I'm trying to compute a simple dot product but leave nonzero values from the original matrix unchanged. A toy example:
import numpy as np
A = np.array([[2, 1, 1, 2],
[0, 2, 1, 0],
[1, 0, 1, 1],
[2, 2, 1, 0]])
B = np.array([[ 0.54331039, 0.41018682, 0.1582158 , 0.3486124 ],
[ 0.68804647, 0.29520239, 0.40654206, 0.20473451],
[ 0.69857579, 0.38958572, 0.30361365, 0.32256483],
[ 0.46195299, 0.79863505, 0.22431876, 0.59054473]])
Desired outcome:
C = np.array([[ 2. , 1. , 1. , 2. ],
[ 2.07466874, 2. , 1. , 0.73203386],
[ 1. , 1.5984076 , 1. , 1. ],
[ 2. , 2. , 1. , 1.42925865]])
The actual matrices in question, however, are sparse and look more like this:
A = sparse.rand(250000, 1700, density=0.001, format='csr')
B = sparse.rand(1700, 1700, density=0.02, format='csr')
One simple way go would be just setting the values using mask index, like that:
mask = A != 0
C = A.dot(B)
C[mask] = A[mask]
However, my original arrays are sparse and quite large, so changing them via index assignment is painfully slow. Conversion to lil matrix helps a bit, but again, conversion itself takes a lot of time.
The other obvious approach, I guess, would be just resort to iteration and skip masked values, but I'd like not to throw away the benefits of numpy/scipy-optimized array multiplication.
Some clarifications: I'm actually interested in some kind of special case, where B is always square, and therefore, A and C are of the same shape. So if there's a solution that doesn't work on arbitrary arrays but fits in my case, that's fine.
UPDATE: Some attempts:
from scipy import sparse
import numpy as np
def naive(A, B):
mask = A != 0
out = A.dot(B).tolil()
out[mask] = A[mask]
return out.tocsr()
def proposed(A, B):
Az = A == 0
R, C = np.where(Az)
out = A.copy()
out[Az] = np.einsum('ij,ji->i', A[R], B[:, C])
return out
%timeit naive(A, B)
1 loops, best of 3: 4.04 s per loop
%timeit proposed(A, B)
/usr/local/lib/python2.7/dist-packages/scipy/sparse/compressed.py:215: SparseEfficiencyWarning: Comparing a sparse matrix with 0 using == is inefficient, try using != instead.
/usr/local/lib/python2.7/dist-packages/scipy/sparse/coo.pyc in __init__(self, arg1, shape, dtype, copy)
173 self.shape = M.shape
174
--> 175 self.row, self.col = M.nonzero()
176 self.data = M[self.row, self.col]
177 self.has_canonical_format = True
MemoryError:
ANOTHER UPDATE:
Couldn't make anything more or less useful out of Cython, at least without going too far away from Python. The idea was to leave the dot product to scipy and just try to set those original values as fast as possible, something like this:
cimport cython
#cython.cdivision(True)
#cython.boundscheck(False)
#cython.wraparound(False)
cpdef coo_replace(int [:] row1, int [:] col1, float [:] data1, int[:] row2, int[:] col2, float[:] data2):
cdef int N = row1.shape[0]
cdef int M = row2.shape[0]
cdef int i, j
cdef dict d = {}
for i in range(M):
d[(row2[i], col2[i])] = data2[i]
for j in range(N):
if (row1[j], col1[j]) in d:
data1[j] = d[(row1[j], col1[j])]
This was a bit better then my pre-first "naive" implementation (using .tolil()), but following hpaulj's approach, lil can be thrown out. Maybe replacing python dict with something like std::map would help.
A possibly cleaner and faster version of your naive code:
In [57]: r,c=A.nonzero() # this uses A.tocoo()
In [58]: C=A*B
In [59]: Cl=C.tolil()
In [60]: Cl[r,c]=A.tolil()[r,c]
In [61]: Cl.tocsr()
C[r,c]=A[r,c] gives an efficiency warning, but I think that's aimed more a people do that kind of assignment in loop.
In [63]: %%timeit C=A*B
...: C[r,c]=A[r,c]
...
