Ndtyjky

Don't really know how to formulate the question...

Suppose I do the following:

    >>> l = []*2

    >>> l

    [, ]

    >>> l[0].append(1)

    >>> l

    [[1], [1]]

Why does 1 gets appended to both lists?

[]*2 is a list of two references to the same list. You are appending to it and then seeing it twice.

Because there is really only one list. Consider this:

>>> l = []

>>> l2 = l*2

>>> l2[0] is l[0]

True

>>> l2[1] is l[0]

True

*2 performed on a list does not copy the list but return a list of length 2 filled with the same reference.

What you probably wanted was this:

>>> l = [ for _ in xrange(2)]

As @Asterisk mentions in a comment, the same behaviour is exposed by all common collections. As a rule of thumb it is therefore best to only use multiplication on immutable types with value-semantics.

Here's how i initialize a list of lists. Rows vary slowest.

nrows = 3; ncols = 5



l_of_ls = [[0]*ncols for i in range(nrows )]



for rix, r in enumerate(l_of_ls):

    for cix, c in enumerate(r):

        print rix, cix, 'val = ',c

RESULT

0 0 val =  0

0 1 val =  0

0 2 val =  0

0 3 val =  0

0 4 val =  0

1 0 val =  0

1 1 val =  0

1 2 val =  0

1 3 val =  0

1 4 val =  0

2 0 val =  0

2 1 val =  0

2 2 val =  0

2 3 val =  0

2 4 val =  0 

Also Worth Noting for Indexing Purposes

for rix in range(nrows):

    for cix in range(ncols):

        print l_of_ls[rix][cix],

    print

RESULT

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0  

Showcasing the difference with memory layout:

listOfLists = [] * 3

listOfListsRange = [ for i in range(0, 3)]