python list by value not by reference [duplicate]

Answer 1

If you want to copy a one-dimensional list, use

b = a[:]

However, if a is a 2-dimensional list, this is not going to work for you. That is, any changes in a will also be reflected in b. In that case, use

b = [[a[x][y] for y in range(len(a[0]))] for x in range(len(a))]

Answer 2

When you do b = a you simply create another pointer to the same memory of a, that's why when you append to b , a changes too.

You need to create copy of a and that's done like this:

b = a[:]

Answer 3

To copy a list you can use list(a) or a[:]. In both cases a new object is created.
These two methods, however, have limitations with collections of mutable objects as inner objects keep their references intact:

>>> a = [[1,2],[3],[4]]

>>> b = a[:]
>>> c = list(a)

>>> c[0].append(9)

>>> a
[[1, 2, 9], [3], [4]]
>>> c
[[1, 2, 9], [3], [4]]
>>> b
[[1, 2, 9], [3], [4]]
>>> 

If you want a full copy of your objects you need copy.deepcopy

>>> from copy import deepcopy
>>> a = [[1,2],[3],[4]]

>>> b = a[:]
>>> c = deepcopy(a)

>>> c[0].append(9)

>>> a
[[1, 2], [3], [4]]
>>> b
[[1, 2], [3], [4]]
>>> c
[[1, 2, 9], [3], [4]]
>>> 

Answer 4

In terms of performance my favorite answer would be:

b.extend(a)

Check how the related alternatives compare with each other in terms of performance:

In [1]: import timeit

In [2]: timeit.timeit('b.extend(a)', setup='b=[];a=range(0,10)', number=100000000)
Out[2]: 9.623248100280762

In [3]: timeit.timeit('b = a[:]', setup='b=[];a=range(0,10)', number=100000000)
Out[3]: 10.84756088256836

In [4]: timeit.timeit('b = list(a)', setup='b=[];a=range(0,10)', number=100000000)
Out[4]: 21.46313500404358

In [5]: timeit.timeit('b = [elem for elem in a]', setup='b=[];a=range(0,10)', number=100000000)
Out[5]: 66.99795293807983

In [6]: timeit.timeit('for elem in a: b.append(elem)', setup='b=[];a=range(0,10)', number=100000000)
Out[6]: 67.9775960445404

In [7]: timeit.timeit('b = deepcopy(a)', setup='from copy import deepcopy; b=[];a=range(0,10)', number=100000000)
Out[7]: 1216.1108016967773

Answer 5

As answered in the official Python FAQ:

b = a[:]

Answer 6

I would recommend the following solution:

b = []
b[:] = a

This will copy all the elements from a to b. The copy will be value copy, not reference copy.