NameError: name 'end' is not defined

Question

lst=[1,2,3,'b',4,5,6,'c','b',4,5,6,2,'c',9,0,1]

def delElement(E1,E2):
    global end
    for i,ele in enumerate(lst):
        if ele==E1:
            start=i
            break
        elif ele== E2:
            end=i
            break
    del lst[start:end+1]
    return(lst)
delElement('b','c')

If you don't hit that line in an elif block, end is never assigned. When you hit an `E1`, you break out of your loop, so you never hit an `E2` — khelwood, Aug 24 '20 at 18:17

score 1 · Answer 1 · answered Aug 24 '20 at 18:25

Even though you have defined that you want to use the global variable end on line 4, this variable does not have any value yet.

Since your for loop breaks after finding the value b in your list, the value end will never be set. Instead, only the value of the variable start will be set, since the first if-statement will evaluate to True first.

After the loop you are trying to delete a range of list items by referencing the variable end, however, since this has not been given a value yet, it will throw an error.

A solution could be to also set the end variable in the first if-statement, depending on your needs. Also, if possible, I would suggest avoiding using global variables like this, as explained in this Stack Overflow post.

rblais · Accepted Answer · 2020-08-25T03:43:12.153

in your current code, at the moment your for loop detects E1, it breaks and go right away to del lst[start:end+1] without defining end. No sure what you want to achieve but keeping the same structure of code, just remove the first break:

def delElement(e1, e2):
        for i, ele in enumerate(lst):
            if ele == e1:
                start = i
            elif ele == e2:
                end = i
                break
        del lst[start:end + 1]
        return lst  

lst = [1, 2, 3, 'b', 4, 5, 6, 'c', 'b', 4, 5, 6, 2, 'c', 9, 0, 1]   
delElement('b', 'c')

I can see many potential error on the results depending of what you want to achieve exactly. Don't hesitate to add details for better answer...

EDIT: An ugly way but workable if your expected result is [1, 2, 3, 9, 0, 1]:

def delElement(elementA, elementB, lst):

    new_lst = []
    feed_new_lst = True  # New list is fed when this is true

    for element in lst:

        if element == elementA:
            feed_new_lst = False  # Stop feeding the new list
            continue

        if feed_new_lst:
            new_lst.append(element)

        if element == elementB:
            feed_new_lst = True  # Continue do feed the new list

    return new_lst

lst = [1, 2, 3, 'b', 4, 5, 6, 'c', 'b', 4, 5, 6, 2, 'c', 9, 0, 1]
lst2 = ['b', 2, 3, 'c', 4, 'b', 6, 'c', 4, 5, 6, 'b', 9, 0, 'c']

print(delElement('b', 'c', lst))
# [1, 2, 3, 9, 0, 1]

print(delElement('b', 'c', lst2))
# [4, 4, 5, 6]

so your expected result (based on your list) would be `[1, 2, 3, 9, 0, 1]` ? — rblais, Aug 25 '20 at 03:22

NameError: name 'end' is not defined

2 Answers2