Why does appending to one list also append to all other lists in my list of lists?

Suppose I do the following:

>>> l = [[]]*2
>>> l
[[], []]
>>> l[0].append(1)
>>> l
[[1], [1]]

Why does 1 get appended to both lists?

Answers:

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Method 1

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

Method 2

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.

Method 3

Showcasing the difference with memory layout:

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

Method 4

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

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0