In Python, creating a new object e.g. with a list comprehension is often better than modifying an existing one:
item_list = ['item', 5, 'foo', 3.14, True]
item_list = [e for e in item_list if e not in ('item', 5)]
... which is equivalent to:
item_list = ['item', 5, 'foo', 3.14, True]
new_list = []
for e in item_list:
if e not in ('item', 5):
new_list.append(e)
item_list = new_list
In case of a big list of filtered out values (here, ('item', 5) is a small set of elements), using a set is faster as the in operation is O(1) time complexity on average. It's also a good idea to build the iterable you're removing first, so that you're not creating it on every iteration of the list comprehension:
unwanted = {'item', 5}
item_list = [e for e in item_list if e not in unwanted]
A bloom filter is also a good solution if memory is not cheap.
Answer from aluriak on Stack OverflowIn Python, creating a new object e.g. with a list comprehension is often better than modifying an existing one:
item_list = ['item', 5, 'foo', 3.14, True]
item_list = [e for e in item_list if e not in ('item', 5)]
... which is equivalent to:
item_list = ['item', 5, 'foo', 3.14, True]
new_list = []
for e in item_list:
if e not in ('item', 5):
new_list.append(e)
item_list = new_list
In case of a big list of filtered out values (here, ('item', 5) is a small set of elements), using a set is faster as the in operation is O(1) time complexity on average. It's also a good idea to build the iterable you're removing first, so that you're not creating it on every iteration of the list comprehension:
unwanted = {'item', 5}
item_list = [e for e in item_list if e not in unwanted]
A bloom filter is also a good solution if memory is not cheap.
You can do it in one line by converting your lists to sets and using set.difference:
item_list = ['item', 5, 'foo', 3.14, True]
list_to_remove = ['item', 5, 'foo']
final_list = list(set(item_list) - set(list_to_remove))
Would give you the following output:
final_list = [3.14, True]
Note: this will remove duplicates in your input list and the elements in the output can be in any order (because sets don't preserve order). It also requires all elements in both of your lists to be hashable.
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Hello, I have created a list. The list has duplicate items, and I want to removed both the items from list.
a = ["Grapes", "Pineapple", "Coconut", "Mango", "Apple", "Banana", "Orange", "Mango"]
I used remove method to remove the items from list. But remove method is removing only first item instead of both.
a.remove("Mango")You don't need to iterate the array. Just:
>>> x = ['[email protected]', '[email protected]']
>>> x
['[email protected]', '[email protected]']
>>> x.remove('[email protected]')
>>> x
['[email protected]']
This will remove the first occurence that matches the string.
EDIT: After your edit, you still don't need to iterate over. Just do:
index = initial_list.index(item1)
del initial_list[index]
del other_list[index]
Using filter() and lambda would provide a neat and terse method of removing unwanted values:
newEmails = list(filter(lambda x : x != '[email protected]', emails))
This does not modify emails. It creates the new list newEmails containing only elements for which the anonymous function returned True.