python - 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 what I have in mind:

>>> remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
[1, 3, 4, 3]

Answer 1

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]

Answer 2

You can use a list comprehension:

def remove_values_from_list(the_list, val):
   return [value for value in the_list if value != val]

x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]

Answer 3

You can use slice assignment if the original list must be modified, while still using an efficient list comprehension (or generator expression).

>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> x[:] = (value for value in x if value != 2)
>>> x
[1, 3, 4, 3]

Answer 4

Repeating the solution of the first post in a more abstract way:

>>> x = [1, 2, 3, 4, 2, 2, 3]
>>> while 2 in x: x.remove(2)
>>> x
[1, 3, 4, 3]

Answer 5

See the simple solution

>>> [i for i in x if i != 2]

This will return a list having all elements of x without 2

Answer 6

All of the answers above (apart from Martin Andersson's) create a new list without the desired items, rather than removing the items from the original list.

>>> import random, timeit
>>> a = list(range(5)) * 1000
>>> random.shuffle(a)

>>> b = a
>>> print(b is a)
True

>>> b = [x for x in b if x != 0]
>>> print(b is a)
False
>>> b.count(0)
0
>>> a.count(0)
1000

>>> b = a
>>> b = filter(lambda a: a != 2, x)
>>> print(b is a)
False

This can be important if you have other references to the list hanging around.

To modify the list in place, use a method like this

>>> def removeall_inplace(x, l):
...     for _ in xrange(l.count(x)):
...         l.remove(x)
...
>>> removeall_inplace(0, b)
>>> b is a
True
>>> a.count(0)
0

As far as speed is concerned, results on my laptop are (all on a 5000 entry list with 1000 entries removed)

• List comprehension - ~400us
• Filter - ~900us
• .remove() loop - 50ms

So the .remove loop is about 100x slower........ Hmmm, maybe a different approach is needed. The fastest I've found is using the list comprehension, but then replace the contents of the original list.

>>> def removeall_replace(x, l):
....    t = [y for y in l if y != x]
....    del l[:]
....    l.extend(t)

• removeall_replace() - 450us

Answer 7

better solution with list comprehension

x = [ i for i in x if i!=2 ]

Answer 8

Numpy approach and timings against a list/array with 1.000.000 elements:

Timings:

In [10]: a.shape
Out[10]: (1000000,)

In [13]: len(lst)
Out[13]: 1000000

In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop

In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop

Conclusion: numpy is 27 times faster (on my notebook) compared to list comprehension approach

PS if you want to convert your regular Python list lst to numpy array:

arr = np.array(lst)

Setup:

import numpy as np
a = np.random.randint(0, 1000, 10**6)

In [10]: a.shape
Out[10]: (1000000,)

In [12]: lst = a.tolist()

In [13]: len(lst)
Out[13]: 1000000

Check:

In [14]: a[a != 2].shape
Out[14]: (998949,)

In [15]: len([x for x in lst if x != 2])
Out[15]: 998949

Answer 9

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 10

a = [1, 2, 2, 3, 1]
to_remove = 1
a = [i for i in a if i != to_remove]
print(a)

Perhaps not the most pythonic but still the easiest for me haha

Answer 11

To remove all duplicate occurrences and leave one in the list:

test = [1, 1, 2, 3]

newlist = list(set(test))

print newlist

[1, 2, 3]

Here is the function I've used for Project Euler:

def removeOccurrences(e):
  return list(set(e))

Answer 12

I believe this is probably faster than any other way if you don't care about the lists order, if you do take care about the final order store the indexes from the original and resort by that.

category_ids.sort()
ones_last_index = category_ids.count('1')
del category_ids[0:ones_last_index]

Answer 13

Let

>>> x = [1, 2, 3, 4, 2, 2, 3]

The simplest and efficient solution as already posted before is

>>> x[:] = [v for v in x if v != 2]
>>> x
[1, 3, 4, 3]

Another possibility which should use less memory but be slower is

>>> for i in range(len(x) - 1, -1, -1):
        if x[i] == 2:
            x.pop(i)  # takes time ~ len(x) - i
>>> x
[1, 3, 4, 3]

Timing results for lists of length 1000 and 100000 with 10% matching entries: 0.16 vs 0.25 ms, and 23 vs 123 ms.