The slowest run took 7.32 times longer than the fastest....
1000 loops, best of 3: 334 µs per loop
In [64]: %%timeit C=A*B
...: Cl=C.tolil()
...: Cl[r,c]=A.tolil()[r,c]
...: Cl.tocsr()
...:
100 loops, best of 3: 2.83 ms per loop
My A is small, only (250,100), but it looks like the round trip to lil isn't a time saver, despite the warning.
Masking with A==0 is bound to give problems when A is sparse
In [66]: Az=A==0
....SparseEfficiencyWarning...
In [67]: r1,c1=Az.nonzero()
Compared to the nonzero r for A, this r1 is much larger - the row index of all zeros in the sparse matrix; everything but the 25 nonzeros.
In [70]: r.shape
Out[70]: (25,)
In [71]: r1.shape
Out[71]: (24975,)
If I index A with that r1 I get a much larger array. In effect I am repeating each row by the number of zeros in it
In [72]: A[r1,:]
Out[72]:
<24975x100 sparse matrix of type '<class 'numpy.float64'>'
with 2473 stored elements in Compressed Sparse Row format>
In [73]: A
Out[73]:
<250x100 sparse matrix of type '<class 'numpy.float64'>'
with 25 stored elements in Compressed Sparse Row format>
I've increased the shape and number of nonzero elements by roughly 100 (the number of columns).
Defining foo, and copying Divakar's tests:
def foo(A,B):
r,c = A.nonzero()
C = A*B
C[r,c] = A[r,c]
return C
In [83]: timeit naive(A,B)
100 loops, best of 3: 2.53 ms per loop
In [84]: timeit proposed(A,B)
/...
SparseEfficiencyWarning)
100 loops, best of 3: 4.48 ms per loop
In [85]: timeit foo(A,B)
...
SparseEfficiencyWarning)
100 loops, best of 3: 2.13 ms per loop
So my version has a modest speed inprovement. As Divakar found out, changing sparsity changes the relative advantages. I expect size to also change them.
The fact that A.nonzero uses the coo format, suggests it might be feasible to construct the new array with that format. A lot of sparse code builds a new matrix via the coo values.
In [97]: Co=C.tocoo()
In [98]: Ao=A.tocoo()
In [99]: r=np.concatenate((Co.row,Ao.row))
In [100]: c=np.concatenate((Co.col,Ao.col))
In [101]: d=np.concatenate((Co.data,Ao.data))
In [102]: r.shape
Out[102]: (79,)
In [103]: C1=sparse.csr_matrix((d,(r,c)),shape=A.shape)
In [104]: C1
Out[104]:
<250x100 sparse matrix of type '<class 'numpy.float64'>'
with 78 stored elements in Compressed Sparse Row format>
This C1 has, I think, the same non-zero elements as the C constructed by other means. But I think one value is different because the r is longer. In this particular example, C and A share one nonzero element, and the coo style of input sums those, where as we'd prefer to have A values overwrite everything.
If you can tolerate this discrepancy, this is a faster way (at least for this test case):
def bar(A,B):
C=A*B
Co=C.tocoo()
Ao=A.tocoo()
r=np.concatenate((Co.row,Ao.row))
c=np.concatenate((Co.col,Ao.col))
d=np.concatenate((Co.data,Ao.data))
return sparse.csr_matrix((d,(r,c)),shape=A.shape)
In [107]: timeit bar(A,B)
1000 loops, best of 3: 1.03 ms per loop
Cracked it! Well, there's a lot of scipy stuffs specific to sparse matrices that I learnt along the way. Here's the implementation that I could muster -
# Find the indices in output array that are to be updated
R,C = ((A!=0).dot(B!=0)).nonzero()
mask = np.asarray(A[R,C]==0).ravel()
R,C = R[mask],C[mask]
# Make a copy of A and get the dot product through sliced rows and columns
# off A and B using the definition of matrix-multiplication
out = A.copy()
out[R,C] = (A[R].multiply(B[:,C].T).sum(1)).ravel()
The most expensive part seems to be element-wise multiplication and summing. On some quick timing tests, it seems that this would be good on a sparse matrices with a high degree of sparsity to beat the original dot-mask-based solution in terms of performance, which I think comes from its focus on memory efficiency.
