Python Forum

Today I meet this exercise:

What this code will output and why?

x = [[]]*3
x[0].append('a')
x[1].append('b')
x[2].append('c')
x[0]=['d']

print(x)

I expected to see 'd b c', but whet i run this program i was surprised. The output is [['d'], ['a', 'b', 'c'], ['a', 'b', 'c']].
Why is this code works this way?

Brcause you are using a 2 dimensional list. Play around with it for a bit and I'm sure you will understand! :)

a = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(a[0])
print(a[0][1])

Try this code and experiment with it by printing, removing, appending etc

Output:
[['d'], ['a', 'b', 'c'], ['a', 'b', 'c']]

x = [[]]*3 this copy 3 list to same memory adress.

>>> x = [[]]*3
>>> [id(i) for i in x]
[57560800, 57560800, 57560800]

Do it like this.

>>> x = [[] for i in range(3)]
>>> [id(i) for i in x]
[66352912, 66412416, 65766344]

Then it look like this.

x = [[] for i in range(3)]
x[0].append('a')
x[1].append('b')
x[2].append('c')
x[0] = ['d']

print(x)

Output:
[['d'], ['b'], ['c']]

Thank you for detailed answer. I will experiment with this python feature.

Valgard

BerlingSwe

dwiga

snippsat

Valgard