## Removing third element from list until there are less than 3 numbers

I have a problem. I need to create a list . And in every iteration i need to remove the third element and print the list without the deleted element.

The thing is. I'm trying to do an algorithm of deleting the third element without remove function or other inbuilt list functions. In my code I covered the following possibilities. If my list has less than 3 elements I print a message saying the list is short. If my list has 3 elements in it I will assign the third element the value of the second element and so on. My problem is when my list has more than 3 elements.

v=[] # list of numbers the user inputs two=[] #list with only 2 elements vector=[] # list with third element deleted when len(v)>3 def create_array(): n=int(input('Digit total elements in dynamic array - ')) for i in range(0,n): element=int(input('Digit element to array - ')) v.append(element) return v print(create_array()) def remove_third_element(): for j in range(0,len(v)): if len(v)<3: # if number of elements in list < 3 there is no element to delete print('There is no element do delete! ') return 0 elif len(v)==3: v[2]==v[1] and v[1]==v[0] two=[v[0],v[1]] return two else: v[0]==v[1] and v[1]==v[2] print(remove_third_element())

elif len(v) > 3: ret = []; for i in range(len(v)): if(i != 2) ret.append(v[i]); return ret

should do the trick

By the way, with this method you can remove you elif len(v) == 3

Also your code :

elif len(v)==3: v[2]==v[1] and v[1]==v[0] two=[v[0],v[1]] return two

won't work because '==' is used as condition in python so it will return a boolean and not assign value. go for

v[2] = v[1] v[1] = v[0]

instead

**Python,** Given a list of numbers, Your task is to remove and print every third number The first third element encountered is 3, after 3 we start the count� We take LinkedList of Integers as an argument, our base case is if list.size() is less than 3 (since we need to remove every 3rd element), because if list has less than 3 elements we can't remove the 3rd one (obviously), then we loop from element at index 2 (third element) and remove it, we also increase our i for 2 every time loop ends so we keep deleting every 3rd element until we reach the end of the list. Then we call our recursive method again and we pass our LinkedList with all the

Here's a Pythonic way of making a new list without the original list's third element.

new_list = old_list[:2] + old_list[3:]

`old_list[:2]`

is shorthand for "until the 2-th index" (so we'll get index 0 and 1) of the `old_list`

.

`old_list[3:]`

is shorthand for, "from 3rd index 'til the end" (so index 3, 4, etc.).

Both return lists; in python, if you add lists, concatenation actually happens.

As an example, if `old_list = [1,2,3,4,5]`

, then `new_list[:2]`

will be `[1,2]`

and `new_list[3:]`

will be `[4,5]`

. So combining that will be `[1,2,4,5]`

.

**Python: Remove and print every third number from a list of numbers ,** Python: Remove and print every third number from a list of numbers until the list becomes empty def remove_nums(int_list): #list starts with 0 index position = 3 - 1 idx = 0 len_list Next: Write a Python program to find unique triplets whose three elements gives the sum of zero from an array of n integers. Approach The index of the list starts from 0 and the first third element will be at position 2. Traverse till the list becomes empty and each time find the index of the next third element and print its corresponding value. After printing reduce the length of the list.

Notice that this statement: `v[2]==v[1] and v[1]==v[0]`

won't assign values!
Operation `==`

return boolean.

Lets say your `v`

looks like this: `v = [1, 1, 3]`

.
Then `v[2]==v[1] and v[1]==v[0]`

gives you result: `False and True`

, and this gives you reulst `False`

. You can check it if you print this `print(v[2]==v[1] and v[1]==v[0])`

.

If you want to assign values, you can use a statement like this: `v[2], v[1] = v[1], v[0]`

.

**Python program to Remove and print every third from list until it ,** Step 1: The index of the list starts from 0 and the first third element will be at Step 3: Traverse till the list becomes empty and each time find the index of the the size of the array ::")) print("Enter the INTEGER number") for i in� If I have a list like: list = {23,21,18,15,13,12,10,9,8,7,7,5} How can I remove numbers from that list so that no number less than 13 is on it without hardcoding, so if the list were different it

def main(): big_list = [ x for x in range(20) ] while len(big_list) > 3: big_list = big_list[:2] + big_list[3:] print(big_list)

**Remove All Occurrences of a Specific Value from a List,** The problem is in the 3rd line: we call List.remove(int), which treats Calling List. remove() shifts all elements after the removed one to smaller While these solutions produce short and clean code, they still have 3. Removing Until the List Changes. List.remove(E element) has a list.remove(number);. You have to pass only the single element which you want to delete. The function requires only one element and can delete the only single element from the list. Use the list variable concatenated with the remove () to remove the list element. See the example given below remove the single element from the list.

**Remove Elements From Lists | Python List remove() Method,** Arrays contain elements of the same data types, whereas Python lists can A single list may contain data types like strings, integers, floating point numbers etc. There are three ways in which you can Remove elements from List: to remove elements from a range of indices, i.e. from index position 1 till� In Python, list's methods clear(), pop(), and remove() are used to remove items (elements) from a list. It is also possible to delete items using del statement by specifying a position or range with an index or slice.Remove all items: clear() Remove an item by index and get its value: pop() Remove a

Given an array, find an element before which all elements are smaller than it, and after which all are greater than it. Return the index of the element if there is such an element, otherwise, return -1.

Python Basic - 1: Exercise-3 with Solution. Write a Python program to remove and print every third number from a list of numbers until the list becomes empty. Pictorial Presentation: Sample Solution: Python Code :

##### Comments

- what is the ret for ?
- ret is an array, the same as the "two" you return when the length is 3
- so you will apend every element except the element in position 2 . I think i get it thanks
- if i put v[2]=v[1]
- and v[1]=v[0] it return equal values for all positions
- but that type of statement is not for switching positions?
- Yes, it is. If you don't want to switch position, then I am not sure what you want to do. Could you explain what you wanted to achieve with this statement?
- i wanted to assign the value in position two to position one
- Basically
`v[2], v[1] = v[1], v[0]`

do same what:`v[2] = v[1], v[1] = v[0]`

. Example:`v = [3, 5, 7]`

, then after operation`v[2], v[1] = v[1], v[0]`

you have:`v = [3, 3, 5]`

.