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Python Python Collections (Retired) Lists Redux Manipulating Lists

Posted by Lakshit Kerai
Lakshit Kerai
3,816 Points

Which 1 are they talking about there are two in the list?

also can we write it in one line like so

the_list = the_list.insert(0, the_list.pop(3))

lists.py 
the_list = ["a", 2, 3, 1, False, [1, 2, 3]]

# Your code goes below here
one =  the_list.pop(3)
the_list = the_list.insert(0, one)

1 Answer

andren
andren
28,558 Points
andren
andren
28,558 Points

They are talking about the first "1", the second "1" is not technically in the list by itself, it is inside another nested list within the list.

And your one line example is mostly correct, the thing is that .pop and .insert modify the list you use them on, they don't return a modified list. Therefore you don't have to assign the list variable to the result of using .pop or .insert on the list like you do. You can just use .pop and .insert without assigning it to a variable like this:

the_list.insert(0, the_list.pop(3))

That code will allow you to pass task 1.

Lakshit Kerai
Lakshit Kerai
3,816 Points

Thanks andren! You cleared a major misunderstanding that I had.