Python array element assignment

Question

I am confused with a simple case about array variable assignment in Python, and hope some one could help me check it.

In my understanding, if a is a list, b just copied the reference of a, and when you edit b, a would be modified as well. Meanwhile, you could use the is operator to check their ids. For example:

a = ["a", ["a", "b"]]
b = a[1]
b.append("c")

Then, it would return True, when I use

In [7]: b is a[1]
Out[7]: True

However, if a and b are an arrays,

import numpy as np
a = np.identity(3)
b = a[0, :]

Afterwards, when I use is to check, it returns False, but when I edited b, a would be modified as well:

In [14]: b is a[1]
Out[14]: False
In [15]: a
Out[15]:
array([[1., 0., 0.],
       [0., 1., 0.],
       [0., 0., 1.]])
In [16]: b
Out[16]: array([0., 1., 0.])
In [17]: b *= 2
In [18]: b
Out[18]: array([0., 2., 0.])
In [19]: a
Out[19]:
array([[1., 0., 0.],
       [0., 2., 0.],
       [0., 0., 1.]])

Basically, I think if is returns False, variables would have different ids and references, which means that they are independent, but it seems wrong right now, could anyone help me to check it?

Many thanks!

Michael · Accepted Answer · 2018-10-23 01:28:47Z

1

Accessing a slice of the array creates a view. b and a[0,:] are distinct views, even though they view the same part of the array a, so their id values are different even though their underlying references are the same.

The id of an object is technically distinct from what it's referencing, so objects with the same id will reference the same data but objects with the same reference don't necessarily have the same id

answered Oct 23, 2018 at 1:28

Michael

2,4249 silver badges13 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Andy Over a year ago

Hi Michael, do you mean that only accessing a slice of array will create a view but not in accessing a slice of list? Meanwhile, I am not sure that how 2 variables with different ids could be modified at same time even though they have same reference. Is it a property of dynamic language?

Michael Over a year ago

Python lists don't have views, that's a Numpy feature. Two variables with the same reference are pointing to the same data in memory. So modifying that data from one reference is the same as modifying that data from the other, even if the two variables have different IDs.

Brian Amadio · Accepted Answer · 2018-10-23 01:15:29Z

0

Numpy arrays work differently than Python lists. In your example, b is a numpy view of the first row of a, which is not the same as a pointer to the first element a.

By the way, you can check the id of each of your variables by doing id(b) or id(a[0]) for example.

answered Oct 23, 2018 at 1:15

Brian Amadio

745 bronze badges

1 Comment

Andy Over a year ago

Hi Brian, I checked them too. I thought is operator has same meaning as checking id of variables. But I am confused that how can b and a[1] be modified together with different ids? Does this problem result from numpy or python?

Collectives™ on Stack Overflow

Python array element assignment

2 Answers 2

2 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related