Why appears few lists after .append?

Question

Here is the python code:

a, b = [], []
for x in range(4):
a.append(x)
b.append(a)
print(b)

Output is:
[[0]]
[[0, 1], [0, 1]]
[[0, 1, 2], [0, 1, 2], [0, 1, 2]]
[[0, 1, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3], [0, 1, 2, 3]]

I suppose the final output like [0, [0, 1], [0, 1, 2], [0, 1, 2, 3]] and don’t understand why there are few equal lists. Could anyone explain, please!

Answer 1

If you append an object to a list, you append no copy but (a reference to) that actual object.

So if you first append a to another list and then append something to a, the a in the list changes, too: because it's the very same a!

And if you append a many times, the name (reference to) a will be in it many times; this just means that whenever your list finds an a, it will look up that very same object a again and again.

If you don't want this behaviour, you have to append an actual copy of a instead of a:

b.append(a[:])

Answer 2

It's happening because the variable "a" is a reference to a list, so each append basically appends the reference to "a" which is constantly changing

edit:
HonFu beat me to it 😅
and explained it more thoroughly

Answer 3

I’ll try a slightly different explanation to that of the other commenters thus far, just on case you find it helpful.

Your empty list b has the list a appended to it 4 times. This means that your list b, at the end of the code, is [a,a,a,a]. Since a, at the end, is [0,1,2,3], that is why it ends up as it does.

In order to get the result you were expecting, simply change line 4 to b.append(list(a)). This appends the list that a is at that moment, rather than a itself.

Hope that helps.

Answer 4

HonFu thank you very match. Your explanation very useful and clearly.
⚡Prometheus ⚡ , Burey , Russ also many thank’s

Answer 5

😎😂

Answer 6

You're appending the same reference again and again.

To solve this, do:
b=[list(range(0,n)) for n in range(4)]