a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

The above shows the following error:

Traceback (most recent call last):
  File "D:\zjm_code\a.py", line 6, in <module>
    b = a.index(6)
ValueError: list.index(x): x not in list

So I have to do this:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

But is there not a simpler way to do this?

To remove an element’s first occurrence in a list, simply use list.remove:

>>> a = ['a', 'b', 'c', 'd']
>>> a.remove('b')
>>> print(a)
['a', 'c', 'd']

Mind that it does not remove all occurrences of your element. Use a list comprehension for that.

>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]

Usually Python will throw an Exception if you tell it to do something it can’t so you’ll have to do either:

if c in a:
    a.remove(c)

or:

try:
    a.remove(c)
except ValueError:
    pass

An Exception isn’t necessarily a bad thing as long as it’s one you’re expecting and handle properly.

You can do

a=[1,2,3,4]
if 6 in a:
    a.remove(6)

but above need to search 6 in list a 2 times, so try except would be faster

try:
    a.remove(6)
except:
    pass

Consider:

a = [1,2,2,3,4,5]

To take out all occurrences, you could use the filter function in python. For example, it would look like:

a = list(filter(lambda x: x!= 2, a))

So, it would keep all elements of a != 2.

To just take out one of the items use

a.remove(2)

Here’s how to do it inplace (without list comprehension):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]

If you know what value to delete, here’s a simple way (as simple as I can think of, anyway):

a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) > 0:
    a.remove(1)

You’ll get [0, 0, 2, 3, 4]

Another possibility is to use a set instead of a list, if a set is applicable in your application.

IE if your data is not ordered, and does not have duplicates, then

my_set=set([3,4,2])
my_set.discard(1)

is error-free.

Often a list is just a handy container for items that are actually unordered. There are questions asking how to remove all occurences of an element from a list. If you don’t want dupes in the first place, once again a set is handy.

my_set.add(3)

doesn’t change my_set from above.

As stated by numerous other answers, list.remove() will work, but throw a ValueError if the item wasn’t in the list. With python 3.4+, there’s an interesting approach to handling this, using the suppress contextmanager:

from contextlib import suppress
with suppress(ValueError):
    a.remove('b')

Finding a value in a list and then deleting that index (if it exists) is easier done by just using list’s remove method:

>>> a = [1, 2, 3, 4]
>>> try:
...   a.remove(6)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 3, 4]
>>> try:
...   a.remove(3)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 4]

If you do this often, you can wrap it up in a function:

def remove_if_exists(L, value):
  try:
    L.remove(value)
  except ValueError:
    pass

This example is fast and will delete all instances of a value from the list:

a = [1,2,3,1,2,3,4]
while True:
    try:
        a.remove(3)
    except:
        break
print a
>>> [1, 2, 1, 2, 4]

In one line:

a.remove('b') if 'b' in a else None

sometimes it usefull.

Even easier:

if 'b' in a: a.remove('b')

If your elements are distinct, then a simple set difference will do.

c = [1,2,3,4,'x',8,6,7,'x',9,'x']
z = list(set(c) - set(['x']))
print z
[1, 2, 3, 4, 6, 7, 8, 9]

We can also use .pop:

>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
...     lst.pop(lst.index(remove_element))
... 
23
>>> lst
[34, 54, 45]
>>> 

Overwrite the list by indexing everything except the elements you wish to remove

>>> s = [5,4,3,2,1]
>>> s[0:2] + s[3:]
[5, 4, 2, 1]

With a for loop and a condition:

def cleaner(seq, value):    
    temp = []                      
    for number in seq:
        if number != value:
            temp.append(number)
    return temp

And if you want to remove some, but not all:

def cleaner(seq, value, occ):
    temp = []
    for number in seq:
        if number == value and occ:
            occ -= 1
            continue
        else:
            temp.append(number)
    return temp

 list1=[1,2,3,3,4,5,6,1,3,4,5]
 n=int(input('enter  number'))
 while n in list1:
    list1.remove(n)
 print(list1)

Say for example, we want to remove all 1’s from x. This is how I would go about it:

x = [1, 2, 3, 1, 2, 3]

Now, this is a practical use of my method:

def Function(List, Unwanted):
    [List.remove(Unwanted) for Item in range(List.count(Unwanted))]
    return List
x = Function(x, 1)
print(x)

And this is my method in a single line:

[x.remove(1) for Item in range(x.count(1))]
print(x)

Both yield this as an output:

[2, 3, 2, 3, 2, 3]

Hope this helps. PS, pleas note that this was written in version 3.6.2, so you might need to adjust it for older versions.

