Remove all occurrences of a value from a list?

Question

In Python remove() will remove the first occurrence of value in a list.

How to remove all occurrences of a value from a list?

This is w

Answer 1

hello =  ['h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']
#chech every item for a match
for item in range(len(hello)-1):
     if hello[item] == ' ': 
#if there is a match, rebuild the list with the list before the item + the list after the item
         hello = hello[:item] + hello [item + 1:]
print hello

['h', 'e', 'l', 'l', 'o', 'w', 'o', 'r', 'l', 'd']

Answer 2

you can do this

while 2 in x:   
    x.remove(2)

Answer 3

for i in range(a.count(' ')):
    a.remove(' ')

Much simpler I believe.

Answer 4

p=[2,3,4,4,4]
p.clear()
print(p)
[]

Only with Python 3

Answer 5

At the cost of readability, I think this version is slightly faster as it doesn't force the while to reexamine the list, thus doing exactly the same work remove has to do anyway:

x = [1, 2, 3, 4, 2, 2, 3]
def remove_values_from_list(the_list, val):
    for i in range(the_list.count(val)):
        the_list.remove(val)

remove_values_from_list(x, 2)

print(x)

Answer 6

Functional approach:

Python 3.x

>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter((2).__ne__, x))
[1, 3, 3, 4]

or

>>> x = [1,2,3,2,2,2,3,4]
>>> list(filter(lambda a: a != 2, x))
[1, 3, 3, 4]

Python 2.x

>>> x = [1,2,3,2,2,2,3,4]
>>> filter(lambda a: a != 2, x)
[1, 3, 3, 4]