如何从列表list1中移除存在于列表list2中的子列表并避免索引越界问题？

The issue with your original code stems from two key problems:

• You’re iterating over the length of list2 instead of list1, which leads to trying to access indices in list1 that don’t exist (since list1 is shorter than list2 in your example).
• Even if you iterated over list1’s length, modifying the list while iterating forward causes index shifting—this can skip elements or trigger out-of-bounds errors as the list shrinks.

A clean, safe way to achieve your goal is to create a new list containing only elements from list1 that are not present in list2. Here’s how to do it with a list comprehension:

list1 = [[6, 0, 3, 1, 5, 7, 8, 2, 4], [1, 5, 0, 4, 6, 2, 7, 8, 3]]
list2 = [[1, 5, 2, 4, 6, 0, 7, 8, 3], [1, 5, 0, 4, 6, 2, 7, 8, 3], [1, 8, 2, 4, 0, 6, 7, 5, 3]]

# Filter list1 to keep elements not found in list2
list1 = [item for item in list1 if item not in list2]

print(list1)  # Output: [[6, 0, 3, 1, 5, 7, 8, 2, 4]]

Why this works: #

• List comprehensions are purpose-built for filtering tasks: they loop through each element in list1, check if it’s absent from list2, and include it in the new list only if the condition holds.
• By creating a new list instead of modifying the original while iterating, you avoid all index-related bugs entirely.

If you need to modify list1 in-place (optional): #

If you prefer to update the original list without creating a new one, iterate backwards over list1’s indices. This way, removing elements doesn’t affect the indices of the items you haven’t checked yet:

# Iterate from the last index to the first
for i in range(len(list1)-1, -1, -1):
    if list1[i] in list2:
        list1.pop(i)

内容的提问来源于stack exchange，提问作者user17277981