Counting longest occurence of repeated sequence in Python

Question

What's the easiest way to count the longest consecutive repeat of a certain character in a string? For example, the longest consecutive repeat of "b" in the following string:

my_str = "abcdefgfaabbbffbbbbbbfgbb"

would be 6, since other consecutive repeats are shorter (3 and 2, respectively.) How can I do this in Python?

thanks.

Answer 1

Here's my really boring, inefficient, straightforward counting method (interjay's is much better). Note, I wrote this in this little text field, which doesn't have an interpreter, so I haven't tested it, and I may have made a really dumb mistake that a proof-read didn't catch.

my_str = "abcdefgfaabbbffbbbbbbfgbb"
last_char = ""
current_seq_len = 0
max_seq_len = 0

for c in mystr:
    if c == last_char:
        current_seq_len += 1
        if current_seq_len > max_seq_len:
            max_seq_len = current_seq_len
    else:
        current_seq_len = 1
        last_char = c

print(max_seq_len)

Answer 2

Here is a one-liner:

max(len(list(y)) for (c,y) in itertools.groupby(my_str) if c=='b')

Explanation:

itertools.groupby will return groups of consecutive identical characters, along with an iterator for all items in that group. For each such iterator, len(list(y)) will give the number of items in the group. Taking the maximum of that (for the given character) will give the required result.

Answer 3

How about a regex example:

import re
my_str = "abcdefgfaabbbffbbbbbbfgbb"
len(max(re.compile("(b+b)").findall(my_str)))    
# max([len(i) for i in re.compile("(b+b)").findall(my_str)]) also works

Edit, Mine vs. interjays

x=timeit.Timer(stmt='import itertools;my_str = "abcdefgfaabbbffbbbbbbfgbb";max(len(list(y)) for (c,y) in itertools.groupby(my_str) if c=="b")')
x.timeit()
22.759046077728271

x=timeit.Timer(stmt='import re;my_str = "abcdefgfaabbbffbbbbbbfgbb";len(max(re.compile("(b+b)").findall(my_str)))')
x.timeit()
8.4770550727844238

Answer 4

Using run-length encoding:

import numpy as NP

signal = NP.array([4,5,6,7,3,4,3,5,5,5,5,3,4,2,8,9,0,1,2,8,8,8,0,9,1,3])

px, = NP.where(NP.ediff1d(signal) != 0)
px = NP.r_[(0, px+1, [len(signal)])]
# collect the run-lengths for each unique item in the signal
rx = [ (m, n, signal[m]) for (m, n) in zip(px[:-1], px[1:]) if (n - m) > 1 ]

# get longest:
rx2 = [ (b-a, c) for (a, b, c) in rx ]
rx2.sort(reverse=True)

# returns: [(4, 5), (3, 8)], ie, '5' occurs 4 times consecutively, '8' occurs 3 times consecutively