The list.pop() method will raise an exception if list length is zero.
if len(lst) > 0
The use of a while loop inside a for loop can be a tricky thing to manage, but first we should determine if that much logic is even needed. What is the most we need to iterate the list one time?
Answer: A single loop.
As for the textual return, that would never be advisable if the return is expected to be a list. It would be feeding garbage back to the caller and raise an exception as soon as a list method was called on that object. It would certainly be messing up the data, nonetheless.
>>> a = "string"
>>> b = a.copy()
Traceback (most recent call last):
File "<pyshell#7>", line 1, in <module>
b = a.copy()
AttributeError: 'str' object has no attribute 'copy'
>>> b = list(a).copy()
['s', 't', 'r', 'i', 'n', 'g']
Here is a list of list methods and built in functions that work on lists, and perhaps other object types, as well.
Python List append()
Add Single Element to The List
Python List extend()
Add Elements of a List to Another List
Python List insert()
Inserts Element to The List
Python List remove()
Removes Element from the List
Python List index()
returns smallest index of element in list
Python List count()
returns occurrences of element in a list
Python List pop()
Removes Element at Given Index
Python List reverse()
Reverses a List
Python List sort()
sorts elements of a list
Python List copy()
Returns Shallow Copy of a List
Python List clear()
Removes all Items from the List
Checks if any Element of an Iterable is True
returns true when all elements in iterable is true
When i is a positional element and we pop that element, all the elements to the right of that get shifted. The next element to be iterated in the list will be from one more to the right that it was expected to be, hence an element is skipped.
4, 8, 10 # i is 4
8, 10 # i is 10
10 # i is None
The iterator moves forwards, same as ever.
You’re moving the values.
Also, you probably don’t want to pop at the front of the list. Do it at the back instead so that you don’t need to move all the other values every time.
(similarly you’re not going to want to copy all-but-the-first every iteration, better to find out where to slice, and then do it once)
…Except that there’s no need to modify the original, the simpler approach is to create a new list with the values to keep.
I guess a lot of people have issues with the ways to solve this exercise. I have used all methods to test my knowledge and all worked: lst[1:], del, lst.remove(), but can not get my peace with .pop() method in for loop as the previous user had issue (it worked in while loop!).
It works in the following code:
for number in lst:
if number % 2 == 0:
print(delete_starting_evens([4, 8, 10, 11, 12, 15]))
# prints: [11, 12, 15]
but here doesn’t return an empty list as it supposed to:
Thank you @mtf for your input - I have used while loop earlier and it worked for me. I was just puzzled why .pop() works in one instance and does not work in another (I would be OK to know if it would not work at all, but it does in the first lst example). When you are a beginner, such things just fry your brain
The key is to keep note of the pointer of a for loop. Think in terms of next being loaded with the index following when reaching the current element, and it does not change when the list is shifted to the left.
I have thought about this and now think I understand it.
Writing the following helped me to understand it.
Is the following a correct illustration of what happens?
lst = [4, 5, 6, 7, 8, 9]
for number in lst:
print(lst) #the order of loops for this is: #loop 1. -> 4 is popped #loop 2. -> 5 moves to the slot vacated by the 4 (as lst is currently empty) #loop 3. -> 5 is popped #loop 4. -> 6 moves to the slot vacated by the 5 #loop 5. -> 6 is popped #loop 6. -> 7 moves to the slot vacated by the 6 #you’re out of loops #lst is now [7, 8, 9]