/Traceback (most recent call last): IndexError: list index out of range

Question

Traceback (most recent call last): IndexError: list index out of range

from sys import argv, exit
import csv
import sys
    
def main ():
    if len(sys.argv) < 3:
        print("Usage: python dna.py data.csv sequence.txt")
        exit(1)
    # open csv_file contain the database 
    with open(sys.argv[1]) as csvfile:
        r = csv.reader(csvfile)
        # open txt_file contain the DNA sequence 
        with open(sys.argv[2]) as txtfile:
            x = txtfile.read()
            y = [v for v in [max_substring(x, seq) for seq in next(r)[1:]] if v >= 0]
        print_match(r, y)

# count the maximum substring           
def max_substring(x, sub_str):
    if not x and not sub_str:
        return -1
    sub_repeat = len(x)*[0]
    for s in range(len(x) - len(sub_str),  -1, -1): 
        a = s + len(sub_str)
        if x[s: a] == sub_str:
            b = len(x) - 1
            c = 1 + sub_repeat[a - 1]
            sub_repeat[s] = 1 if  a > b else c
    return max(sub_repeat)

# Make some assertions
assert max_substring('abcabbcab', 'ab') == 1
assert max_substring('', '') == -1
assert max_substring('abcabbcab', 'abcabbcab') == 1
    
# print the result
def print_match(r,y):
    for line in r:
        if [int(val) for val in line[1:]] == y:
            print(line[0]) 
            return
    print("No match")
# run the functions & class
if __name__ == "__main__" :
    main()

Traceback (most recent call last): IndexError: list index out of range

[new output error [1]: https://i.stack.imgur.com/lKKD8.png

Answer 1

You error can happen in two cases.

Empty Inputs

Just try calling: max_substring('', '') to reproduce the error (i.e. both parameters to the function are the empty string).

In that case:

• len(x) and len(sub_str) are both 0; therefore, len(x) - len(sub_str) is 0, and sub_repeat ends up being another empty string.
• list(range(0, -1, -1)) is just [0]
• Therefore, in the expression a = s + len(sub_str), s is 0, and so is a.
• x[s: a], being x[0:0] is the empty string. Therefore [s: a] == sub_str is True, and the if statement is entered.
• In the expression c = 1 + sub_repeat[a], since a is 0, it is trying to access the first index of sub_repeat. But because sub_repeat is an empty string, this results in the IndexError.

Non-Empty Input

• Repeat this for non-empty parameter values, and you will see that when the substring occurs at the very end, sub_repeat[a] doesn't exist because the length of sub_repeat is only a-1. I don't know if you want to change it to c = 1 + sub_repeat[a - 1] or sub_repeat = (1 + len(x))*[0], but both of those would get rid of the error.

Possible Solution

To avoid the error, you can add a check to prevent this case of both a and sub_str both being empty, and conditionally return the expected value for that edge case, and also avoid the second edge case above (by changing to c = 1 + sub_repeat[a - 1] or sub_repeat = (1 + len(x))*[0]). I'm not sure which one is appropriate, without knowing the purpose of this function.

    if not x and not sub_str:
        return -1
    sub_repeat = len(x)*[0]
    for s in range(len(x) - len(sub_str),  -1, -1): 
        a = s + len(sub_str)
        if x[s: a] == sub_str:
            b = len(x) - 1
            c = 1 + sub_repeat[a - 1]
            sub_repeat[s] = 1 if  a > b else c
    return max(sub_repeat)

# Make some assertions
assert max_substring('abcabbcab', 'ab') == 1
assert max_substring('', '') == -1
assert max_substring('abcabbcab', 'abcabbcab') == 1

The calling code will need to handle this new return value and ignore those cases where both inputs where empty:

y = [v for v in [max_substring(x, seq) for seq in next(r)[1:]] if v >= 0]

Note that by the time print_match(r, y) is called, that r will have no more items left to iterate over, due to the list comprehension, so you may have another problem.