Here is Python code written to perform the following operations:

1. Find each occurrence of a three-letter sequence in a character array (e.g. a sequence such as ('a','b','c')), including overlapping sequences of up to 2 shared characters.
2. Count the characters between the start of each sequence and the start of all identical sequences following it.
3. Every time the same number of characters results from #2, increment a counter for that specific number of characters (regardless of which character sequence caused it).
4. Return a dictionary containing all accumulated counters for character distances.

The code is shown below.

counts = {}
# repeating groups of three identical chars in a row
for i in range(len(array)-2):
    for j in range(i+1,len(array)-2):
        if ((array[i] == array[j]) & (array[i+1] == array[j+1]) & (array[i+2] == array[j+2])):
            if counts.has_key(j-i) == False:
                counts[j-i] = 1
            else:
                counts[j-i] += 1

This code was originally written in another programming language, but I would like to apply any optimizations or improvements available in Python. Thank you for your time.

Here is Python code written to perform the following operations:

1. Find each occurrence of a three-letter sequence in a character array (e.g. a sequence such as ('a','b','c')), including overlapping sequences of up to 2 shared characters.
2. Count the characters between the start of each sequence and the start of all identical sequences following it.
3. Every time the same number of characters results from #2, increment a counter for that specific number of characters (regardless of which character sequence caused it).
4. Return a dictionary containing all accumulated counters for character distances.

counts = {}
# repeating groups of three identical chars in a row
for i in range(len(array)-2):
    for j in range(i+1,len(array)-2):
        if ((array[i] == array[j]) & (array[i+1] == array[j+1]) & (array[i+2] == array[j+2])):
            if counts.has_key(j-i) == False:
                counts[j-i] = 1
            else:
                counts[j-i] += 1

This code was originally written in another programming language, but I would like to apply any optimizations or improvements available in Python.

i have the code in python, which parses an array (containing characters), searching for repeating groups of characters (in this case repeating groups of the same 3 chars) and counting the occurences:

counts = {}
# repeating groups of three identical chars in a row
for i in range(len(array)-2):
    for j in range(i+1,len(array)-2):
        if ((array[i] == array[j]) & (array[i+1] == array[j+1]) & (array[i+2] == array[j+2])):
            if counts.has_key(j-i) == False:
                counts[j-i] = 1
            else:
                counts[j-i] += 1

But i think, that the code should be improved in python some way (originally was written in another language). if is there some way, please help me to improve this. thanks

Here is Python code written to perform the following operations:

1. Find each occurrence of a three-letter sequence in a character array (e.g. a sequence such as ('a','b','c')), including overlapping sequences of up to 2 shared characters.
2. Count the characters between the start of each sequence and the start of all identical sequences following it.
3. Every time the same number of characters results from #2, increment a counter for that specific number of characters (regardless of which character sequence caused it).
4. Return a dictionary containing all accumulated counters for character distances.

The code is shown below.

counts = {}
# repeating groups of three identical chars in a row
for i in range(len(array)-2):
    for j in range(i+1,len(array)-2):
        if ((array[i] == array[j]) & (array[i+1] == array[j+1]) & (array[i+2] == array[j+2])):
            if counts.has_key(j-i) == False:
                counts[j-i] = 1
            else:
                counts[j-i] += 1

This code was originally written in another programming language, but I would like to apply any optimizations or improvements available in Python. Thank you for your time.