BA1_A - count pattern in text.py

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Thu Dec 19 02:18:03 2019

@author: jasonmoggridge

BA1A

Compute the Number of Times a Pattern Appears in a Text

This is the first problem in a collection of "code challenges" to
 accompany Bioinformatics Algorithms: An Active-Learning Approach
 by Phillip Compeau & Pavel Pevzner.

A k-mer is a string of length k. We define Count(Text, Pattern)
 as the number of times that a k-mer Pattern appears as a substring of Text.

 For example,

    Count(ACAACTATGCATACTATCGGGAACTATCCT,ACTAT)=3.
    
    We note that Count(CGATATATCCATAG, ATA) is equal to 3 
    (not 2) since we should account for overlapping occurrences
    of Pattern in Text.

To compute Count(Text, Pattern), our plan is to “slide a window”
 down Text, checking whether each k-mer substring of Text matches
 Pattern. We will therefore refer to the k-mer starting at position
 i of Text as Text(i, k). Throughout this book, we will often use
 0-based indexing, meaning that we count starting at 0 instead of 1.
 In this case, Text begins at position 0 and ends at position:
     |Text| − 1 (|Text| denotes the number of symbols in Text). 
     
     For example, if Text = GACCATACTG, 
     then Text(4, 3) = ATA. 
     Note that the last k-mer of Text begins at position 
     |Text| − k, e.g., the last 3-mer of GACCATACTG starts
     at position 10 − 3 = 7. This discussion results in the
     following pseudocode for computing Count(Text, Pattern).

    PatternCount(Text, Pattern)
        count ← 0
        for i ← 0 to |Text| − |Pattern|
            if Text(i, |Pattern|) = Pattern
                count ← count + 1
        return count
        
        
Implement PatternCount

Given: {DNA strings}} Text and Pattern.

Return: Count(Text, Pattern).

Sample Dataset
GCGCG
GCG
Sample Output
2

"""

text = 'CCGGAAGTAAGGAAGTAGGAAGTAGGAAGTATGGAAGTAAGCGAGGAAGTAGGAAGTAGGAAGTATAGGAAGTATTGGAAGTATCGGAAGTAGGAAGTAGGAAGTACGAACCAAGGAAGTAGGAAGTAGGAAGTACTGGAAGTATCGGAAGTACCGGGGAAGTAGGGAAGTAAGGAAGTAACGGAAGTATGGAAGTAGGAAGTAAGCGGAAGTATATGGGAAGTAACGATATGCCATTCTCGGAAGTACAGTAGGAAGTAGGAAGTAGGAAGTACCTGGAAGTAGTCACGGGTGGAAGTACGGAAGTAGGAAGTAGGGAAGTAGGTTCTTCGGAAGTACCTTCGTGGGAAGTAGGAAGTACGGAAGTATGGAAGTAGGAAGTAGCGGAAGTAATGCCGGAAGTAGGAAGTATGAGGAAGTAGGAAGTAGGGGGAAGTATTGGGAAGTAGGGAAGTATGGGAAGTACATGGAAGTAGGAAGTATGGAAGTACTGGGAAGTAGGAAGTATAACGGAAGTAGACGGAAGTAGGAAGTAGGAAGTATTAGGAAGTAGTCAGGGGAAGTATAGGAAGTAACGGAAGTACGGAAGTAGGAAGTAGGAAGTAGGGAAGTATATTGCCAGGAAGTATTCGAGGAAGTAAAGCGGAAGTAGGAAGTATTGGAAGTATGGAAGTAGGAAGTAGGGAAGTATTGGAAGTAGGAGGAAGTAATCTGGAAGTATTGATACTGGGGAAGTAGGAAGTATCGGAAGTAACTTGGAAGTAGGGAAGTAAGGAAGTATTGTGGAAGTAATGGAAGTAACTCAGGAAGTATGGAAGTAGGAAGTAGAGCGAAGTATTCAGGAAGTAACTAAGGAAGTAGTAAGTGGAAGTAGGAAGTATAAGGGAAGTATTCGGAAGTAGGAAGTACTGGAAGTACTAGGAAGTA'
pattern = 'GGAAGTAGG'
count = 0

for i in range(len(text)-len(pattern)+1):
    print('i:',i)
    print(text[i:i+len(pattern)])
    print(pattern)
    
    if text[i:i+len(pattern)] == pattern:
        count+=1

print(count)