Answer 14

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

Much simpler I believe.

Answer 15

Remove all occurrences of a value from a Python list

lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list():
    for list in lists:
      if(list!=7):
         print(list)
remove_values_from_list()

Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11

Alternatively,

lists = [6.9,7,8.9,3,5,4.9,1,2.9,7,9,12.9,10.9,11,7]
def remove_values_from_list(remove):
    for list in lists:
      if(list!=remove):
        print(list)
remove_values_from_list(7)

Result: 6.9 8.9 3 5 4.9 1 2.9 9 12.9 10.9 11

Answer 16

If you didn't have built-in filter or didn't want to use extra space and you need a linear solution...

def remove_all(A, v):
    k = 0
    n = len(A)
    for i in range(n):
        if A[i] !=  v:
            A[k] = A[i]
            k += 1

    A = A[:k]

Answer 17

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 18

I just did this for a list. I am just a beginner. A slightly more advanced programmer can surely write a function like this.

for i in range(len(spam)):
    spam.remove('cat')
    if 'cat' not in spam:
         print('All instances of ' + 'cat ' + 'have been removed')
         break

Answer 19

We can also do in-place remove all using either del or pop:

import random

def remove_values_from_list(lst, target):
    if type(lst) != list:
        return lst

    i = 0
    while i < len(lst):
        if lst[i] == target:
            lst.pop(i)  # length decreased by 1 already
        else:
            i += 1

    return lst

remove_values_from_list(None, 2)
remove_values_from_list([], 2)
remove_values_from_list([1, 2, 3, 4, 2, 2, 3], 2)
lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)], 2)
print(len(lst))

Now for the efficiency:

In [21]: %timeit -n1 -r1 x = random.randrange(0,10)
1 loop, best of 1: 43.5 us per loop

In [22]: %timeit -n1 -r1 lst = [random.randrange(0, 10) for x in range(1000000)]
g1 loop, best of 1: 660 ms per loop

In [23]: %timeit -n1 -r1 lst = remove_values_from_list([random.randrange(0, 10) for x in range(1000000)]
    ...: , random.randrange(0,10))
1 loop, best of 1: 11.5 s per loop

In [27]: %timeit -n1 -r1 x = random.randrange(0,10); lst = [a for a in [random.randrange(0, 10) for x in
    ...:  range(1000000)] if x != a]
1 loop, best of 1: 710 ms per loop

As we see that in-place version remove_values_from_list() does not require any extra memory, but it does take so much more time to run:

• 11 seconds for inplace remove values
• 710 milli seconds for list comprehensions, which allocates a new list in memory

Answer 20

No one has posted an optimal answer for time and space complexity, so I thought I would give it a shot. Here is a solution that removes all occurrences of a specific value without creating a new array and at an efficient time complexity. The drawback is that the elements do not maintain order.

Time complexity: O(n)
Additional space complexity: O(1)

def main():
    test_case([1, 2, 3, 4, 2, 2, 3], 2)     # [1, 3, 3, 4]
    test_case([3, 3, 3], 3)                 # []
    test_case([1, 1, 1], 3)                 # [1, 1, 1]


def test_case(test_val, remove_val):
    remove_element_in_place(test_val, remove_val)
    print(test_val)


def remove_element_in_place(my_list, remove_value):
    length_my_list = len(my_list)
    swap_idx = length_my_list - 1

    for idx in range(length_my_list - 1, -1, -1):
        if my_list[idx] == remove_value:
            my_list[idx], my_list[swap_idx] = my_list[swap_idx], my_list[idx]
            swap_idx -= 1

    for pop_idx in range(length_my_list - swap_idx - 1):
        my_list.pop() # O(1) operation


if __name__ == '__main__':
    main()

Answer 21

About the speed!

import time
s_time = time.time()

print 'start'
a = range(100000000)
del a[:]
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 3.25

s_time = time.time()
print 'start'
a = range(100000000)
a = []
print 'finished in %0.2f' % (time.time() - s_time)
# start
# finished in 2.11

Answer 22

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

Only with Python 3

Answer 23

What's wrong with:

Motor=['1','2','2']
For i in Motor:
       If i  != '2':
       Print(i)
Print(motor)

Using anaconda