Multi-dimensional list initialization trouble

[jonkje]

Hello I found this very strange; is it a bug, is it a "feature", am I
being naughty or what?

foo = [[0, 0], [0, 0]]
baz = [ [0]*2 ] * 2
foo [[0, 0], [0, 0]] baz [[0, 0], [0, 0]] foo[0][0]=1
baz[0][0]=1
foo [[1, 0], [0, 0]] baz

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

Why on earth does foo and baz behave differently??

Btw.:
Python 2.4.1 (#1, Apr 10 2005, 22:30:36)
[GCC 3.3.5] on linux2

--- Jon Øyvind

[Scott David Daniels]

jo****@gmail.com wrote:

Hello I found this very strange; is it a bug, is it a "feature", am I
being naughty or what?

foo = [[0, 0], [0, 0]]
baz = [ [0]*2 ] * 2...
Why on earth does foo and baz behave differently??

This is a frequently made mistake.
try also:

bumble = [[0]*2 for 0 in xrange(2)]
Think hard about why that might be.
Then try:
[id(x) for x in foo]
[id(x) for x in baz]
[id(x) for x in bumble]

Now check your upper scalp for lightbulbs.

--Scott David Daniels
sc***********@acm.org

[Fredrik Lundh]

jo****@gmail.com wrote:

Hello I found this very strange; is it a bug, is it a "feature", am I
being naughty or what?

the repeat operator (*) creates a new list with references to the same
inner objects, so you end up with a list containing multiple references
to the same list. also see:

http://pyfaq.infogami.com/how-do-i-c...mensional-list

</F>

[trebucket]

An expression like this creates a list of integers:

[0] * 2 [0, 0]

But an expression like this creates list of references to the list
named `foo': foo = [0, 0]
baz = [foo] * 2 [foo, foo]

So, setting baz[0][0] = 1, is really setting foo[0] = 1. There is only
one instance of foo, but you have multiple references.

Try a list comprehension to get the result you want: foo = [[0 for ii in range(2)] for jj in range(2)]