Runtime test
Function definitions -
def naive(A, B):
mask = A != 0
out = A.dot(B).tolil()
out[mask] = A[mask]
return out.tocsr()
def proposed(A, B):
R,C = ((A!=0).dot(B!=0)).nonzero()
mask = np.asarray(A[R,C]==0).ravel()
R,C = R[mask],C[mask]
out = A.copy()
out[R,C] = (A[R].multiply(B[:,C].T).sum(1)).ravel()
return out
Timings -
In [57]: # Input matrices
...: M,N = 25000, 170
...: A = sparse.rand(M, N, density=0.001, format='csr')
...: B = sparse.rand(N, N, density=0.02, format='csr')
...:
In [58]: %timeit naive(A, B)
10 loops, best of 3: 92.2 ms per loop
In [59]: %timeit proposed(A, B)
10 loops, best of 3: 132 ms per loop
In [60]: # Input matrices with increased sparse-ness
...: M,N = 25000, 170
...: A = sparse.rand(M, N, density=0.0001, format='csr')
...: B = sparse.rand(N, N, density=0.002, format='csr')
...:
In [61]: %timeit naive(A, B)
10 loops, best of 3: 78.1 ms per loop
In [62]: %timeit proposed(A, B)
100 loops, best of 3: 8.03 ms per loop
Python isn't my main language, but I thought this was an interesting problem and I wanted to give this a stab :)
Preliminaries:
import numpy
import scipy.sparse
# example matrices and sparse versions
A = numpy.array([[1, 2, 0, 1], [1, 0, 1, 2], [0, 1, 2 ,1], [1, 2, 1, 0]])
B = numpy.array([[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]])
A_s = scipy.sparse.lil_matrix(A)
B_s = scipy.sparse.lil_matrix(B)
So you want to convert the original problem of:
C = A.dot(B)
C[A.nonzero()] = A[A.nonzero()]
To something sparse-y.
Just to get this out of the way, the direct "sparse" translation of the above is:
C_s = A_s.dot(B_s)
C_s[A_s.nonzero()] = A_s[A_s.nonzero()]
But it sounds like you're not happy about this, as it calculates all the dot products first, which you worry might be inefficient.
So, your question is, if you find the zeros first, and only evaluate dot products on those elements, will that be faster? I.e. for a dense matrix this could be something like:
Xs, Ys = numpy.nonzero(A==0)
D = A[:]
D[Xs, Ys] = map ( lambda x,y: A[x,:].dot(B[:,y]), Xs, Ys)
Let's translate this to a sparse matrix. My main stumbling block here was finding the "Zero" indices; since A_s==0 doesn't make sense for sparse matrices, I found them this way:
Xmax, Ymax = A_s.shape
DenseSize = Xmax * Ymax
Xgrid, Ygrid = numpy.mgrid[0:Xmax, 0:Ymax]
Ygrid = Ygrid.reshape([DenseSize,1])[:,0]
Xgrid = Xgrid.reshape([DenseSize,1])[:,0]
AllIndices = numpy.array([Xgrid, Ygrid])
NonzeroIndices = numpy.array(A_s.nonzero())
ZeroIndices = numpy.array([x for x in AllIndices.T.tolist() if x not in NonzeroIndices.T.tolist()]).T
If you know of a better / faster way, by all means try it. Once we have the Zero indices, we can do a similar mapping as before:
D_s = A_s[:]
D_s[ZeroIndices[0], ZeroIndices[1]] = map ( lambda x, y : A_s[x,:].dot(B[:,y])[0], ZeroIndices[0], ZeroIndices[1] )
which gives you your sparse matrix result.
Now I don't know if this is faster or not. I mostly took a stab because it was an interesting problem, and to see if I could do it in python. In fact I suspect it might not be faster than direct whole-matrix dotproduct, because it uses listcomprehensions and mapping on a large dataset (like you say, you expect a lot of zeros). But it is an answer to your question of "how can I only calculate dot products for the zero values without doing multiplying the matrices as a whole". I'd be interested to see if you do try this how it compares in terms of speed on your datasets.
