ericmjl ericmjl - 6 months ago 12
Python Question

Code optimization - number of function calls in Python

I'd like to know how I might be able to transform this problem to reduce the overhead of the

np.sum()
function calls in my code.

I have an
input
matrix, say of
shape=(1000, 36)
. Each row represents a node in a graph. I have an operation that I am doing, which is iterating over each row and doing an element-wise addition to a variable number of other rows. Those "other" rows are defined in a dictionary
nodes_nbrs
that records, for each row, a list of rows that must be summed together. An example is as such:

nodes_nbrs = {0: [0, 1],
1: [1, 0, 2],
2: [2, 1],
...}


Here, node
0
would be transformed into the sum of nodes
0
and
1
. Node
1
would be transformed into the sum of nodes
1
,
0
, and
2
. And so on for the rest of the nodes.

The current (and naive) way I currently have implemented is as such. I first instantiate a zero array of the final shape that I want, and then iterate over each key-value pair in the
nodes_nbrs
dictionary:

output = np.zeros(shape=input.shape)
for k, v in nodes_nbrs.items():
output[k] = np.sum(input[v], axis=0)


This code is all cool and fine in small tests (
shape=(1000, 36)
), but on larger tests (
shape=(~1E(5-6), 36)
), it takes ~2-3 seconds to complete. I end up having to do this operation thousands of times, so I'm trying to see if there's a more optimized way of doing this.

After doing line profiling, I noticed that the key killer here is calling the
np.sum
function over and over, which takes about 50% of the total time. Is there a way I can eliminate this overhead? Or is there another way I can optimize this?




Apart from that, here is a list of things I have done, and (very briefly) their results:


  • A
    cython
    version: eliminates the
    for
    loop type checking overhead, 30% reduction in time taken. With the
    cython
    version,
    np.sum
    takes about 80% of the total wall clock time, rather than 50%.

  • Pre-declare
    np.sum
    as a variable
    npsum
    , and then call
    npsum
    inside the
    for
    loop. No difference with original.

  • Replace
    np.sum
    with
    np.add.reduce
    , and assign that to the variable
    npsum
    , and then call
    npsum
    inside the
    for
    loop. ~10% reduction in wall clock time, but then incompatible with
    autograd
    (explanation below in sparse matrices bullet point).

  • numba
    JIT-ing: did not attempt more than adding decorator. No improvement, but didn't try harder.

  • Convert the
    nodes_nbrs
    dictionary into a dense
    numpy
    binary array (1s and 0s), and then do a single
    np.dot
    operation. Good in theory, bad in practice because it would require a square matrix of
    shape=(10^n, 10^n)
    , which is quadratic in memory usage.



Things I have not tried, but am hesitant to do so:


  • scipy
    sparse matrices: I am using
    autograd
    , which does not support automatic differentiation of the
    dot
    operation for
    scipy
    sparse matrices.






For those who are curious, this is essentially a convolution operation on graph-structured data. Kinda fun developing this for grad school, but also somewhat frustrating being at the cutting edge of knowledge.

Answer

If scipy.sparse is not an option, one way you might approach this would be to massage your data so that you can use vectorized functions to do everything in the compiled layer. If you change your neighbors dictionary into a two-dimensional array with appropriate flags for missing values, you can use np.take to extract the data you want and then do a single sum() call.

Here's an example of what I have in mind:

import numpy as np

def make_data(N=100):
    X = np.random.randint(1, 20, (N, 36))
    connections = np.random.randint(2, 5, N)
    nbrs = {i: list(np.random.choice(N, c))
            for i, c in enumerate(connections)}
    return X, nbrs

def original_solution(X, nbrs):
    output = np.zeros(shape=X.shape)
    for k, v in nbrs.items():
        output[k] = np.sum(X[v], axis=0)
    return output

def vectorized_solution(X, nbrs):
    # Make neighbors all the same length, filling with -1
    new_nbrs = np.full((X.shape[0], max(map(len, nbrs.values()))), -1, dtype=int)
    for i, v in nbrs.items():
        new_nbrs[i, :len(v)] = v

    # add a row of zeros to X
    new_X = np.vstack([X, 0 * X[0]])

    # compute the sums
    return new_X.take(new_nbrs, 0).sum(1)

Now we can confirm that the results match:

>>> X, nbrs = make_data(100)
>>> np.allclose(original_solution(X, nbrs),
                vectorized_solution(X, nbrs))
True

And we can time things to see the speedup:

X, nbrs = make_data(1000)
%timeit original_solution(X, nbrs)
%timeit vectorized_solution(X, nbrs)
# 100 loops, best of 3: 13.7 ms per loop
# 100 loops, best of 3: 1.89 ms per loop

Going up to larger sizes:

X, nbrs = make_data(100000)
%timeit original_solution(X, nbrs)
%timeit vectorized_solution(X, nbrs)
1 loop, best of 3: 1.42 s per loop
1 loop, best of 3: 249 ms per loop

It's about a factor of 5-10 faster, which may be good enough for your purposes (though this will heavily depend on the exact characteristics of your nbrs dictionary).


Edit: Just for fun, I tried a couple other approaches, one using numpy.add.reduceat, one using pandas.groupby, and one using scipy.sparse. It seems that the vectorized approach I originally proposed above is probably the best bet. Here they are for reference:

from itertools import chain

def reduceat_solution(X, nbrs):
    ind, j = np.transpose([[i, len(v)] for i, v in nbrs.items()])
    i = list(chain(*(nbrs[i] for i in ind)))
    j = np.concatenate([[0], np.cumsum(j)[:-1]])
    return np.add.reduceat(X[i], j)[ind]

np.allclose(original_solution(X, nbrs),
            reduceat_solution(X, nbrs))
# True

-

import pandas as pd

def groupby_solution(X, nbrs):
    i, j = np.transpose([[k, vi] for k, v in nbrs.items() for vi in v])
    return pd.groupby(pd.DataFrame(X[j]), i).sum().values

np.allclose(original_solution(X, nbrs),
            groupby_solution(X, nbrs))
# True

-

from scipy.sparse import csr_matrix
from itertools import chain

def sparse_solution(X, nbrs):
    items = (([i]*len(col), col, [1]*len(col)) for i, col in nbrs.items())
    rows, cols, data = (np.array(list(chain(*a))) for a in zip(*items))
    M = csr_matrix((data, (rows, cols)))
    return M.dot(X)

np.allclose(original_solution(X, nbrs),
            sparse_solution(X, nbrs))
# True

And all the timings together:

X, nbrs = make_data(100000)
%timeit original_solution(X, nbrs)
%timeit vectorized_solution(X, nbrs)
%timeit reduceat_solution(X, nbrs)
%timeit groupby_solution(X, nbrs)
%timeit sparse_solution(X, nbrs)
# 1 loop, best of 3: 1.46 s per loop
# 1 loop, best of 3: 268 ms per loop
# 1 loop, best of 3: 416 ms per loop
# 1 loop, best of 3: 657 ms per loop
# 1 loop, best of 3: 282 ms per loop
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