Python 3 , list of lists initialisation

Question

take a look at this code : https://code.sololearn.com/c076UQaTbPk5
Can you help me figure out why the output differs?

Answer 1

This resembles your situation:
https://www.sololearn.com/Discuss/1327452/?ref=app

Basically when you [...]*3 all the lists inside "a" share the same space in memory so they're the same thing, you can check that by comparing id(a[0]), id(a[1]), id(a[2]) (which I shamefully didn't myself)

So when you append '1', '2', ... to each element of "a" they also append to the other lists inside "a", and there's your unexpected behaviour.

Also, note:
>>> a = []
>>> b = a
>>> a.append(1)
>>> b.append(2)
>>> a
[1, 2]
>>> b
[1, 2]

Answer 2

Thank you very much Zuke , your explanation was very clear :) and thanks to all you guys answering

Answer 3

it looks like this statement
a = [[]]*5
use the same object 5 times.
just do this test

print(a[0] is a[1])#output True

but
b = [[]for _ in range(5)]
creates 5 different objects.

print(b[0] is b[1]) #output False

Answer 4

a = [list()]*5 # this behaves strangely when appended
b = [[]for _ in range(5)] # this behaves as expected
print(a)
print(b)
print(a == b) # Till here everything is equal

# a == b because a and b contains both 5 empty lists, but a containd 5 times the same list object (same reference in memory), while b contains 5 different lists (different references in memory).

c = []
for i in range(5):
c.append("12345")
for j in range(5):
a[j].append(c[i][j])
# here you access the same list at each iteration (a[n] == a[m])

c = []
for i in range(5):
c.append("12345")
for j in range(5):
b[j].append(c[i][j])
# here you access a different list at each iteration

print(a)
print(b)
print(a == b) # different result

Answer 5

Game Play hi(?)

Answer 6

Hello