CODEWITHSUNDEEP
pythonnumpymemorympi4py
hzfmer
hzfmer

Reputation: 55

How is the memory allocated for numpy arrays in python?

I tried to understand the difference caused by numpy "2D" arrays, that is, numpy.zeros((3, )), numpy.zeros((3, 1)), numpy.zeros((1, 3)).

I used id to look at the memory allocation for each element. But I found some weird outputs in iPython console.

a = np.zeros((1, 3))
In [174]: id(a[0, 0])
Out[174]: 4491074656

In [175]: id(a[0, 1])
Out[175]: 4491074680

In [176]: id(a[0, 2])
Out[176]: 4491074704

In [177]: id(a[0, 0])
Out[177]: 4491074728

In [178]: id(a[0, 1])
Out[178]: 4491074800

In [179]: id(a)
Out[179]: 4492226688

In [180]: id(a[0, 1])
Out[180]: 4491074752

The memories of the elements are

  1. not consecutive
  2. changing without reassignment

Moreover, the elements in the array of shape (1, 3) seem to be of successive memory at first, but it's not even the case for other shapes, like

In [186]: a = np.zeros((3, ))

In [187]: id(a)
Out[187]: 4490927280

In [188]: id(a[0])
Out[188]: 4491075040

In [189]: id(a[1])
Out[189]: 4491074968

In [191]: a = np.random.rand(4, 1)

In [192]: id(a)
Out[192]: 4491777648

In [193]: id(a[0])
Out[193]: 4491413504

In [194]: id(a[1])
Out[194]: 4479900048

In [195]: id(a[2])
Out[195]: 4491648416

I am actually not quite sure whether id is suitable to check memory in Python. From my knowledge I guess there is no easy way to get the physical address of variables in Python.

Just like C or Java, I expect the elements in such "2D" arrays should be consecutive in memory, which seems not to be true. Besides, the results of id are keeping changing, which really confuses me.

I am interested in this because I am using mpi4py a little bit, and I wanna figure out how the variables are sent/received between CPUs.

Upvotes: 4

Views: 5541

Answers (1)

HYRY
HYRY

Reputation: 97301

Numpy array saves its data in a memory area seperated from the object itself. As following image shows:

enter image description here

To get the address of the data you need to create views of the array and check the ctypes.data attribute which is the address of the first data element:

import numpy as np
a = np.zeros((3, 2))
print(a.ctypes.data)
print(a[0:1, 0].ctypes.data)
print(a[0:1, 1].ctypes.data)
print(a[1:2, 0].ctypes.data)
print(a[1:2, 1].ctypes.data)

Upvotes: 5

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