Is there a simple way to delete a list element by value?
this has nothing to do with deleting a value in a list, since your code does not reach the del statement. Maybe you should retitle it «how do I get the index of a value that is not in a list. Obvious answer — you can’t.
@Dave Well, not really. He wants to delete an item from the list regardless of whether it exists or not, not to get the index for a nonexistent item. The question is well asked.
26 Answers 26
To remove the first occurrence of an element, use list.remove :
>>> xs = ['a', 'b', 'c', 'd'] >>> xs.remove('b') >>> print(xs) ['a', 'c', 'd'] To remove all occurrences of an element, use a list comprehension:
>>> xs = ['a', 'b', 'c', 'd', 'b', 'b', 'b', 'b'] >>> xs = [x for x in xs if x != 'b'] >>> print(xs) ['a', 'c', 'd'] The list comprehension change the list reference so if there is a copy of the reference somewhere, the removal will not follow.
I do not imagine why it should be more or less than O(n). I think that it uses an ordinary search algorithm. No additional information about any sorting of the list!
Usually Python will throw an Exception if you tell it to do something it can’t so you’ll have to do either:
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.
Prevention is better than cure. If you can check for exceptional conditions first (example a), you should.
Whilst this is true in other languages, in Python it is «easier to ask for forgiveness than permission.» docs.python.org/2/glossary.html#term-eafp
@Gusdor: if the list is shared between threads then a.remove(c) might fail anyway despite the if c in a check ( a could be modified in another thread after the c in a check but before the a.remove(c) call). try/except or locks could be used to avoid the race condition.
@J.F.Sebastian if a list is shared between threads and you are not applying critical sections then you have bigger problems.
@Gusdor, the Pythonique idiom is to try without checking and catch the exception if it occurs. It is more efficient (only one lookup instead of two), albeit a little uglier
but above need to search 6 in list a 2 times, so try except would be faster
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
@OlafDietsche In Python 3.x, it returns a filter object (in 2.x, it returns a list), so I have to cast «a» to a list for it to have any functionality.
I would never use filter with a lambda and wrapped in list like that. Just use a list comprehension: a = [x for x in a if x != 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:] Very clever — I really like this — unfortunately it seems to be as inefficient as the most popular answer. gil’s solution is actually much faster for giant lists with only a few occurrences of the value you wish to remove.
@Larold The fastest way would be a separate question. My money are on list comprehension in the general case. This solution should perform really well if the value is frequent in the input list and list comprehension is not used. Try Pypy, numpy data in Cython. [@Gill’s answer is O(n*n) unnecessarily (compare 1e6 and 1e12 – you don’t want to risk the latter). while 1: L.remove(value) and return on ValueError might work well with a few value s or small lists in CPython.
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') 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) Of course @MadPhysicist is right, and TerminalDilettante’s version is a lot more pythonic, even if we don’t care about performance. 2013 was I just began to learn Python and nowadays I’m quite often ashamed of what I wrote back then.
Nevertheless, this changes the list in place without copying it which may be desirable in specific cases.
a.remove('b') if 'b' in a else None 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) 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.
doesn’t change my_set from above.
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] 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] 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 >>> lst = [23,34,54,45] >>> remove_element = 23 >>> if remove_element in lst: . lst.pop(lst.index(remove_element)) . 23 >>> lst [34, 54, 45] >>> Many of the answers here involve creating a new list. This involves copying all the data from the old list to the new list (except for the removed items). If your list is huge, you may not be able to afford it (or you should not want to).
In these cases, it is much faster to alter the list in place. If you have to remove more than 1 element from the list it can be tricky. Suppose you loop over the list, and remove an item, then the list changes and a standard for-loop will not take this into account. The result of the loop may not be what you expected.
a = [0, 1, 2, 3, 4, 5] for i in a: a.remove(i) # Remove all items print(a) Out: [1, 3, 5] A simple solution is to loop through the list in reverse order. In this case you get:
a = [0, 1, 2, 3, 4, 5] for i in reversed(a): a.remove(i) # Remove all items print(a) Out: [] Then, if you need to only remove elements having some specific values, you can simply put an if statement in the loop resulting in:
a = [0, 1, 2, 3, 4, 5] for i in reversed(a): if i == 2 or i == 3: # Remove all items having value 2 or 3. a.remove(i) print(a) Out: [0, 1, 4, 5] 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) 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] >>> s = [5,4,3,2,1] >>> i = s.index(3) >>> s[:i] + s[i+1:] [5, 4, 2, 1] When nums is the list and c is the value to be removed:
To remove the first occurrence of c in the list, just do:
To remove all occurrences of c from the list do:
while c in nums: nums.remove(c) Adding the exception handling would be the best practice, but I mainly wanted to show how to remove all occurrences of an element from the list.
Say for example, we want to remove all 1’s from x. This is how I would go about it:
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:
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.
In general, list-comprehension should not be used just for the side-effects, in this case — removing elements from another list. It is better to just use a regular loop and avoid the unnecessary extra list creation
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'] 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
c = len(list)-1 for element in (reversed(list)): if condition(element): del list[c] c -= 1 print(list) 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) 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() Benchmark of some of the simplest method:
import random from copy import copy sample = random.sample(range(100000), 10000) remove = random.sample(range(100000), 1000) %%timeit sample1 = copy(sample) remove1 = copy(remove) for i in reversed(sample1): if i in remove1: sample1.remove(i) # 271 ms ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) # remove all instances %%timeit sample1 = copy(sample) remove1 = copy(remove) filtered = list(filter(lambda x: x not in remove1, sample1)) # 280 ms ± 18.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) # remove all instances %%timeit sample1 = copy(sample) remove1 = copy(remove) filtered = [ele for ele in sample1 if ele not in remove1] # 293 ms ± 72.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) # remove all instances %%timeit sample1 = copy(sample) remove1 = copy(remove) for val in remove1: if val in sample1: sample1.remove(val) # 558 ms ± 40.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) # only remove first occurrence %%timeit sample1 = copy(sample) remove1 = copy(remove) for val in remove1: try: sample1.remove(val) except: pass # 609 ms ± 11.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) # only remove first occurrence