EDIT: I'm putting below an example "block processing" version based on the above, which I think should allow you to process your large dataset without problems. Let me know if it works.
import numpy
import scipy.sparse
# example matrices and sparse versions
A = numpy.array([[1, 2, 0, 1], [1, 0, 1, 2], [0, 1, 2 ,1], [1, 2, 1, 0]])
B = numpy.array([[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]])
A_s = scipy.sparse.lil_matrix(A)
B_s = scipy.sparse.lil_matrix(B)
# Choose a grid division (i.e. how many processing blocks you want to create)
BlockGrid = numpy.array([2,2])
D_s = A_s[:] # initialise from A
Xmax, Ymax = A_s.shape
BaseBSiz = numpy.array([Xmax, Ymax]) / BlockGrid
for BIndX in range(0, Xmax, BlockGrid[0]):
for BIndY in range(0, Ymax, BlockGrid[1]):
BSizX, BSizY = D_s[ BIndX : BIndX + BaseBSiz[0], BIndY : BIndY + BaseBSiz[1] ].shape
Xgrid, Ygrid = numpy.mgrid[BIndX : BIndX + BSizX, BIndY : BIndY + BSizY]
Xgrid = Xgrid.reshape([BSizX*BSizY,1])[:,0]
Ygrid = Ygrid.reshape([BSizX*BSizY,1])[:,0]
AllInd = numpy.array([Xgrid, Ygrid]).T
NZeroInd = numpy.array(A_s[Xgrid, Ygrid].reshape((BSizX,BSizY)).nonzero()).T + numpy.array([[BIndX],[BIndY]]).T
ZeroInd = numpy.array([x for x in AllInd.tolist() if x not in NZeroInd.tolist()]).T
#
# Replace zero-values in current block
D_s[ZeroInd[0], ZeroInd[1]] = map ( lambda x, y : A_s[x,:].dot(B[:,y])[0], ZeroInd[0], ZeroInd[1] )
Related
I have two very large numpy arrays, which are both 3D. I need to find an efficient way to check if they are overlapping, because turning them both into sets first takes too long. I tried to use another solution I found here for this same problem but for 2D arrays, but I didn't manage to make it work for 3D.
Here is the solution for 2D:
nrows, ncols = A.shape
dtype={'names':['f{}'.format(i) for i in range(ndep)],
'formats':ndep * [A.dtype]}
C = np.intersect1d(A.view(dtype).view(dtype), B.view(dtype).view(dtype))
# This last bit is optional if you're okay with "C" being a structured array...
C = C.view(A.dtype).reshape(-1, ndep)
(where A and B are the 2D arrays)
I need to find the number of overlapping numpy arrays, but not the specific ones.
We could leverage views using a helper function that I have used across few Q&As. To get the presence of subarrays, we could use np.isin on the views or use a more laborious one with np.searchsorted.
Approach #1 : Using np.isin -
# https://stackoverflow.com/a/45313353/ #Divakar
def view1D(a, b): # a, b are arrays
a = np.ascontiguousarray(a)
b = np.ascontiguousarray(b)
void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
return a.view(void_dt).ravel(), b.view(void_dt).ravel()
def isin_nd(a,b):
# a,b are the 3D input arrays to give us "isin-like" functionality across them
A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1))
return np.isin(A,B)
Approach #2 : We could also leverage np.searchsorted upon the views -
def isin_nd_searchsorted(a,b):
# a,b are the 3D input arrays
A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1))
sidx = A.argsort()
sorted_index = np.searchsorted(A,B,sorter=sidx)
sorted_index[sorted_index==len(A)] = len(A)-1
idx = sidx[sorted_index]
return A[idx] == B
So, these two solutions give us the mask of presence of each of the subarrays from a in b. Hence, to get our desired count, it would be - isin_nd(a,b).sum() or isin_nd_searchsorted(a,b).sum().
