python – How to initialize a list of empty lists?

Question:

Using Python 2.7.12

I need to create a list like this:

lista = [[],[],[],[],.........,[]]

This list needs to have a very large number of lists within it (hence the ……..). I found the following way to do this out there:

lista = [[]]*n

Where "n" is the number of sub-lists. But with that I come to a problem. Below is an example (with a much smaller number of sub-lists for illustration).

lista = [[]]*3
lista[0].append(1)
lista[1].append(2)
print lista

The output should be:

[[1], [2], []]

But the output of this code is:

[[1, 2], [1, 2], [1, 2]]

I have no idea what it is, nor do I know/found another way to create such a list.

Answer:

The problem with the code you tried to do:

lista = [[]]*n

The object that will be repeated, [] , is initialized only once, when its reference is defined and this is used in the other positions. To demonstrate this, just scroll through the list and display the id value:

lista = [[]]*3

for l in lista:
    print id(l)

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The three values ​​will be the same as in:

47056207983464
47056207983464
47056207983464

To better demonstrate what happens, just check the opcode executed, with the help of the dis module:

>>> print dis.dis('[[]]*3')
  1           0 BUILD_LIST               0
              2 BUILD_LIST               1
              4 LOAD_CONST               0 (3)
              6 BINARY_MULTIPLY
              8 RETURN_VALUE

Note that the BUILD_LIST operation is executed twice, once for the internal list and once for the external one; then constant 3 is loaded and the values ​​are multiplied. That is, only one reference to the internal list is created, which is multiplied by 3.

To work around this problem, you can use the list comprehension :

lista = [[] for _ in xrange(n)]

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Thus, n distinct references are defined.

For the same solution in Python 3, just replace the xrange function with range .

This even happens with all mutable types in Python. For example, if you have a class Foo and want to create a list of instances, you can't do:

lista = [Foo()]*3

Making it even clearer that Foo will only be instantiated once and the created object will be multiplied by 3.