Maybe your solutions works with ints, but It Doesnt work for me with dictionarys.

In one hand, remove() has not worked for me. But maybe it works with basic Types. I guess the code bellow is also the way to remove items from objects list.

In the other hand, ‘del’ has not worked properly either. In my case, using python 3.6: when I try to delete an element from a list in a ‘for’ bucle with ‘del’ command, python changes the index in the process and bucle stops prematurely before time. It only works if You delete element by element in reversed order. In this way you dont change the pending elements array index when you are going through it

Then, Im used:

c = len(list)-1
for element in (reversed(list)):
    if condition(element):
        del list[c]
    c -= 1
print(list)

where ‘list’ is like [{‘key1′:value1’},{‘key2’:value2}, {‘key3’:value3}, …]

Also You can do more pythonic using enumerate:

for i, element in enumerate(reversed(list)):
    if condition(element):
        del list[(i+1)*-1]
print(list)

arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]

This removes all instances of "-v" from the array sys.argv, and does not complain if no instances were found:

while "-v" in sys.argv:
    sys.argv.remove('-v')

You can see the code in action, in a file called speechToText.py:

$ python speechToText.py -v
['speechToText.py']

$ python speechToText.py -x
['speechToText.py', '-x']

$ python speechToText.py -v -v
['speechToText.py']

$ python speechToText.py -v -v -x
['speechToText.py', '-x']

this is my answer, just use while and for

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()

>>> a=[1,2,3]
>>> a.remove(2)
>>> a
[1, 3]
>>> a=[1,2,3]
>>> del a[1]
>>> a
[1, 3]
>>> a= [1,2,3]
>>> a.pop(1)
2
>>> a
[1, 3]
>>> 

Is there any difference between the above three methods to remove an element from a list?

Yes, remove removes the first matching value, not a specific index:

>>> a = [0, 2, 3, 2]
>>> a.remove(2)
>>> a
[0, 3, 2]

del removes the item at a specific index:

>>> a = [9, 8, 7, 6]
>>> del a[1]
>>> a
[9, 7, 6]

and pop removes the item at a specific index and returns it.

>>> a = [4, 3, 5]
>>> a.pop(1)
3
>>> a
[4, 5]

Their error modes are different too:

>>> a = [4, 5, 6]
>>> a.remove(7)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: list.remove(x): x not in list
>>> del a[7]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list assignment index out of range
>>> a.pop(7)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: pop index out of range

Use del to remove an element by index, pop() to remove it by index if you need the returned value, and remove() to delete an element by value. The latter requires searching the list, and raises ValueError if no such value occurs in the list.

When deleting index i from a list of n elements, the computational complexities of these methods are

del     O(n - i)
pop     O(n - i)
remove  O(n)

Since no-one else has mentioned it, note that del (unlike pop) allows the removal of a range of indexes because of list slicing:

>>> lst = [3, 2, 2, 1]
>>> del lst[1:]
>>> lst
[3]

This also allows avoidance of an IndexError if the index is not in the list:

>>> lst = [3, 2, 2, 1]
>>> del lst[10:]
>>> lst
[3, 2, 2, 1]

Already answered quite well by others. This one from my end :)

Evidently, pop is the only one which returns the value, and remove is the only one which searches the object, while del limits itself to a simple deletion.

Many best explanations are here but I will try my best to simplify more.

Among all these methods, reverse & pop are postfix while delete is prefix.

remove(): It used to remove first occurrence of element

remove(i) => first occurrence of i value

>>> a = [0, 2, 3, 2, 1, 4, 6, 5, 7]
>>> a.remove(2)   # where i = 2
>>> a
[0, 3, 2, 1, 4, 6, 5, 7]

pop(): It used to remove element if:

unspecified

pop() => from end of list

>>>a.pop()
>>>a
[0, 3, 2, 1, 4, 6, 5]

specified

pop(index) => of index

>>>a.pop(2)
>>>a
[0, 3, 1, 4, 6, 5]

WARNING: Dangerous Method Ahead

delete(): Its a prefix method.