Sample run -
In [71]: # Setup with 3 common "subarrays"
...: np.random.seed(0)
...: a = np.random.randint(0,9,(10,4,5))
...: b = np.random.randint(0,9,(7,4,5))
...:
...: b[1] = a[4]
...: b[3] = a[2]
...: b[6] = a[0]
In [72]: isin_nd(a,b).sum()
Out[72]: 3
In [73]: isin_nd_searchsorted(a,b).sum()
Out[73]: 3
Timings on large arrays -
In [74]: # Setup
...: np.random.seed(0)
...: a = np.random.randint(0,9,(100,100,100))
...: b = np.random.randint(0,9,(100,100,100))
...: idxa = np.random.choice(range(len(a)), len(a)//2, replace=False)
...: idxb = np.random.choice(range(len(b)), len(b)//2, replace=False)
...: a[idxa] = b[idxb]
# Verify output
In [82]: np.allclose(isin_nd(a,b),isin_nd_searchsorted(a,b))
Out[82]: True
In [75]: %timeit isin_nd(a,b).sum()
10 loops, best of 3: 31.2 ms per loop
In [76]: %timeit isin_nd_searchsorted(a,b).sum()
100 loops, best of 3: 1.98 ms per loop
I have an extremely sparse structured matrix. My matrix has exactly one non zero entry per column. But its huge(10k*1M) and given in following form(uisng random values for example)
rows = np.random.randint(0, 10000, 1000000)
values = np.random.randint(0,10,1000000)
where rows gives us the row number for nonzero entry in each column. I want fast matrix multiplication with S and I am doing following right now - I convert this form to a sparse matrix(S) and do S.dot(X) for multiplication with matrix X(which can be sparse or dense).
S=scipy.sparse.csr_matrix( (values, (rows, scipy.arange(1000000))), shape = (10000,1000000))
For X of size 1M * 2500 and nnz(X)=8M this takes 178ms to create S and 255 ms to apply it. So my question is this what is the best way of doing SX (where X could be sparse or dense) given my S is as described. Since creating S is itself very time consuming I was thinking of something adhoc. I did try creating something using loops but its not even close.
Simple looping procedure looks something like this
SX = np.zeros((rows.size,X.shape[1]))
for i in range(X.shape[0]):
SX[rows[i],:]+=values[i]*X[i,:]
return SX
Can we make this efficient?
Any suggestions are greatly appreciated. Thanks
Approach #1
Given that there's exactly one entry per column in the first input, we could use np.bincount using inputs - rows, values and X and thus also avoids creating sparse matrix S -
def sparse_matrix_mult(rows, values, X):
nrows = rows.max()+1
ncols = X.shape[1]
nelem = nrows * ncols
ids = rows + nrows*np.arange(ncols)[:,None]
sums = np.bincount(ids.ravel(), (X.T*values).ravel(), minlength=nelem)
out = sums.reshape(ncols,-1).T
return out
Sample run -
In [746]: import numpy as np
...: from scipy.sparse import csr_matrix
...: import scipy as sp
...:
In [747]: np.random.seed(1234)
...: m,n = 3,4
...: rows = np.random.randint(0, m, n)
...: values = np.random.randint(2,10,n)
...: X = np.random.randint(2, 10, (n,5))
...:
In [748]: S = csr_matrix( (values, (rows, sp.arange(n))), shape = (m,n))
In [749]: S.dot(X)
Out[749]:
array([[42, 27, 45, 78, 87],
[24, 18, 18, 12, 24],
[18, 6, 8, 16, 10]])
In [750]: sparse_matrix_mult(rows, values, X)
Out[750]:
array([[ 42., 27., 45., 78., 87.],
[ 24., 18., 18., 12., 24.],
[ 18., 6., 8., 16., 10.]])
