Python：将所有列表项与字符串的子字符串进行比较

Question

All the items in the list should be compared to the every 50 long substring of a string. The code I have written is working smaller string lengths but if string is very large(eg:8800) its not. Can anyone suggest a better way or debug the code?

Code:

a_str = 'CGGACTCGACAGATGTGAAGAACGACAATGTGAAGACTCGACACGACAGAGTGAAGAGAAGAGGAAACATTGTAA'
a = 0
b = 5
c = 50
leng = len(a_str)
lengb = leng - b + 1
list1 = []
list2 = []
list3 = []
list4 = []
for i in a_str[a:lengb]:
    findstr = a_str[a:b]
    if findstr not in list2:
        count = a_str.count(findstr)
        list1 = [m.start() for m in re.finditer(findstr, a_str)]
        last = list1[-1]
        first = list1[0]
        diff = last - first
        if diff > 45:
            count = count - 1
        if count > 3:
            list2.append(findstr)
            list3.append(list1)
    a += 1
    b += 1

a = 0
dictionary = dict(zip(list2, list3))
for j in list2:
    for k in a_str[a:c]:
        if c < leng:
            str1 = a_str[a:c]
            if str1.count(j) == 4:
                list4.append(j)
    a += 1
    c += 1

print(list4)

For a string which is 8800, b=10, count1=17, and c=588 long c is taking value only till 1161 during looping

I need substring of length 5 repeated 4 times in a window length of 50(ie; for every 50 characters of the main string)

Answer 1

I used comprehensions and sets to create a more understandable function.

def find_four_substrings(a_str, sub_len=5, window=50, occurs=4):
    '''
    Given a string of any length return the set of substrings
    of sub_length (default is 5) that exists exactly occurs 
    (default 4) times in the string, for a window (default 50)
    '''
    return set(a_str[i:i+sub_len] for i in range(len(a_str) - sub_len) 
                if a_str.count(a_str[i:i+sub_len], i, window) == occurs)

and

a_str = 'CGGACTCGACAGATGTGAAGAACGACAATGTGAAGACTCGACACGACAGAGTGAAGAGAAGAGGAAACATTGTAA'
print(find_four_substrings(a_str))

returns

set(['CGACA'])

Answer 2

This finds all substrings of length 5 that are repeated at least 4 or more times (not overlapping) within 50 characters. The resulting list does not have duplicates.

a_str = 'CGGACTCGACAGATGTGAAGAACGACAATGTGAAGACTCGACACGACAGAGTGAAGAGAAGAGGAAACATTGTAA'
b = 5      #length of substring
c = 50     #length of window
repeat = 4 #minimum number of repetitions

substrings = list({
    a_str[i:i+b]
    for i in range(len(a_str) - b)
    if a_str.count(a_str[i:i+b], i+b, i+c) >= repeat - 1
})
print(substrings)

I believe this is what you want. Let me know if otherwise.

['CGACA', 'GAAGA']