Feb 21, 2024 · For pop. pop takes the index as a parameter and removes the element at that index. Unlike del, pop when called on list object returns the value at that index >>> a = [1, 5, 3, 4, 7, 8] >>> a.pop(3) # Will return the value at index 3 4 >>> a [1, 5, 3, 7, 8] For remove. remove takes the parameter value and remove that value from the list.

Apr 19, 2020 · whatever you pass in a.pop("here") is the index of the List so . a.pop(a[0]) # pops a[0] which is 1 therefore it becomes a.pop(1) # this will pop the first index of the list similarly in your second part . a.pop(0) # will remove the 0th index of the list.

Jun 25, 2014 · where as pop(0) means it removes the element in the index that is first element of the list. as per the Docs . list.pop([i]): Remove the item at the given position in the list, and return it. If no index is specified, a.pop() removes and returns the last item in the list.

In the below random array: a = [[1,2,3,4], [6,7,8,9]] Could you please tell me how to remove element at a specific position.

Feb 8, 2018 · Since you pop from the end, you won't get an IndexError, because a list_reverseiterator starts indexing at alist[len(alist) - 1] until it gets to alist[0]. I probably wouldn't feel comfortable relying ont this behavior, but it is documented, and it does work:

Aug 23, 2019 · It looks like you want to remove and process the 200th character of large_string until pop throws "IndexError: pop index out of range" – Jacob Wan Commented Jan 17, 2015 at 5:07

Jun 19, 2020 · If i use .pop() within a conditional statement, does the pop still execute even if the condition is not met? e.g. if something != stack.pop(): return False else: return True Say the condition is not met and does not return False, then is the last element of [stack] is still popped? Edit: Thanks all for clarifying this.

Jan 2, 2010 · In other words, the pop function takes at least two arguments. The first two get assigned the names self and key ; and the rest are stuffed into a tuple called args . What's happening on the next line when *args is passed along in the call to self.data.pop is the inverse of this - the tuple *args is expanded to of positional parameters which ...

Oct 12, 2008 · To clarify the N in the Big-O notation is NOT the index of the element being returned, but a function bounding the running time of the algorithm with O(1) being constant time - i.e., it doesn't depend on the size of the list. O(N) means that the bounding function is some constant times the size of the list, f(n) = c*n + b. –

You need to call .pop() on the inner list then: fruit_list[0].pop() Your outer fruit_list contains more list objects. If you want to pop an item from one of those lists, call .pop() directly on such a list. A quick demo: