Finding if a value is found within a list in python [duplicate]

.index() will always raise a ValueError If the item is not found, it does not return None or -1

You can do it by following these points.

1. Use in instead of index to avoid ValueError. It returns True or False

2. Use remove() which is clearer and safer than index + del.

3. No need for else: pass — It’s redundant.

a_list = ['apple', 'banana', 'cherry']
another_list = ['orange', 'apple', 'blueberry']
different_list = ['strawberry', 'pear', 'peach']
all_lists = [a_list, another_list, different_list]

for sublist in all_lists:        # using `in` instead of `index`
    if "apple" in sublist:
        sublist.remove("apple")  # Safely removes 'apple' if it exists

print(all_lists)

It looks like there might be a small typo in your loop, and you must have had there aSet instead of a set, please double-check that to avoid syntax errors.

Also, be careful with using .index() in a condition. Since .index() returns the position of the element (which can be 0), using it directly in an if statement can be tricky because 0 is considered False in Python. Instead, it’s safer to check if the element exists with the in operator first. examples of how .index() works

aList = ['apple', 'banana', 'cherry'] 
anotherList = ['orange', 'apple', 'blueberry'] 
adifferentList = ['strawberry', 'pear', 'peach']
aSet = [aList, anotherList, adifferentList]

for sublist in aSet:
    if "apple" in sublist:
        temp = sublist.index("apple")
        del sublist[temp]

And operator in - Returns True if a sequence with the specified value is present in the object

The way you approach this could depend on whether it's possible that "apple" occurs more than once in one of the sub-lists. If you want to remove any/all occurrences then you could do this:

all_lists = [
    ["apple", "banana", "cherry"],
    ["orange", "apple", "blueberry", "apple"],
    ["strawberry", "pear", "peach"],
]

for list_ in all_lists:
    list_[:] = [e for e in list_ if e != "apple"]

print(*all_lists, sep="\n")

Note that the 2nd sub-list has 2 occurrences of "apple". Thus, the output will be:

['banana', 'cherry']
['orange', 'blueberry']
['strawberry', 'pear', 'peach']

Here is an alternative solution using a list comprehension:

aList = ['apple', 'banana', 'cherry']
anotherList = ['orange', 'apple', 'blueberry', 'apple']
adifferentList = ['strawberry', 'pear', 'peach']
aSet = [aList, anotherList, adifferentList]

aSet = [[item for item in sublist if item != 'apple'] for sublist in aSet]
print(aSet)