Keep an eye on two different syntax for same method: [] and (). It possesses power to:

1.Delete index

del a[index] => used to delete index and its associated value just like pop.

>>>del a[1]
>>>a
[0, 1, 4, 6, 5]

2.Delete values in range [index 1:index N]

del a[0:3] => multiple values in range

>>>del a[0:3]
>>>a
[6, 5]

3.Last but not list, to delete whole list in one shot

del (a) => as said above.

>>>del (a)
>>>a

Hope this clarifies the confusion if any.

pop – Takes Index and returns Value

remove – Takes value, removes first occurrence, and returns nothing

delete – Takes index, removes value at that index, and returns nothing

Any operation/function on different data structures is defined for particular actions. Here in your case i.e. removing an element, delete, Pop and remove. (If you consider sets, Add another operation – discard) Other confusing case is while adding. Insert/Append. For Demonstration, Let us Implement deque. deque is a hybrid linear data structure, where you can add elements / remove elements from both ends.(Rear and front Ends)

class Deque(object):

  def __init__(self):

    self.items=[]

  def addFront(self,item):

    return self.items.insert(0,item)
  def addRear(self,item):

    return self.items.append(item)
  def deleteFront(self):

    return self.items.pop(0)
  def deleteRear(self):
    return self.items.pop()
  def returnAll(self):

    return self.items[:]

In here, see the operations:

def deleteFront(self):

    return self.items.pop(0)
def deleteRear(self):
    return self.items.pop()

Operations have to return something. So, pop – With and without an index. If I don’t want to return the value: del self.items[0]

Delete by value not Index:

• remove :

  list_ez=[1,2,3,4,5,6,7,8]
  for i in list_ez:
      if i%2==0:
          list_ez.remove(i)
  print list_ez
  

Returns [1,3,5,7]

let us consider the case of sets.

set_ez=set_ez=set(range(10))

set_ez.remove(11)

# Gives Key Value Error. 
##KeyError: 11

set_ez.discard(11)

# Does Not return any errors.

While pop and delete both take indices to remove an element as stated in above comments. A key difference is the time complexity for them. The time complexity for pop() with no index is O(1) but is not the same case for deletion of last element.

If your use case is always to delete the last element, it’s always preferable to use pop() over delete(). For more explanation on time complexities, you can refer to https://www.ics.uci.edu/~pattis/ICS-33/lectures/complexitypython.txt

The remove operation on a list is given a value to remove. It searches the list to find an item with that value and deletes the first matching item it finds. It is an error if there is no matching item, raises a ValueError.

>>> x = [1, 0, 0, 0, 3, 4, 5]
>>> x.remove(4)
>>> x
[1, 0, 0, 0, 3, 5]
>>> del x[7]
Traceback (most recent call last):
  File "<pyshell#1>", line 1, in <module>
    del x[7]
IndexError: list assignment index out of range

The del statement can be used to delete an entire list. If you have a specific list item as your argument to del (e.g. listname[7] to specifically reference the 8th item in the list), it’ll just delete that item. It is even possible to delete a “slice” from a list. It is an error if there index out of range, raises a IndexError.

>>> x = [1, 2, 3, 4]
>>> del x[3]
>>> x
[1, 2, 3]
>>> del x[4]
Traceback (most recent call last):
  File "<pyshell#1>", line 1, in <module>
    del x[4]
IndexError: list assignment index out of range

The usual use of pop is to delete the last item from a list as you use the list as a stack. Unlike del, pop returns the value that it popped off the list. You can optionally give an index value to pop and pop from other than the end of the list (e.g listname.pop(0) will delete the first item from the list and return that first item as its result). You can use this to make the list behave like a queue, but there are library routines available that can provide queue operations with better performance than pop(0) does. It is an error if there index out of range, raises a IndexError.

>>> x = [1, 2, 3] 
>>> x.pop(2) 
3 
>>> x 
[1, 2]
>>> x.pop(4)
Traceback (most recent call last):
  File "<pyshell#1>", line 1, in <module>
    x.pop(4)
IndexError: pop index out of range

See collections.deque for more details.

Remove basically works on the value . Delete and pop work on the index

Remove basically removes the first matching value. Delete deletes the item from a specific index Pop basically takes an index and returns the value at that index. Next time you print the list the value doesnt appear.

You can also use remove to remove a value by index as well.

n = [1, 3, 5]

n.remove(n[1])

n would then refer to [1, 5]