Different behavior between the normal python list and the numpy array object

Question

Straight to the point. The next statements have the effect of swapping the contents of the two list elements of the python 2D-list:

a = [[1,2,3],
     [4,5,6]]
b = [[7,8,9],
     [10,11,12]]

tmp = a[1]
a[1] = b[1]
b[1] = tmp

output:

a = [[1,2,3],
     [10,11,12]]

b = [[7,8,9],
     [4,5,6]]

For the numpy array type this would not have happened.

a = np.array([[1,2,3],
              [4,5,6]])
b = np.array([[7,8,9],
              [10,11,12]])

tmp = a[1]
a[1] = b[1]
b[1] = tmp

output:

a = array([[ 1,  2,  3],
           [10, 11, 12]])

b = array([[ 7,  8,  9],
           [10, 11, 12]])

Why is that so?

Mike Müller · Accepted Answer · 2018-02-22 05:46:51Z

3

The list of lists are nested objects. The NumPy array is not nested but 2D. There is no notion of 2D with lists. So you have a list in a list. Whereas in NumPy it is just one object.

NumPy always returns a view when indexing and the result of indexing is another array. You need to make explicit copies:

tmp = a[1].copy()
a[1] = b[1].copy()
b[1] = tmp

to get it swapped:

>>> a
array([[ 1,  2,  3],
       [10, 11, 12]])
>>> b
array([[7, 8, 9],
       [4, 5, 6]])

Rule of thumb:

Indexing of list in list == reference to sub-list
Indexing of NumPy 2D array along first dimension == view, i.e. shared data

edited Feb 22, 2018 at 5:46

answered Feb 21, 2018 at 23:55

Mike Müller

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4 Comments

user2357112 Over a year ago

Slicing a list produces a shallow copy, but indexing does something even shallower, sometimes referred to as a "reference copy" or "not a copy".

Mike Müller Over a year ago

@user2357112 You are right. Improved my answer accordingly.

Lockon2000 Over a year ago

@user2357112 could you please elaborate on that a little more. As I understand this, a reference copy is some kind of a shallow copy (a shallow copy that can and will change the actual object if changed, i.e. not a lazy copy). So why do you make it sound like they are two different things? Am I missing something?

user2357112 Over a year ago

@Lockon2000: Slicing a list produces a new list object with references to the same elements the original list had references to. Indexing a list doesn't produce any new objects at all; it just gives you a reference to the object the corresponding list cell had a reference to.

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