python pattern count Find a “hidden message” in the replication origin

Question

The question ask to find a “hidden message” in the replication origin.

Input: A string Text (representing the replication origin of a genome).

Output: A hidden message in Text.

Translate to computational language,

Input: Strings Text and Pattern.

Output: Count(Text, Pattern).

For example,

Count(ACAACTATGCATACTATCGGGAACTATCCT, ACTAT) = 3.

In theory, we should account for overlapping occurrences of Pattern in Text right? So one way to do it is to screen down from first element to the length of text-length of the pattern we are looking for?

Here's the pseudo code i come up with,

def PatternCount(Text, Pattern):
    count = 0
    for i = 0 to len(Text)-len(Pattern):
        if Text(i, len(Pattern)) = Pattern:
            count = count + 1
    return count

Any suggestion? I'm new to python. Thanks in advance.

possible duplicate of [How can I find the number of overlapping sequences in a String with Python?](http://stackoverflow.com/questions/6844005/how-can-i-find-the-number-of-overlapping-sequences-in-a-string-with-python) — Stefano Sanfilippo, Nov 02 '14 at 19:24
A similar question and answer (if you want non-overlapping) can be found here: http://stackoverflow.com/questions/22566503/count-the-number-of-occurrences-of-a-word-in-a-string — tvandenbrande, Nov 02 '14 at 19:25
@tvandenbrande it seems that the OP wants to count all overlapping sequences. — Stefano Sanfilippo, Nov 02 '14 at 19:26
OP, have a look to the possible duplicate and to https://stackoverflow.com/q/19302525 — Stefano Sanfilippo, Nov 02 '14 at 19:27

score 1 · Answer 1 · answered Nov 02 '14 at 19:54

This is what I came up with:

def pattern_count(text, pattern):
    count = 0
    for i in range(0, len(text) - len(pattern) + 1):
        if text[i : len(pattern) + i] == pattern:
            count += 1
    return count

We're using string slicing (text[i : len(pattern) + i]) to check if the sub-string matches the pattern.

Input: text = "abc123!@#654abcabc" and pattern = "abc" Output: 3

score 0 · Answer 2 · answered Nov 03 '14 at 20:40

0

import re
print len(re.findall("abc", "abc123!@#654abcabc"))

answered Nov 03 '14 at 20:40

Stylize

1,058
5
16
32

score 0 · Answer 3 · answered Jan 13 '18 at 05:31

0

I think a more "pythonic" solution would be to use list comprehensions.

def pattern_count(text, pattern):
    return len([x for x in range(len(text) - len(pattern)+1) if pattern in text[x:len(pattern)+x]])

answered Jan 13 '18 at 05:31

Alexsh

67
7

python pattern count Find a “hidden message” in the replication origin

3 Answers3