Approach #2
Using np.add.reduceat to replace np.bincount -
def sparse_matrix_mult_v2(rows, values, X):
nrows = rows.max()+1
ncols = X.shape[1]
scaled_ar = X*values[:,None]
sidx = rows.argsort()
rows_s = rows[sidx]
cut_idx = np.concatenate(([0],np.flatnonzero(rows_s[1:] != rows_s[:-1])+1))
sums = np.add.reduceat(scaled_ar[sidx],cut_idx,axis=0)
out = np.empty((nrows, ncols),dtype=sums.dtype)
row_idx = rows_s[cut_idx]
out[row_idx] = sums
return out
Runtime test
I could not run it on the sizes mentioned in the question, as those were too big for me to handle. So, on reduced datasets, here's what I am getting -
In [149]: m,n = 1000, 100000
...: rows = np.random.randint(0, m, n)
...: values = np.random.randint(2,10,n)
...: X = np.random.randint(2, 10, (n,2500))
...:
In [150]: S = csr_matrix( (values, (rows, sp.arange(n))), shape = (m,n))
In [151]: %timeit csr_matrix( (values, (rows, sp.arange(n))), shape = (m,n))
100 loops, best of 3: 16.1 ms per loop
In [152]: %timeit S.dot(X)
1 loop, best of 3: 193 ms per loop
In [153]: %timeit sparse_matrix_mult(rows, values, X)
1 loop, best of 3: 4.4 s per loop
In [154]: %timeit sparse_matrix_mult_v2(rows, values, X)
1 loop, best of 3: 2.81 s per loop
So, the proposed methods don't seem to over-power numpy.dot on performance, but they should be good on memory efficiency.
For sparse X
For sparse X, we need some modifications, as listed in the modified method listed below -
from scipy.sparse import find
def sparse_matrix_mult_sparseX(rows, values, Xs): # Xs is sparse
nrows = rows.max()+1
ncols = Xs.shape[1]
nelem = nrows * ncols
scaled_vals = Xs.multiply(values[:,None])
r,c,v = find(scaled_vals)
ids = rows[r] + c*nrows
sums = np.bincount(ids, v, minlength=nelem)
out = sums.reshape(ncols,-1).T
return out
Inspired by this post Fastest way to sum over rows of sparse matrix, I have found the best way to do this is to write loops and port things to numba. Here is the
`
#njit
def sparse_mul(SX,row,col,data,values,row_map):
N = len(data)
for idx in range(N):
SX[row_map[row[idx]],col[idx]]+=data[idx]*values[row[idx]]
return SX
X_coo=X.tocoo()
s=row_map.max()+1
SX = np.zeros((s,X.shape[1]))
sparse_mul(SX,X_coo.row,X_coo.col,X_coo.data,values,row_map)`
Here row_map is the rows in the question. On X of size (1M* 1K), 1% sparsity and with s=10K this performs twice as good as forming sparse matrix from row_map and doing S.dot(A).
As I recall, Knuth TAOP talks about representing a sparse matrix instead as a linked list of (for your app) non-zero values. Maybe something like that? Then traverse the linked list rather than the entire array by each dimension.
Suppose I have an N-dimensional numpy array x and an (N-1)-dimensional index array m (for example, m = x.argmax(axis=-1)). I'd like to construct (N-1) dimensional array y such that y[i_1, ..., i_N-1] = x[i_1, ..., i_N-1, m[i_1, ..., i_N-1]] (for the argmax example above it would be equivalent to y = x.max(axis=-1)).
For N=3 I could achieve what I want by
y = x[np.arange(x.shape[0])[:, np.newaxis], np.arange(x.shape[1]), m]
The question is, how do I do this for an arbitrary N?
you can use indices :
firstdims=np.indices(x.shape[:-1])
And add yours :
ind=tuple(firstdims)+(m,)
Then x[ind] is what you want.
