How to force a list to a fixed size?

In Python 3, I want to create a list that will contain the last 5 variables entered into it.

Here is an example:

>>>l = []
>>>l.append('apple')
>>>l.append('orange')
>>>l.append('grape')
>>>l.append('banana')
>>>l.append('mango')
>>>print(l)
['apple','orange','grape','banana','mango']
>>>l.append('kiwi')
>>>print(l)
['orange','grape','banana','mango','kiwi'] #only 5 items in list

So, in python, is there any way to achieve what is demonstrated above? The variable does not need to be a list, I just used it as an example.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

You might want to use a collections.deque object with the maxlen constructor argument instead:

>>>l = collections.deque(maxlen=5)
>>>l.append('apple')
>>>l.append('orange')
>>>l.append('grape')
>>>l.append('banana')
>>>l.append('mango')
>>>print(l)
deque(['apple','orange','grape','banana','mango'], maxlen=5)
>>>l.append('kiwi')
>>>print(l)
deque(['orange','grape','banana','mango','kiwi'], maxlen=5) #only 5 items in list

Method 2

I ran into this same issue… maxlen=5 from deque was NOT a supported option due to access speed / reliability issues.

SIMPLE Solution:

l = []
l.append(x)                         # add 'x' to right side of list
l = l[-5:]                          # maxlen=5

After you append, just redefine ‘l’ as the most recent five elements of ‘l’.

print(l)

Call it Done.

For your purposes you could stop right there… but I needed a popleft(). Whereas pop() removes an item from the right where it was just appended… pop(0) removes it from the left:

if len(l) == 5:                     # if the length of list 'l' has reached 5 
    right_in_left_out = l.pop(0)    # l.popleft()
else:                               #
    right_in_left_out = None        # return 'None' if not fully populated

Hat tip to James at Tradewave.net

No need for class functions or deque.

Further… to append left and pop right:

l = []
l.insert(0, x)                      # l.appendleft(x)
l = l[-5:]                          # maxlen=5

Would be your appendleft() equivalent should you want to front load your list without using deque

Finally, if you choose to append from the left…

if len(l) == 5:                     # if the length of list 'l' has reached 5 
    left_in_right_out = l.pop()     # pop() from right side
else:                               #
    left_in_right_out = None        # return 'None' if not fully populated

Method 3

You could subclass list

>>> class L(list):
...     def append(self, item):
...         list.append(self, item)
...         if len(self) > 5: del self[0]
... 
>>> l = L()
>>> l.append('apple')
>>> l.append('orange')
>>> l.append('grape')
>>> l.append('banana')
>>> l.append('mango')
>>> print(l)
['apple', 'orange', 'grape', 'banana', 'mango']
>>> l.append('kiwi')
>>> print(l)
['orange', 'grape', 'banana', 'mango', 'kiwi']
>>>

Method 4

deque is slow for random access and does not support slicing. Following on gnibbler’s suggestion, I put together a complete list subclass.

However, it is designed to “roll” right-to-left only. For example, insert() on a “full” list will have no effect.

class LimitedList(list):

    # Read-only
    @property
    def maxLen(self):
        return self._maxLen

    def __init__(self, *args, **kwargs):
        self._maxLen = kwargs.pop("maxLen")
        list.__init__(self, *args, **kwargs)

    def _truncate(self):
        """Called by various methods to reinforce the maximum length."""
        dif = len(self)-self._maxLen
        if dif > 0:
            self[:dif]=[]

    def append(self, x):
        list.append(self, x)
        self._truncate()

    def insert(self, *args):
        list.insert(self, *args)
        self._truncate()

    def extend(self, x):
        list.extend(self, x)
        self._truncate()

    def __setitem__(self, *args):
        list.__setitem__(self, *args)
        self._truncate()

    def __setslice__(self, *args):
        list.__setslice__(self, *args)
        self._truncate()

Method 5

You could use a capped collection in PyMongo – it’s overkill, but it does the job nicely:

import pymongo

#create collection
db.createCollection("my_capped_list",{capped:True, max:5})

#do inserts ...

#Read list
l = list(db.my_capped_list.find())

Hence any time you call my_capped_list, you will retrieve the last 5 elements inserted.

Method 6

Most often when you need such a kind of facility, you would write a function which takes the list and then returns the last five elements.

>>> l = range(10)
>>> l[-5:]

But if you really want a custom list, having a cap on five elements, you can override the built-in list and it’s methods, you would do something like this, for all it’s methods.

class fivelist(list):
    def __init__(self, items):
        list.__init__(self, items[-5:])

    def insert(self, i, x):
        list.insert(self, i, x)
        return self[-5:]

    def __getitem__(self, i):
        if i > 4:
           raise IndexError
        return list.__getitem__(self, i)

    def __setitem__(self, i, x):
        if 0<= i <= 4:
          return list.__setitem__(self, i, x)
        else:
          raise IndexError

Method 7

It can be as simple as the below solution

lst = []
arr_size = int(input("Enter the array size "))
while len(lst) != arr_size:
    arr_elem= int(input("Enter the array element "))
    lst.append(arr_elem)

sum_of_elements = sum(lst)

print("Sum is {0}".format(sum_of_elements))

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0