记忆效率稀疏对称矩阵计算

Question

I have to perform a large number of such calculations:

X.dot(Y).dot(Xt)

X = 1 xn matrix

Y = symmetric nxn matrix, each element one of 5 values (eg 0, 0.25, 0.5, 0.75, 1)

Xt = nx 1 matrix, transpose of X, ie X[np.newaxis].T

X and Y are dense. The problem I have is for large n, I cannot store and use matrix Y due to memory issues. I am limited to using one machine, so distributed calculations are not an option.

It occurred to me that Y has 2 features which theoretically can reduce the amount of memory required to store Y:

1. Elements of Y are covered by a small list of values.
2. Y is symmetric.

How can I implement this in practice? I have looked up storage of symmetric matrices, but as far as I am aware all numpy matrix multiplications require "unpacking" the symmetry to produce a regular nxn matrix.

I understand numpy is designed for in-memory calculations, so I'm open to alternative python-based solutions not reliant on numpy. I'm also open to sacrificing speed for memory-efficiency.

Answer 1

UPDATE: I found using a format that crams 3 matrix elements into one byte is actually quite fast. In the example below the speed penalty is less than 2x compared to direct multiplication using @ while the space saving is more than 20x .

>>> Y = np.random.randint(0, 5, (3000, 3000), dtype = np.int8)
>>> i, j = np.triu_indices(3000, 1)
>>> Y[i, j] = Y[j, i]
>>> values = np.array([0.3, 0.5, 0.6, 0.9, 2.0])
>>> Ycmp = (np.reshape(Y, (1000, 3, 3000)) * np.array([25, 5, 1], dtype=np.int8)[None, :, None]).sum(axis=1, dtype=np.int8)
>>> 
>>> full = values[Y]
>>> x @ full @ x
1972379.8153972814
>>> 
>>> vtable = values[np.transpose(np.unravel_index(np.arange(125), (5,5,5)))]
>>> np.dot(np.concatenate([(vtable * np.bincount(row, x, minlength=125)[:, None]).sum(axis=0) for row in Ycmp]), x)
1972379.8153972814
>>> 
>>> timeit('x @ full @ x', globals=globals(), number=100)
0.7130507210385986
>>> timeit('np.dot(np.concatenate([(vtable * np.bincount(row, x, minlength=125)[:, None]).sum(axis=0) for row in Ycmp]), x)', globals=globals(), number=100)
1.3755558349657804

The solutions below are slower and less memory efficient. I'll leave them merely for reference.

If you can afford half a byte per matrix element, then you can use np.bincount like so:

>>> Y = np.random.randint(0, 5, (1000, 1000), dtype = np.int8)
>>> i, j = np.triu_indices(1000, 1)
>>> Y[i, j] = Y[j, i]
>>> values = np.array([0.3, 0.5, 0.6, 0.9, 2.0])
>>> full = values[Y]
>>> # full would correspond to your original matrix,
>>> # Y is the 'compressed' version
>>>
>>> x = np.random.random((1000,))
>>>
>>> # direct method for reference 
>>> x @ full @ x
217515.13954751115
>>> 
>>> # memory saving version
>>> np.dot([(values * np.bincount(row, x)).sum() for row in Y], x)
217515.13954751118
>>>
>>> # to save another almost 50% exploit symmetry
>>> upper = Y[i, j]
>>> diag = np.diagonal(Y)
>>> 
>>> boundaries = np.r_[0, np.cumsum(np.arange(999, 0, -1))]
>>> (values*np.bincount(diag, x*x)).sum() + 2 * np.dot([(values*np.bincount(upper[boundaries[i]:boundaries[i+1]], x[i+1:],minlength=5)).sum() for i in range(999)], x[:-1])
217515.13954751115

Answer 2

Each row of Y , if represented as a numpy.array of datatype int as suggested in @PaulPanzer's answer, can be compressed to occupy less memory: In fact, you can store 27 elements with 64 bit, because 64 / log2(5) = 27.56...

First, compress:

import numpy as np

row = np.random.randint(5, size=100)

# pad with zeros to length that is multiple of 27
if len(row)%27:
    row_pad = np.append(row, np.zeros(27 - len(row)%27, dtype=int))
else:
    row_pad = row

row_compr = []
y_compr = 0
for i, y in enumerate(row_pad):
    if i > 0 and i % 27 == 0:
        row_compr.append(y_compr)
        y_compr = 0
    y_compr *= 5
    y_compr += y

# append last 
row_compr.append(y_compr)
row_compr = np.array(row_compr, dtype=np.int64)

Then, decompress:

row_decompr = []
for y_compr in row_compr:
    y_block = np.zeros(shape=27, dtype=np.uint8)
    for i in range(27):
        y_block[26-i] = y_compr % 5
        y_compr = int(y_compr // 5)
    row_decompr.append(y_block)

row_decompr = np.array(row_decompr).flatten()[:len(row)]

The decompressed array coincides with the original row of Y :

assert np.allclose(row, row_decompr)