Why does assiging a list element to a variable work in a different manner here?

Question

I have this piece of code:

lst = [[1,1], [2,1],[3,1]]
n = len(lst)
head = lst[n - 1]

head[0] += 1

lst.append(head)
del lst[0]

print(lst)

And I'm expecting this code to print: [[2,1], [3,1], [4,1]] But it's printing: [[2, 1], [4, 1], [4, 1]]. I don't understand why. Please help me.

Answer 1

When you write head = lst[n - 1] , that sets head to the last element of lst by reference. This means that the pair with values [3,1] is shared by both variables. If you want to not alter the pair in the original list, make sure that head copies the data.

head = lst[n - 1].copy()

Answer 2

A quick fix for your need is

inc_lst = [ [ x[0]+1 , x[1] ] for x in lst]

Hopefully it will help

Answer 3

When you use

head = lst[n - 1]

you got reference [3,1] as head, when you execute

head[0] += 1

you changed [[1,1], [2,1], [3,1]] to [[1,1],[2,1], [4,1]]

When you make append, lst got another copy of head.

Therefore, after del lst[0], you got the result [[2,1], [4,1], [4,1]]

Answer 4

1. lst = [[1,1], [2,1],[3,1]]
2. n = len(lst) - n will be equal 3
3. head = lst[n - 1] => head = lst[2] => head = [3, 1]
4. head[0] += 1 it change head, head = [4, 1]
5. lst.append(head) , remember lst = [[1, 1], [2, 1], [4, 1]] and head = [4, 1] => [[1, 1], [2, 1], [4, 1], [4, 1]]
6. del lst[0] => [[2, 1], [4, 1], [4, 1]]

Anyway, when head was [3, 1] then you change head and it automatically change lst[n - 1]

Also, please try using http://www.pythontutor.com/visualize.html

A very useful tool if need to understand how code works step by step.

Answer 5

It's actually a very common problem among beginners in programming (such as myself). The other questions already answered the question, but I just wanted to mention that this is a case of "value semantics vs reference semantics". It can be usefull to shortly read up on this topic in order to avoid this kind of mistakes in the future.