Small refactor suggestion in ApproximatePatternCount.py #1

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oekk · 2019-11-17T15:12:35+01:00

oekk commented

2019-11-17 15:12:35 +01:00

Since the code under the if condition is the same under the else statement, I believe you can change this:

def ApproximatePatternCount(Pattern, Text, d):
    count = 0
    for i in range(len(Text)-len(Pattern)+1):
        if Text[i:i+len(Pattern)] == Pattern:
            count += 1
        elif HammingDistance(Text[i:i+len(Pattern)], Pattern) <= d:
            count += 1
    return count

for this:

def ApproximatePatternCount(Pattern, Text, d):
    count = 0
    for i in range(len(Text)-len(Pattern)+1):
        if Text[i:i+len(Pattern)] == Pattern or HammingDistance(Text[i:i+len(Pattern)], Pattern) <= d:
            count += 1
    return count

In [ApproximatePatternCount.py](https://coolneng.duckdns.org/gitea/coolneng/biology-meets-programming/src/branch/master/Code/ApproximatePatternCount.py) Since the code under the if condition is the same under the else statement, I believe you can change this: def ApproximatePatternCount(Pattern, Text, d): count = 0 for i in range(len(Text)-len(Pattern)+1): if Text[i:i+len(Pattern)] == Pattern: count += 1 elif HammingDistance(Text[i:i+len(Pattern)], Pattern) <= d: count += 1 return count for this: def ApproximatePatternCount(Pattern, Text, d): count = 0 for i in range(len(Text)-len(Pattern)+1): if Text[i:i+len(Pattern)] == Pattern or HammingDistance(Text[i:i+len(Pattern)], Pattern) <= d: count += 1 return count

coolneng commented

2019-11-17 15:49:46 +01:00

Good catch, will update it right now

coolneng referenced this issue from a commit

2019-11-17 15:51:41 +01:00

Refactor ApproximatePatternCount fixes: #1

coolneng closed this issue

2019-11-17 15:51:41 +01:00

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