python - The last list elements and conditional loops -

mylist="'a','b','c'"  count=0 i=0  while count< len(mylist):     if mylist[i]==mylist[i+1]:         print mylist[i]     count +=1     +=1 

error:

file "<string>", line 6, in <module> indexerror: string index out of range 

i'm assuming when gets last (nth) element can't find n+1 compare to, gives me error.

interestingly, think i've done before , not had problem on larger list: here example (with credit raymond hettinger fixing up)

list=['a','a','x','c','e','e','f','f','f']  i=0 count = 0  while count < len(list)-2:     if list[i] == list[i+1]:         if list [i+1] != list [i+2]:             print list[i]             i+=1             count +=1         else:             print "no"             count += 1     else:            +=1         count += 1 

for crawling through list in way i've attempted, there fix don't go "out of range?" plan implement on large list, i'll have check if "list[i]==list[i+16]", example. in future, add on conditions "if int(mylist[i+3])-int(mylist[i+7])>10: newerlist.append[mylist[i]". it's important solve problem.

i thought inserting break statement, unsuccessful.

i know not efficient, i'm @ point it's understand best.

edit:

right, new information in op, becomes simpler. use the itertools grouper() recipe group data each person tuples:

import itertools  def grouper(iterable, n, fillvalue=none):     """collect data fixed-length chunks or blocks"""     # grouper('abcdefg', 3, 'x') --> abc def gxx"     args = [iter(iterable)] * n     return itertools.zip_longest(*args, fillvalue=fillvalue)  data = ['john', 'sally', '5', '10', '11', '4', 'john', 'sally', '3', '7', '7', '10', 'bill', 'hallie', '4', '6', '2', '1']  grouper(data, 6) 

now data looks like:

[     ('john', 'sally', '5', '10', '11', '4'),      ('john', 'sally', '3', '7', '7', '10'),      ('bill', 'hallie', '4', '6', '2', '1') ] 

which should easy work with, comparison.

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old answer:

if need make more arbitrary links, rather checking continuous values:

def offset_iter(iterable, n):     offset = iter(iterable)     consume(offset, n)     return offset  data = ['a', 'a', 'x', 'c', 'e', 'e', 'f', 'f', 'f']  offset_3 = offset_iter(data, 3)  item, plus_3 in zip(data, offset_3): #naturally, itertools.izip() in 2.x     print(item, plus_3)                  #if memory usage important. 

naturally, want use semantically valid names. advantage method works arbitrary iterables, not lists, , efficient , readable, without ugly, inefficient iteration index. if need continue checking once offset values have run out (for other conditions, say) use itertools.zip_longest() (itertools.izip_longest() in 2.x).

using the consume() recipe itertools.

import itertools import collections  def consume(iterator, n):     """advance iterator n-steps ahead. if n none, consume entirely."""     # use functions consume iterators @ c speed.     if n none:         # feed entire iterator zero-length deque         collections.deque(iterator, maxlen=0)     else:         # advance empty slice starting @ position n         next(itertools.islice(iterator, n, n), none) 

i would, however, question if need re-examine data structure in case.

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original answer:

i'm not sure aim is, gather want itertools.groupby():

>>> import itertools >>> data = ['a', 'a', 'x', 'c', 'e', 'e', 'f', 'f', 'f'] >>> grouped = itertools.groupby(data) >>> [(key, len(list(items))) key, items in grouped] [('a', 2), ('x', 1), ('c', 1), ('e', 2), ('f', 3)] 

you can use work out when there (arbitrarily large) runs of repeated items. it's worth noting can provide itertools.groupby() key argument group them based on factor want, not equality.