CODEWITHSUNDEEP
pythonnumpy
hlin117
hlin117

Reputation: 22270

Numpy: Memory allocations on each operation?

Does numpy allocate new matrices for every operation you perform on a matrix?

For example:

A = np.random.rand(10, 20)
A = 2 * A  # Operation 1: Is a copy of A made, and a reference assigned to A?
B = 2 * A  # Operation 2: Does B get a completely different copy of A?
C = A      # Operation 3: Does C get a reference to A?

And slice operations:

A[0, :] = 3

How about chained operations?

D = A * B * C  # Elementwise multiplication, if A * B allocates memory, does 
               # (A * B) * C allocate another patch of memory?

Numpy's a fantastic library, but I just want to know what happens under the hood. My intuition says that slice operations modify the memory view in place, but I don't know about assignments.

Upvotes: 5

Views: 1611

Answers (2)

hpaulj
hpaulj

Reputation: 231385

Keep in mind that a numpy array is a Python object. Python creates and deletes objects continually. The array has attributes shown in the .FLAGS and .__array_interface__ dictionaries, things like the shape and dtype. The attribute that takes up (potentially) a lot of memory is the data buffer. It may be a few bytes long, or may be MB.

Where possible numpy operations try to avoid copying the data buffer. When indexing, it will return a view if possible. I think the documentation compares views and copies well enough.

But views are different from Python references. A shared reference means two variables (or pointers in a list or dictionary) point to the same Python object. A view is a different array object, but one which shares the data buffer with another array. A copy has its own data buffer.

In your examples:

A = np.random.rand(10, 20)

A is a variable pointing to an array object. That object has a data buffer with 200 floats (200*8 bytes).

A = 2 * A  # Operation 1: Is a copy of A made, and a reference assigned to A?

2*A creates a new object, with a new data buffer. None of its data values can be shared with the original A. A=... reassigns the A variable. The old A object is 'lost', and eventually memory is garbage collected.

B = 2 * A  # Operation 2: Does B get a completely different copy of A?

This 2*A operates on the new A array. The object is assigned to B. A remains unchanged.

C = A      # Operation 3: Does C get a reference to A?

Yes, this is just normal Python assignment. C refers to the same object as A. id(C)==id(A).

B = A[1,:]  #  B is a view

B is a reference to a new array object. But that object shares the data buffer with A. That's because the desired values can be found in the buffer by just starting at a different point, and using a different shape.

A[0, :] = 3

This LHS slice will change a subset of the values of A. It is similar to:

B = A[0, :]
B = 3

But there are subtile differences betwee LHS and RHS slices. On the LHS you have to pay more attention to when you get a copy as opposed to a view. I've seen this especially with expressions like A[idx1,:][:,idx2] = 3.

D = A * B * C

The details of how many intermediate copies are made in a calculation like this are buried in the numpy C code. It's safest to assume that it does something like:

temp1 = A*B
temp2 = temp1*C
D = temp2
(temp1 goes to garbage)

For ordinary calculations it isn't worth worrying about those details. If you are really pushing for speed you could do a timeit on alternatives. And occasionally we get SO questions about operations giving memory errors. Do a search to get more details on those.

Upvotes: 4

RobertB
RobertB

Reputation: 1929

Yes it creates new arrays. Except C. C and A point to same memory.

You can test all of this yourself. Try the id(A) command to see where in memory A is pointing. Also, just create a smaller structure and modify parts of it and then see if A, B, and/or C are also updated.

Upvotes: 3

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