To remove an item from a list, we have two options. One is using del mylist[i] where i is the index. Other is call mylist.remove(i) method where i is item in the list.

Generally, you would remove item from list if a certain condition is satisfied. Assuming that we want to delete even numbers from mylist, the iteration with index results in error

Example: Remove Items From List using del

mylist=[5,3,7,8,20,15,2,6,10,1]
l=len(mylist)
for i in range(l):
    if (mylist[i]%2==0):
        del mylist[i]

Output

IndexError: list index out of range

This is because list object is dynamically resized as items get deleted. Hence this method for removing item while iterating list doesn't work. A workaround for this could be to employ a decrementing loop. In this example, list size is 10, hence decrement index from 9 to 0

Example: Remove Items From List using del

mylist=[5,3,7,8,20,15,2,6,10,1]
for i in range(len(mylist)-1, -1, -1):
    if mylist[i]%2==0:
        del mylist[i]
print (mylist)

Output

[5, 3, 7, 15, 1]

Calling remove() method of List object also gives incorrect result

Example: Remove Items From List using remove()

mylist=[5,3,7,8,20,15,2,6,10,1]
for i in mylist:
    if (i%2==0):
        mylist.remove(i)
print (mylist)

Output

[5, 3, 7, 20, 15, 6, 1]

We can see that even numbers 20 and 6 are not deleted. This is because when i becomes 8, it is removed and items on right move one place to left and 20 becomes current item before next iteration. Similarly after 2 is removed, 6 becomes current item and escapes condition for removal.

The answer to this strange situation is using list comprehension

Example: Remove List Items using List Comprehension

mylist=[5,3,7,8,20,15,2,6,10,1]
mylist=[i for i in mylist if i%2!=0]
print(mylist)

Output

[5, 3, 7, 15, 1]

We can also use built-in filter() function. First argument is a function itself which is applied to each item in list and returns an iterator containing those items of list for which argument function evaluates to true.

Example:Remove List Items using filter()

mylist=[5,3,7,8,20,15,2,6,10,1]
mylist=list(filter(lambda x:x%2!=0, mylist))
print (mylist)

Output

[5, 3, 7, 15, 1]

Another way is to use filterfalse() function defined in itertools module.

Example: Remove Items using filterfalse()

from itertools import filterfalse
mylist=[5,3,7,8,20,15,2,6,10,1]
def iseven(i):
    if i%2==0:
        return i
mylist=list(filterfalse(iseven, mylist))
print (mylist)

Output

[5, 3, 7, 15, 1]

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Malhar Lathkar

Author

Malhar Lathkar is an independent software professional having 30+ years of experience in various technologies such as Python, Java, Android, PHP, and Databases. He is an author of Python Data Persistence: With SQL and NOSQL Databases