Python output

Question

Expected output : [[8],[9]]
My output : [[8,9],[8,9]]

ans=[[]]*2
ans[0].append(8)
ans[1].append(9)
print(ans)
Why is the output wrong? What should be done to get the expected output.

Answer 1

This is because when you do the "*2", what you're duplicating is in fact the reference to the list, because lists are mutable types. Therefore, any change to one is a change to another.

Answer 2

Jay Matthews can you explain it more precisely. Why isn't my approach correct.

Answer 3

ans=[[],[]]
ans[0].append(8)
ans[1].append(9)
print(ans)

Answer 4

I'm not good at Python, but I found that ans[0] and ans[1] refer to the same object. It might be the reason of the output. I don't know why though.

https://code.sololearn.com/cnk27rmgF36s/?ref=app

Answer 5

this is working
A=[]
B=[]
C=[]
A.append(8)
B.append(9)
C.append(A)
C.append(B)
Print(C)

Answer 6

I have written a mini tutorial about that sort of problems.

After reading it, you should understand what's going wrong here.

https://code.sololearn.com/c89ejW97QsTN/?ref=app

Answer 7

to fix this, you can do:
a1=[]
ans=[a1[:], a1[:]]
ans[0]. append(8)
ans[1]. append(9)
print(ans)#[[8], [9]]

Answer 8

ans =[ [[]],[[]] ]

ans[0].append(8) is applied to both tables so you have: ans = [ [8,[]],[8,[]] ]
at the next step you replace the scond element of each table (which are tables) by 9
so you get [[8,9],[8,9]]

Answer 9

ans [[]] * 2 is basically the same as:

a1 = []
ans = [a1, a1]

To fix this, you can do:
a1 = []
ans = [a1[:], a1[:]]
ans[0].append(8)
ans[1].append(9)
print(ans) #[[8],[9]]