If x not in list python

I have two lists:

mylist = ['total','age','gender','region','sex']
checklist = ['total','civic']

I have to work with some code I have inherited which looks like this:

for item in mylist:
    if item in checklist:
        do something:

How can I work with the code above to tell me that 'civic' is not in mylist?.

This would've been the ideal way to do it but I cant use it, don't ask me why.

for item in checklist:
    if item not in mylist:
        print item

Outcome:

civic

Paco

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asked Apr 3, 2014 at 9:50

Boosted_d16Boosted_d16

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1

Your code should work, but you can also try:

    if not item in mylist :

answered Dec 8, 2014 at 21:50

1

How about this?

for item in mylist:
    if item in checklist:
        pass
    else:
       # do something
       print item

answered Apr 3, 2014 at 9:53

Santosh GhimireSantosh Ghimire

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2

if I got it right, you can try

for item in [x for x in checklist if x not in mylist]:
    print (item)

answered Jun 8, 2018 at 21:50

Yury WalletYury Wallet

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You better do this syntax

if not (item in mylist):  
    Code inside the if

answered Oct 13, 2016 at 17:44

1

ValueError: list.remove(x): x not in list (Python) #

The Python "ValueError: list.remove(x): x not in list" occurs when we call the remove() method with a value that does not exist in the list. To solve the error, check if the value exists in the list before removing it, or use a try/except block.

Here is an example of how the error occurs.

Copied!
my_list = ['apple', 'banana', 'kiwi']

# ⛔️ ValueError: list.remove(x): x not in list
my_list.remove('melon')

We passed a value that is not in the list to the remove() method which caused the error.

One way to solve the error is to check if the value is present in the list before passing it to the remove() method.

Copied!
my_list = ['apple', 'banana', 'kiwi']


if 'melon' in my_list:
    my_list.remove('melon')
    print(my_list)
else:
    # 👇️ this runs
    print('value is not in the list')

The in operator tests for membership. For example, x in l evaluates to True if x is a member of l, otherwise it evaluates to False.

If you use a for loop, make sure to iterate over a copy of the list if you need to remove any items.

Copied!
my_list = ['apple', 'banana', 'kiwi']


# ✅ iterate over copy
for i in my_list.copy():
    my_list.remove(i)

print(my_list)  # 👉️ []

We used the copy() method to create a shallow copy of the list when iterating.

This is necessary because mutating the list while iterating over it leads to confusing behavior.

The list.remove() method removes the first item from the list whose value is equal to the passed in argument.

Copied!
my_list = ['a', 'b', 'c']

my_list.remove('a')

print(my_list)  # 👉️ ['b', 'c']

The method raises a ValueError if there is no such item.

The remove() method mutates the original list and returns None.

You can also use a try/except statement to handle the error in case the value is not present in the list.

Copied!
my_list = ['a', 'b', 'c']

try:
    my_list.remove('r')
except ValueError:
    print('Item not in list')

print(my_list)  # 👉️ ['a', 'b', 'c']

We call the remove() method on the list and if a ValueError is raised, the except block is run.

If you have a two-dimensional list, make sure you are calling the remove() method on the correct list.

Copied!
my_list = [['a', 'b'], ['c', 'd']]

my_list[0].remove('b')

print(my_list)  # 👉️ [['a'], ['c', 'd']]

We accessed the list item at index 0 and called the remove() method on it.

Had we called the remove() method on the outer list, we would get a ValueError because it doesn't contain the string "b".

Conclusion #

The Python "ValueError: list.remove(x): x not in list" occurs when we call the remove() method with a value that does not exist in the list. To solve the error, check if the value exists in the list before removing it, or use a try/except block.

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