CODEWITHSUNDEEP
python
The T
The T

Reputation: 347

List index out of range after using shuffle function in Python

It is important to let the readers know that I sometimes have to delete items in my list of lists before calling the function.

Error Code

Traceback (most recent call last):
  File "C:\Users\User\Desktop\permdebunk.py", line 195, in <module>
    shuff(c,len(c))
  File "C:\Users\User\Desktop\permdebunk.py", line 181, in shuff
    c[i],c[j] = c[j],c[i]
IndexError: list index out of range

Python's random.shuffle() is insufficient for my needs if it will not perform all possible permutations. So I did some google searching and found this function that shuffles a list that supposedly circumvents the problems stated above about Python's random.shuffle()

def shuff(c, n):
    for i in range(n-1,0,-1):
        random.seed()
        j = random.randint(0,i+1)
        c[i],c[j] = c[j],c[i]

The problem seems to occur after I append c[0] to c and then del c[0] to push into the list. I have over 200+ lines of code as an experiment into an NP hard problem.Please look at the first log in the link. And remember to check line 195 and 181

Edit: The link didn't have line numbers so use ctrl + f to find the word shuff

Question

What ways can I "push" into the list without getting any index errors? And, why does python throw that error?

Upvotes: 0

Views: 218

Answers (1)

RootTwo
RootTwo

Reputation: 4418

random.randint() includes both endpoints. So when i = n - 1, randint(0, i + 1) could return (n - 1) + 1, which is n. c[n] is out of range of the list.

I should think changing the randint() call to randint(0, i) should work.

Upvotes: 3

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