List created by multiplication behaves different

[xintx-ua]

BPO	12597
Nosy	@bitdancer, @sandrotosi

^{Note: these values reflect the state of the issue at the time it was migrated and might not reflect the current state.}

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GitHub fields:

assignee = None
closed_at = <Date 2011-07-20.17:07:10.009>
created_at = <Date 2011-07-20.16:38:58.825>
labels = ['type-bug', 'invalid']
title = 'List created by multiplication behaves different'
updated_at = <Date 2011-07-20.18:15:22.518>
user = 'https://bugs.python.org/xintx-ua'

bugs.python.org fields:

activity = <Date 2011-07-20.18:15:22.518>
actor = 'r.david.murray'
assignee = 'none'
closed = True
closed_date = <Date 2011-07-20.17:07:10.009>
closer = 'sandro.tosi'
components = []
creation = <Date 2011-07-20.16:38:58.825>
creator = 'xintx-ua'
dependencies = []
files = []
hgrepos = []
issue_num = 12597
keywords = []
message_count = 4.0
messages = ['140753', '140754', '140756', '140760']
nosy_count = 3.0
nosy_names = ['r.david.murray', 'sandro.tosi', 'xintx-ua']
pr_nums = []
priority = 'normal'
resolution = 'not a bug'
stage = 'resolved'
status = 'closed'
superseder = None
type = 'behavior'
url = 'https://bugs.python.org/issue12597'
versions = ['Python 2.6', 'Python 3.1', 'Python 2.7', 'Python 3.2', 'Python 3.3']

[xintx-ua]

Next code:

def ill(row):
  row[1]=1
list_manual=[[0,0,0],[0,0,0],[0,0,0]]
list_generated=[[0,0,0]]*3
ill(list_manual[1])
print(list_manual)
ill(list_generated[1])
print(list_generated)

Will output:

[[0, 0, 0], [0, 1, 0], [0, 0, 0]]
[[0, 1, 0], [0, 1, 0], [0, 1, 0]]

Is it a bug?

[sandrotosi]

Hi,
no it's not a bug :)

What you actually get with [[0]]*3 is a list of 3 references to the same list, [0]:

>>> a = [[0], [0], [0]]
>>> b = [[0]]*3
>>> for i in a: print(id(i))
... 
139807725184032
139807725300280
139807725300520
>>> for i in b: print(id(i))
... 
139807725324016
139807725324016
139807725324016

[xintx-ua]

So the workaround is

for i in range(len(list_generated)):
  list_generated[i]=list(tuple(list_generated[i]))

[bitdancer]

I'm not sure what your example is trying to achieve, but the list(tuple(...)) looks redundant.

Your initial example could be written:

  b = [[0] for i in range(3)]