In [228]: allclose(x.max(-1),x[ind])
Out[228]: True
Here's one approach using reshaping and linear indexing to handle multi-dimensional arrays of arbitrary dimensions -
shp = x.shape[:-1]
n_ele = np.prod(shp)
y_out = x.reshape(n_ele,-1)[np.arange(n_ele),m.ravel()].reshape(shp)
Let's take a sample case with a ndarray of 6 dimensions and let's say we are using m = x.argmax(axis=-1) to index into the last dimension. So, the output would be x.max(-1). Let's verify this for the proposed solution -
In [121]: x = np.random.randint(0,9,(4,5,3,3,2,4))
In [122]: m = x.argmax(axis=-1)
In [123]: shp = x.shape[:-1]
...: n_ele = np.prod(shp)
...: y_out = x.reshape(n_ele,-1)[np.arange(n_ele),m.ravel()].reshape(shp)
...:
In [124]: np.allclose(x.max(-1),y_out)
Out[124]: True
I liked #B. M.'s solution for its elegance. So, here's a runtime test to benchmark these two -
def reshape_based(x,m):
shp = x.shape[:-1]
n_ele = np.prod(shp)
return x.reshape(n_ele,-1)[np.arange(n_ele),m.ravel()].reshape(shp)
def indices_based(x,m): ## #B. M.'s solution
firstdims=np.indices(x.shape[:-1])
ind=tuple(firstdims)+(m,)
return x[ind]
Timings -
In [152]: x = np.random.randint(0,9,(4,5,3,3,4,3,6,2,4,2,5))
...: m = x.argmax(axis=-1)
...:
In [153]: %timeit indices_based(x,m)
10 loops, best of 3: 30.2 ms per loop
In [154]: %timeit reshape_based(x,m)
100 loops, best of 3: 5.14 ms per loop
I want to extract multiple slices from the same 1D numpy array, where the slice indices are drawn from a random distribution. Basically, I want to achieve the following:
import numpy as np
import numpy.random
# generate some 1D data
data = np.random.randn(500)
# window size (slices are 2*winsize long)
winsize = 60
# number of slices to take from the data
inds_size = (100, 200)
# get random integers that function as indices into the data
inds = np.random.randint(low=winsize, high=len(data)-winsize, size=inds_size)
# now I want to extract slices of data, running from inds[0,0]-60 to inds[0,0]+60
sliced_data = np.zeros( (winsize*2,) + inds_size )
for k in range(inds_size[0]):
for l in range(inds_size[1]):
sliced_data[:,k,l] = data[inds[k,l]-winsize:inds[k,l]+winsize]
# sliced_data.shape is now (120, 100, 200)
The above nested loop works fine, but is very slow. In my real code, I will need to do this thousands of times, for data arrays a lot bigger than these. Is there any way to do this more efficiently?
Note that inds will always be 2D in my case, but after getting the slices I will always be summing over one of these two dimensions, so an approach that only accumulates the sum across the one dimension would be fine.
I found this question and this answer which seem almost the same. However, the question is only about a 1D indexing vector (as opposed to my 2D). Also, the answer lacks a bit of context, as I don't really understand how the suggested as_strided works. Since my problem does not seem uncommon, I thought I'd ask again in the hope of a more explanatory answer rather than just code.
Using as_strided in this way appears to be somewhat faster than Divakar's approach (20 ms vs 35 ms here), although memory usage might be an issue.
data_wins = as_strided(data, shape=(data.size - 2*winsize + 1, 2*winsize), strides=(8, 8))
inds = np.random.randint(low=0, high=data.size - 2*winsize, size=inds_size)
sliced = data_wins[inds]
sliced = sliced.transpose((2, 0, 1)) # to use the same index order as before
Strides are the steps in bytes for the index in each dimension. For example, with an array of shape (x, y, z) and a data type of size d (8 for float64), the strides will ordinarily be (y*z*d, z*d, d), so that the second index steps over whole rows of z items. Setting both values to 8, data_wins[i, j] and data_wins[j, i] will refer to the same memory location.
