Consecutive Character Sequences

[Walter Brunswick]

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.

[Peter Hansen]

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.

Why would it return 1? Is that instead of True, or does it represent a
count of items, or the position of something in the list, or what?

-Peter

[mensanator]

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

"""

[George Sakkis]

"Walter Brunswick" <wa*************@sympatico.ca> 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.

If you're in 2.4, use itertools.groupby:

import itertools as it

def hasConsequent(aString, minConsequent):
for _,group in it.groupby(aString):
if len(list(group)) >= minConsequent:
return True
return False
George

[Aries Sun]

I have tested George's solutions, it seems not complete. When pass (s,
3) to the function hasConsequent(), it returns the wrong result.

The following is my approach. The performence may be not so good. There
must be better ones.

from re import findall
def hasConsequent(aString, minConsequent): for ch in aString:
result = findall(ch*minConsequent, aString)
if len(result) >= 1:
return True
return False
hasConsequent(s, 2) True hasConsequent(s, 3) True hasConsequent(s, 4)

False

Aries Sun

[George Sakkis]

"Aries Sun" <su*******@gmail.com> wrote:

I have tested George's solutions, it seems not complete. When pass (s,
3) to the function hasConsequent(), it returns the wrong result.

What are you talking about ? I get the correct answer for

hasConsequent("taaypiqee88adbbba", 3)

True
George

[Aries Sun]

Hi George,
I used Python 2.4.1, the following are the command lines.
But the reslut was still False. Is there anything wrong with below
codes?

import itertools as it
def hasConsequent(aString, minConsequent): for _,group in it.groupby(aString):
if len(list(group)) >= minConsequent:
return True
return False

hasConsequent("taaypiqee88adbbba", 3) False

Regards,

Aries

[George Sakkis]

"Aries Sun" <su*******@gmail.com> wrote:

Hi George,
I used Python 2.4.1, the following are the command lines.
But the reslut was still False. Is there anything wrong with below
codes?hasConsequent("taaypiqee88adbbba", 3)

All indentation was lost in your message, so I'm not quite sure; here it is again, just in case:

import itertools as it

def hasConsequent(aString, minConsequent):
for _,group in it.groupby(aString):
if len(list(group)) >= minConsequent:
return True
return False

Run it from a file instead of the command line and see if you get the same result.

I'm using 2.4:

sys.version

'2.4 (#1, Mar 29 2005, 15:15:45) \n[GCC 3.3.3 (cygwin special)]'

I would be very surprised if something so crucial changed between 2.4.0 and 2.4.1. What does[list(group) for _,group in it.groupby("taaypiqee88adbbba")]
return to you ?
George

[Sion Arrowsmith]

Aries Sun <su*******@gmail.com> wrote:

I used Python 2.4.1, the following are the command lines.
But the reslut was still False. Is there anything wrong with below
codes?

import itertools as it
def hasConsequent(aString, minConsequent):

for _,group in it.groupby(aString):
if len(list(group)) >= minConsequent:
return True
return False

Yes: return False is at the wrong indentation level.

--
\S -- si***@chiark.greenend.org.uk -- http://www.chaos.org.uk/~sion/
___ | "Frankly I have no feelings towards penguins one way or the other"
\X/ | -- Arthur C. Clarke
her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump

[Walter Brunswick]

Thank you me********@aol.com <me********@aol.com>, George Sakkis <gs*****@rutgers.edu> and Aries Sun <su*******@gmail.com> for your
code contributions. I haven't tested them yet, but they look alright and enough for me to work with.
Thanks to everyone else for your comments.

W. Brunswick.

[Aries Sun]

Hi George,

Here's the result:

[list(group) for _,group in it.groupby("taaypiqee88adbbba")] [['t'], ['a', 'a'], ['y'], ['p'], ['i'], ['q'], ['e', 'e'], ['8', '8'],
['a'], ['d'], ['b', 'b', 'b'], ['a']] [len(list(group)) for _,group in it.groupby("taaypiqee88adbbba")] [1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3, 1]

According to the above output, your solution is correct.

Regards,
Aries