How to check how many of the last elements in a list are equal in python?

Question

I am appending integers to a list one by one (using a loop) as so:

A.append(x) where x is an integer, which eventually gives for example:

A = [4, 8, 2, 4, 3, 7, 7, 7]

During every loop, ie just after each integer is added to the end of the array, I would like to check whether the same integer has been added a certain number of times (say, 3, in the example below) and throw an exception if so.

Pseudo code:

if somewayofcheckingA == 3:
    raise Exception("Lots of the latest integers are similar")

I could just do the below, but if I wanted to check for say, 100 repeats, then obviously the code would become a mess.

if A[-1] == A[-2] and A[-2] == A[-3]:
    raise Exception("Lots of the latest integers are similar")

Thanks!

Answer 1

Passing a list to set() will return a set with all unique values in the list. You can use slice notation to get a list of the last n values using the following

n = 3
if len(A) >= n and len(set(A[-n:])) == 1:
    raise Exception("Lots of the latest integers are similar")

Answer 2

if you just want to check just last 3 then this would do it.

limit = 3
if len(set(A[-limit:])) == 1:
    raise Exception("Lots of the latest integers are similar")

Answer 3

You could use collections.Counter() to count how many times the last element appears. For example:

occurrences = collections.Counter(A)
if occurrences[A[-1]] >= 3:
   raise Exception("Lots of the latest integers are similar")

Or an even simpler way

if A.count(A[-1]) >= 3:
   raise Exception("Lots of the latest integers are similar")

**THIS CODE CHECK FOR OCCURRENCES OF THE LAST ELEMENT AT ANY OTHER INDEX OF THE LIST

Answer 4

I did this for you. And I think this is some kind of pythonic.

class CustomList(list):
    def __init__(self, seq=()):
        self.last_equal_items = []
        super().__init__(seq)

    def append(self, some_item):
        if self.last_equal_items and some_item != self.last_equal_items[-1]:
            self.last_equal_items = []
        self.last_equal_items.append(some_item)
        if len(self.last_equal_items) >= 3:
            raise ValueError("Last equal items larger that 3")
        else:
            super(CustomList, self).append(some_item)

test = CustomList([])
test.append(1)
test.append(1)
test.append(1)
test.append(2)
test.append(1)
print(test)

You can just use CustomList like list . And it will alert when you insert the third equal value.

Answer 5

lists = [1,4,3,3];
def somewayofcheckingA(lists, a):
    lists.reverse()
    i = 0
    k = lists[0]
    count = 0
    while i < a:
        if(lists[i] == k):
            count= count+1
        i = i+1
    return count
        
print(test(lists, 3))

where lists is the list and a is the number of times you want to check

This answer is easy to follow and utilizes basic looping and conditional statements, which I think you should master before venturing into the other proposed solutions, which are more pythonic, but where you might get lost in the mix.