>>> import numpy as np
>>> from numpy.lib.stride_tricks import as_strided
>>> a = np.arange(10, dtype=np.int8)
>>> as_strided(a, shape=(3, 10 - 2), strides=(1, 1))
array([[0, 1, 2, 3, 4, 5, 6, 7],
[1, 2, 3, 4, 5, 6, 7, 8],
[2, 3, 4, 5, 6, 7, 8, 9]], dtype=int8)
Here's a vectorized approach using broadcasting -
# Get 3D offsetting array and add to inds for all indices
allinds = inds + np.arange(-60,60)[:,None,None]
# Index into data with all indices for desired output
sliced_dataout = data[allinds]
Runtime test -
In [20]: # generate some 1D data
...: data = np.random.randn(500)
...:
...: # window size (slices are 2*winsize long)
...: winsize = 60
...:
...: # number of slices to take from the data
...: inds_size = (100, 200)
...:
...: # get random integers that function as indices into the data
...: inds=np.random.randint(low=winsize,high=len(data)-winsize, size=inds_size)
...:
In [21]: %%timeit
...: sliced_data = np.zeros( (winsize*2,) + inds_size )
...: for k in range(inds_size[0]):
...: for l in range(inds_size[1]):
...: sliced_data[:,k,l] = data[inds[k,l]-winsize:inds[k,l]+winsize]
...:
10 loops, best of 3: 66.9 ms per loop
In [22]: %%timeit
...: allinds = inds + np.arange(-60,60)[:,None,None]
...: sliced_dataout = data[allinds]
...:
10 loops, best of 3: 24.1 ms per loop
Memory consumption : Compromise solution
If memory consumption is an issue, here's a compromise solution with one loop -
sliced_dataout = np.zeros( (winsize*2,) + inds_size )
for k in range(sliced_data.shape[0]):
sliced_dataout[k] = data[inds-winsize+k]
Given a list of rotation angles (lets say about the X axis):
import numpy as np
x_axis_rotations = np.radians([0,10,32,44,165])
I can create an array of matrices matching these angles by doing so:
matrices = []
for angle in x_axis_rotations:
matrices.append(np.asarray([[1 , 0 , 0],[0, np.cos(angle), -np.sin(angle)], [0, np.sin(angle), np.cos(angle)]]))
matrices = np.array(matrices)
This will work but it doesn't take advantage of numpy's strengths for dealing with large arrays... So if my array of angles is in the millions, doing it this way won't be very fast.
Is there a better (faster) way to do create an array of transform matrices from an array of inputs?
Here's a direct and simple approach:
c = np.cos(x_axis_rotations)
s = np.sin(x_axis_rotations)
matrices = np.zeros((len(x_axis_rotations), 3, 3))
matrices[:, 0, 0] = 1
matrices[:, 1, 1] = c
matrices[:, 1, 2] = -s
matrices[:, 2, 1] = s
matrices[:, 2, 2] = c
timings, for the curious:
In [30]: angles = 2 * np.pi * np.random.rand(1000)
In [31]: timeit OP(angles)
100 loops, best of 3: 5.46 ms per loop
In [32]: timeit askewchan(angles)
10000 loops, best of 3: 39.6 µs per loop
In [33]: timeit divakar(angles)
10000 loops, best of 3: 93.8 µs per loop
In [34]: timeit divakar_oneline(angles)
10000 loops, best of 3: 56.1 µs per loop
In [35]: timeit divakar_combine(angles)
10000 loops, best of 3: 43.9 µs per loop
All are much faster than your loop, so use whichever you like the most :)
You can use linear indexing to help out, like so -
# Get cosine and sine values in one-go
cosv = np.cos(x_axis_rotations)
sinv = np.sin(x_axis_rotations)
# Get size parameter
N = x_axis_rotations.size
# Initialize output array
out = np.zeros((N,3,3))
# Set the first element in each 3D slice as 1
out[:,0,0] = 1
# Calculate the first of positions where cosine valued elements are to be put
idx1 = 4 + 9*np.arange(N)[:,None]
# One by one put those 4 values in 2x2 blocks across all 3D slices
out.ravel()[idx1] = cosv
out.ravel()[idx1+1] = -sinv
out.ravel()[idx1+3] = sinv
out.ravel()[idx1+4] = cosv
Alternatively, you can set the elements in one-go after you have initialized the output array with zeros and set the first element in each slice as 1, like so -
out.reshape(N,-1)[:,[4,5,7,8]] = np.column_stack((cosv,-sinv,sinv,cosv))
Between the above mentioned two approaches, two more middleground approaches could evolve, again put right after initializing with zeros and setting the first element in each 3D slice as 1, like so -
out.reshape(N,-1)[:,[4,8]] = cosv[:,None]
out.reshape(N,-1)[:,[5,7]] = np.column_stack((-sinv[:,None],sinv[:,None]))
The last one would be -
out.reshape(N,-1)[:,[4,8]] = cosv[:,None]
out.reshape(N,-1)[:,5] = -sinv
out.reshape(N,-1)[:,7] = sinv