Walter Brunswick wrote:
> Is there any way to [efficiently] iterate through a sequence of characters to
> find N [or more] consecutive equivalent characters?
>
> So, for example, the string "taaypiqee88adbbba" would return 1 if the number
> (of consequtive characters) supplied in the parameters
> of the function call was 2 or 3, because "a", "e", 8, and "b" is repeated 2
> or 3 times.
>
> Thanks for any assistance.
> W. Brunswick.
def rep(n):
# current repetition count
the_rep = 1
# previous character inspected
last_c = ''
# history of repetions
max_rep = {}
for c in s:
# duplicate character found?
if c==last_c:
# count how many consecutive dups
the_rep += 1
# repetition (if any) ended, save previous rep count
else:
# has this count occured before?
if max_rep.has_key(the_rep):
# if so, track how many times it has
max_rep[the_rep] += 1
# otherwise, add this rep count to history
else:
max_rep[the_rep] = 1
# reset rep count to look for next block
the_rep = 1
# save current character to compare to next character
last_c = c
# check that last character in string wasn't part of a block
if max_rep.has_key(the_rep):
max_rep[the_rep] += 1
else:
max_rep[the_rep] = 1
# finally, did the block size we asked for ever occur?
if max_rep.has_key(n):
return 1
else:
return 0
s = 'taaypiqee88adbbba'
for i in range(9):
print rep(i),
"""
0 1 1 1 0 0 0 0 0
"""
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