Why does NumPy array not update original array using predictor
Question:
I am learning NumPy (and Python) and working through some exercises regarding arrays. I had a question that came up that I could not comprehend. I understand that the following code should update the original array that b is pointed to. The code is as follows.
a = np.zeros((4,5))
b = a[1:3,0:3]
b = b + 100
print('a = ', a)
print('b = ', b)
The output is:
a = [[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
b = [[100. 100. 100.]
[100. 100. 100.]]
Why does a not update along with b? I understand that a pointer object should update the original with edits. I’m assuming it is due to the syntax that 100 is added to b. The code below updates, as I thought, with the original array changing. What is the difference?
a = np.zeros((4,5))
b = a[1:3,0:3]
b[0, 0] = 100
print('a = ', a)
Output:
a = [[ 0. 0. 0. 0. 0.]
[100. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]]
Answers:
Python does not have pointers, at least not in the sense of other languages such as C.
What is happening here is that when you index an array, i.e. a[1:3,0:3], you create a view of a part of the data in a.
Numpy docs on what views are
In your first example, you then evaluate b + 100, which creates a new array.
This behavior is not changed by the fact that you assign the result to the same variable name b.
In your second example b[0, 0] = 100, you modify b in-place.
Since b is just a view of a, you also modify corresponding entries of a.
You could rewrite your first example to do b += 100 instead, which again modifies b in-place and will also change a
In the first code block, you are setting the value of b equal to what it was before + and adding 100 to all of its value.
b = b + 100
In the second code block, you are setting the value of the number in the [0, 0] position equal to 100. Since b is just a pointer to a, it also modifies the value of a
b[0, 0] = 100
The difference between the two is that in the first one you are cloning b then modify it (then assign it to b). Where in the second one, you do not clone it, and just modifies it.
I am learning NumPy (and Python) and working through some exercises regarding arrays. I had a question that came up that I could not comprehend. I understand that the following code should update the original array that b is pointed to. The code is as follows.
a = np.zeros((4,5))
b = a[1:3,0:3]
b = b + 100
print('a = ', a)
print('b = ', b)
The output is:
a = [[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0.]]
b = [[100. 100. 100.]
[100. 100. 100.]]
Why does a not update along with b? I understand that a pointer object should update the original with edits. I’m assuming it is due to the syntax that 100 is added to b. The code below updates, as I thought, with the original array changing. What is the difference?
a = np.zeros((4,5))
b = a[1:3,0:3]
b[0, 0] = 100
print('a = ', a)
Output:
a = [[ 0. 0. 0. 0. 0.]
[100. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]]
Python does not have pointers, at least not in the sense of other languages such as C.
What is happening here is that when you index an array, i.e. a[1:3,0:3], you create a view of a part of the data in a.
Numpy docs on what views are
In your first example, you then evaluate b + 100, which creates a new array.
This behavior is not changed by the fact that you assign the result to the same variable name b.
In your second example b[0, 0] = 100, you modify b in-place.
Since b is just a view of a, you also modify corresponding entries of a.
You could rewrite your first example to do b += 100 instead, which again modifies b in-place and will also change a
In the first code block, you are setting the value of b equal to what it was before + and adding 100 to all of its value.
b = b + 100
In the second code block, you are setting the value of the number in the [0, 0] position equal to 100. Since b is just a pointer to a, it also modifies the value of a
b[0, 0] = 100
The difference between the two is that in the first one you are cloning b then modify it (then assign it to b). Where in the second one, you do not clone it, and just modifies it.