702
inrow 11.8577565327           
zip 16.6348844045
izip 15.1468557105

Conclusion:

Poke's solution is the fastest solution in comparison to other alternatives.

Answer 2

This sounds like homework, so I'll just outline what I would do:

1. Iterate over a, but keep the index of each element in a variable. enumerate() will be useful.
2. Inside of your for loop, start a while loop from the current item's index.
3. Repeat the loop as long as the next element is the same as the previous (or the original). break will be useful here.
4. Count the number of times that loop repeats (you'll need some counter variable for this).
5. Append the item to your result if your counter variable is >= 2.