My answer is not exactly to your question but after you read this, I hope you can decide which type you need to choose for your needs.
Python’s lists are variable-length arrays, not Lisp-style linked lists. The implementation uses a contiguous array of references to other objects, and keeps a pointer to this array.
This makes indexing a list a[i] an operation whose cost is independent of the size of the list or the value of the index.
When items are appended or inserted, the array of references is resized. Some algorithm is applied to improve the performance of appending items repeatedly; when the array must be grown, some extra space is allocated so the next few times don’t require an actual resize i.e over-allocation. More Information
Removing vs Pop vs Delete:
At first glance it looks like all of them are doing the same thing.
Under the hood its behaving different.
removing : remove an element from the list by iterating from 0 index till the first match of the element is found. taking more time to iterate if the element is at the end.
pop : removing element from the list by using the index. taking less time.
del : is a python statement that removes a name from a namespace, or an item from a dictionary, or an item from a list by using the index.
REMOVE:
- it removes the first occurence of value.
- raises ValueError if the value is not present.
- it takes only one argument, so you can't remove multiple value in one shot.
POP:
- remove and return item at index (default last).
- Raises IndexError if list is empty or index is out of range.
- it takes only one argument, so you can't remove multiple value in one shot.
DEL:
- remove the item at index and return nothing.
- it can remove slices from a list or can clear the whole list.
Benchmark:
Worst case : deleting from the end of the list.
yopy:-> python -m timeit "x=range(1000)" "x.pop(999)"
100000 loops, best of 3: 10 usec per loop
yopy:-> python -m timeit "x=range(1000)" "x.remove(999)"
10000 loops, best of 3: 31.3 usec per loop
yopy:-> python -m timeit "x=range(1000)" "del x[999]"
100000 loops, best of 3: 9.86 usec per loop
yopy:->
Best case: begining of the list.
yopy:-> python -m timeit "x=range(1000)" "x.remove(1)"
100000 loops, best of 3: 10.3 usec per loop
yopy:-> python -m timeit "x=range(1000)" "x.pop(1)"
100000 loops, best of 3: 10.4 usec per loop
yopy:-> python -m timeit "x=range(1000)" "del x[1]"
100000 loops, best of 3: 10.4 usec per loop
yopy:->
Point to be noted:
if array grows or shrinks in the middle
- Realloc still depends on total length.
- But, All the trailing elements have to be copied
So, now I hope you can decide what you need to choose for your needs.
Answer from James Sapam on Stack OverflowMy answer is not exactly to your question but after you read this, I hope you can decide which type you need to choose for your needs.
Python’s lists are variable-length arrays, not Lisp-style linked lists. The implementation uses a contiguous array of references to other objects, and keeps a pointer to this array.
This makes indexing a list a[i] an operation whose cost is independent of the size of the list or the value of the index.
When items are appended or inserted, the array of references is resized. Some algorithm is applied to improve the performance of appending items repeatedly; when the array must be grown, some extra space is allocated so the next few times don’t require an actual resize i.e over-allocation. More Information
Removing vs Pop vs Delete:
At first glance it looks like all of them are doing the same thing.
Under the hood its behaving different.
removing : remove an element from the list by iterating from 0 index till the first match of the element is found. taking more time to iterate if the element is at the end.
pop : removing element from the list by using the index. taking less time.
del : is a python statement that removes a name from a namespace, or an item from a dictionary, or an item from a list by using the index.
REMOVE:
- it removes the first occurence of value.
- raises ValueError if the value is not present.
- it takes only one argument, so you can't remove multiple value in one shot.
POP:
- remove and return item at index (default last).
- Raises IndexError if list is empty or index is out of range.
- it takes only one argument, so you can't remove multiple value in one shot.
DEL:
- remove the item at index and return nothing.
- it can remove slices from a list or can clear the whole list.
Benchmark:
Worst case : deleting from the end of the list.
yopy:-> python -m timeit "x=range(1000)" "x.pop(999)"
100000 loops, best of 3: 10 usec per loop
yopy:-> python -m timeit "x=range(1000)" "x.remove(999)"
10000 loops, best of 3: 31.3 usec per loop
yopy:-> python -m timeit "x=range(1000)" "del x[999]"
100000 loops, best of 3: 9.86 usec per loop
yopy:->
Best case: begining of the list.
yopy:-> python -m timeit "x=range(1000)" "x.remove(1)"
100000 loops, best of 3: 10.3 usec per loop
yopy:-> python -m timeit "x=range(1000)" "x.pop(1)"
100000 loops, best of 3: 10.4 usec per loop
yopy:-> python -m timeit "x=range(1000)" "del x[1]"
100000 loops, best of 3: 10.4 usec per loop
yopy:->
Point to be noted:
if array grows or shrinks in the middle
- Realloc still depends on total length.
- But, All the trailing elements have to be copied
So, now I hope you can decide what you need to choose for your needs.
Use a list comprehension:
Scenario 1:
[item for item in my_list if 1 <= item <=5 ]
Scenario 2:
to_be_removed = {'a', '1', 2}
[item for item in my_list if item not in to_be_removed ]
Scenario 3:
[item for item in my_list if some_condition()]
Scenario 4(Nested list comprehension):
[[item for item in seq if some_condition] for seq in my_list]
Note that if you want to remove just one item then list.remove, list.pop and del are definitely going to be very fast, but using these methods while iterating over the the list can result in unexpected output.
Related: Loop “Forgets” to Remove Some Items
python - How to remove an element from a list by index - Stack Overflow
Removing a specific element from a list
[Question] Removing objects from a list while iterating over it
Removing comma from a tuple
Use del and specify the index of the element you want to delete:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> del a[-1]
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8]
Also supports slices:
>>> del a[2:4]
>>> a
[0, 1, 4, 5, 6, 7, 8, 9]
Here is the section from the tutorial.
You probably want pop:
a = ['a', 'b', 'c', 'd']
a.pop(1)
# now a is ['a', 'c', 'd']
By default, pop without any arguments removes the last item:
a = ['a', 'b', 'c', 'd']
a.pop()
# now a is ['a', 'b', 'c']
I have a function that receives two lists. The goal is for the function to return lst1 without the elements on lst2, however I'm having trouble with that. What can I do?
I do have to say that I cannot use any while or for loop, only recursion (that's the hard part)
def minus(lst1,lst2):
element = lst2.pop()
if element is in lst1:
#how can I find the position of the elements in lst1 that is equal to "element"?
else:
if len(lst2) == 0:
return lst1
else:
return minus(lst1,lst2)