Using Python (3.7.7) and numpy (1.17.4), I am working with medium sized 2d numpy arrays (from 5000x80 up to 200,000x120). For a given array, I want to calculate the Hadamard product between all possbible uniqe pairs of column-vectors of that array.
I have:
A A
[a,b,c,d] [a,b,c,d]
[1,2,3,4] [1,2,3,4]
[4,5,6,7] * [4,5,6,7]
[7,8,9,1] [7,8,9,1]
and I want to get:
[a*b, ac, ad, bc, bd, cd]
[ 2., 3., 4., 6., 8., 12.]
[20., 24., 28., 30., 35., 42.]
[56., 63., 7., 72., 8., 9.]
I already have a solution from a colleague using np.kron which I adapated a bit:
def hadamard_kron(A: np.ndarray) -> :
"""Returns the hadamard products of all unique pairs of all columns,
and return indices signifying which columns constitute a given pair.
"""
n = raw_inputs.shape[0]
ind1 = (np.kron(np.arange(0, n).reshape((n, 1)), np.ones((n, 1)))).squeeze().astype(int)
ind2 = (np.kron(np.ones((n, 1)), np.arange(0, n).reshape((n, 1)))).squeeze().astype(int)
xmat2 = np.kron(raw_inputs, np.ones((n, 1))) * np.kron(np.ones((n, 1)), raw_inputs)
hadamard_inputs = xmat2[ind2 > ind1, :]
ind1_ = ind1[ind1 < ind2]
ind2_ = ind2[ind1 < ind2]
return hadamard_A, ind1_, ind2_
hadamard_A, first_pair_members, second_pair_members = hadamard_kron(a.transpose())
Note that hadamard_A is what I want, but transposed (which is also what I want for further processing). Also, ind1_ (ind2_) gives the indices for the objects which feature as the first (second) element in the pair for which the hadamard product is calculated. I need those as well.
However, I feel this code is too inefficient: it takes to long and since I call this function several times during my algorithm, I was wondering whether there is a cleverer solution? Am I overlooking some numpy/scipy tools I could cleverly combine for this task?
Thanks all! :)
Approach #1
Simplest one with np.triu_indices
-
In [45]: a
Out[45]:
array([[1, 2, 3, 4],
[4, 5, 6, 7],
[7, 8, 9, 1]])
In [46]: r,c = np.triu_indices(a.shape[1],1)
In [47]: a[:,c]*a[:,r]
Out[47]:
array([[ 2, 3, 4, 6, 8, 12],
[20, 24, 28, 30, 35, 42],
[56, 63, 7, 72, 8, 9]])
Approach #2
Memory-efficient one for large arrays -
m,n = a.shape
s = np.r_[0,np.arange(n-1,-1,-1).cumsum()]
out = np.empty((m, n*(n-1)//2), dtype=a.dtype)
for i,(s0,s1) in enumerate(zip(s[:-1], s[1:])):
out[:,s0:s1] = a[:,i,None] * a[:,i+1:]
Approach #3
Masking based one -
m,n = a.shape
mask = ~np.tri(n,dtype=bool)
m3D = np.broadcast_to(mask, (m,n,n))
b1 = np.broadcast_to(a[...,None], (m,n,n))
b2 = np.broadcast_to(a[:,None,:], (m,n,n))
out = (b1[m3D]* b2[m3D]).reshape(m,-1)
Approach #4
Extend approach #2 for a numba
one -
from numba import njit
def numba_app(a):
m,n = a.shape
out = np.empty((m, n*(n-1)//2), dtype=a.dtype)
return numba_func(a,out,m,n)
@njit
def numba_func(a,out,m,n):
for p in range(m):
I = 0
for i in range(n):
for j in range(i+1,n):
out[p,I] = a[p,i] * a[p,j]
I += 1
return out
Then, leverage parallel
processing (as pointed out in comments by @max9111), like so -
from numba import prange
def numba_app_parallel(a):
m,n = a.shape
out = np.empty((m, n*(n-1)//2), dtype=a.dtype)
return numba_func_parallel(a,out,m,n)
@njit(parallel=True)
def numba_func_parallel(a,out,m,n):
for p in prange(m):
I = 0
for i in range(n):
for j in range(i+1,n):
out[p,I] = a[p,i] * a[p,j]
I += 1
return out
Using benchit
package (few benchmarking tools packaged together; disclaimer: I am its author) to benchmark proposed solutions.
import benchit
in_ = [np.random.rand(5000, 80), np.random.rand(10000, 100), np.random.rand(20000, 120)]
funcs = [ehsan, app1, app2, app3, numba_app, numba_app_parallel]
t = benchit.timings(funcs, in_, indexby='shape')
t.rank()
t.plot(logx=False, save='timings.png')
Conclusion : Numba
ones seem to be doing pretty well and app2
among NumPy ones.