rxu rxu - 6 months ago 15
Python Question

Directly use Intel mkl library on Scipy sparse matrix to calculate A dot A.T with less memory

I want to call mkl.mkl_scsrmultcsr from python. The goal is to calculate a sparse matrix C in compressed sparse row format. Sparse matrix C is the matrix product between A and transpose of A, where A is also a sparse matrix in csr format. When calculating C = A dot (A.T) with scipy, scipy seems (?) to allocate new memory for holding transpose of A (A.T), and definitely allocates memory for a new C matrix (This means I can't use an existing C matrix). So, I want to try to use the mkl c function directly to decrease memory usage.

I compiled scipy with mkl. But the following seems faster than scipy for like 5-10 times on the actual data that I work with.

Here is an answer that works for another mkl function. In that answer, mkl function was faster by 4 times.

The following version finally works after working on it for 4 days. I want to know if I wasted any memory. Would ctype make any copy of the numpy array? Is that that csr->csc conversion in the test function necessary? The intel c function can calculate (A.T) dot A, or A dot A, but not A dot (A.T).

Thanks again.

from ctypes import *
import scipy.sparse as spsp
import numpy as np
import multiprocessing as mp
# June 2nd 2016 version.

# Load the share library
mkl = cdll.LoadLibrary("libmkl_rt.so")

def get_csr_handle(A,clear=False):
if clear == True:
A.indptr[:] = 0
A.indices[:] = 0
A.data[:] = 0

a_pointer = A.data.ctypes.data_as(POINTER(c_float))
# Array containing non-zero elements of the matrix A.
# This corresponds to data array of csr_matrix
# Its length is equal to #non zero elements in A
# (Can this be longer than actual #non-zero elements?)
assert A.data.ctypes.data % 16 == 0 # Check alignment

ja_pointer = A.indices.ctypes.data_as(POINTER(c_int))
# Array of column indices of all non-zero elements of A.
# This corresponds to the indices array of csr_matrix
assert A.indices.ctypes.data % 16 == 0 # Check alignment

ia_pointer = A.indptr.ctypes.data_as(POINTER(c_int))
# Array of length m+1.
# a[ia[i]:ia[i+1]] is the value of nonzero entries of
# the ith row of A.
# ja[ia[i]:ia[i+1]] is the column indices of nonzero
# entries of the ith row of A
# This corresponds to the indptr array of csr_matrix
assert A.indptr.ctypes.data % 16 == 0 # Check alignment

A_data_size = A.data.size
A_indices_size = A.indices.size
A_indptr_size = A.indptr.size

return (a_pointer, ja_pointer, ia_pointer, A)

def csr_dot_csr_t(A_handle, C_handle, nz=None):
# Calculate (A.T).dot(A) and put result into C
#
# This uses one-based indexing
#
# Both C.data and A.data must be in np.float32 type.
#
# Number of nonzero elements in C must be greater than
# or equal to the size of C.data
#
# size of C.indptr must be greater than or equal to
# 1 + (num rows of A).
#
# C_data = np.zeros((nz), dtype=np.single)
# C_indices = np.zeros((nz), dtype=np.int32)
# C_indptr = np.zeros((m+1),dtype=np.int32)

#assert len(c_pointer._obj) >= 1 + A_shape[0]


(a_pointer, ja_pointer, ia_pointer, A) = A_handle
(c_pointer, jc_pointer, ic_pointer, C) = C_handle
#print "CCC",C
#assert type(C.data[0]) == np.float32
#assert type(A.data[0]) == np.float32
#assert C.indptr.size >= A.shape[0] + 1

#CC = A.dot(A.T)
#assert C.data.size >= nz
#assert C.indices.size >= nz

trans_pointer = byref(c_char('T'))
sort_pointer = byref(c_int(0))

(m, n) = A.shape
sort_pointer = byref(c_int(0))
m_pointer = byref(c_int(m)) # Number of rows of matrix A
n_pointer = byref(c_int(n)) # Number of columns of matrix A
k_pointer = byref(c_int(n)) # Number of columns of matrix B
# should be n when trans='T'
# Otherwise, I guess should be m
###
b_pointer = a_pointer
jb_pointer = ja_pointer
ib_pointer = ia_pointer
###
if nz == None:
nz = n*n #*n # m*m # Number of nonzero elements expected
# probably can use lower value for sparse
# matrices.
nzmax_pointer = byref(c_int(nz))
# length of arrays c and jc. (which are data and
# indices of csr_matrix). So this is the number of
# nonzero elements of matrix C
#
# This parameter is used only if request=0.
# The routine stops calculation if the number of
# elements in the result matrix C exceeds the
# specified value of nzmax.

info = c_int(-3)
info_pointer = byref(info)
request_pointer_list = [byref(c_int(0)), byref(c_int(1)), byref(c_int(2))]
return_list = []
for ii in [0]:
request_pointer = request_pointer_list[ii]
ret = mkl.mkl_scsrmultcsr(trans_pointer, request_pointer, sort_pointer,
m_pointer, n_pointer, k_pointer,
a_pointer, ja_pointer, ia_pointer,
b_pointer, jb_pointer, ib_pointer,
c_pointer, jc_pointer, ic_pointer,
nzmax_pointer, info_pointer)
info_val = info.value
return_list += [ (ret,info_val) ]
return return_list


def show_csr_internal(A, indent=4):
# Print data, indptr, and indices
# of a scipy csr_matrix A
name = ['data', 'indptr', 'indices']
mat = [A.data, A.indptr, A.indices]
for i in range(3):
str_print = ' '*indent+name[i]+':\n%s'%mat[i]
str_print = str_print.replace('\n', '\n'+' '*indent*2)
print(str_print)

def fix_for_scipy(C,A):
n = A.shape[1]
print "fix n", n
nz = C.indptr[n] - 1 # -1 as this is still one based indexing.
print "fix nz", nz
data = C.data[:nz]

C.indptr[:n+1] -= 1
indptr = C.indptr[:n+1]
C.indices[:nz] -= 1
indices = C.indices[:nz]
return spsp.csr_matrix( (data, indices, indptr), shape=(n,n))

def test():
AA= [[1,0,0,1],
[1,0,1,0],
[0,0,1,0]]
AA = np.random.choice([0,1], size=(3,750000), replace=True, p=[0.99,0.01])
A_original = spsp.csr_matrix(AA)
#A = spsp.csr_matrix(A_original, dtype=np.float32)
A = A_original.astype(np.float32).tocsc()
#A_original = A.todense()
A = spsp.csr_matrix( (A.data, A.indices, A.indptr) )
print "A:"
show_csr_internal(A)
print A.todense()
A.indptr += 1 # convert to 1-based indexing
A.indices += 1 # convert to 1-based indexing
A_ptrs = get_csr_handle(A)

C = spsp.csr_matrix( np.ones((3,3)), dtype=np.float32)
#C.data = C.data[:16].view()
#C.indptr = C.indptr
C_ptrs = get_csr_handle(C, clear=True)

print "C:"
show_csr_internal(C)
print "=call mkl function="
return_list= csr_dot_csr_t(A_ptrs, C_ptrs)
print "(ret, info):", return_list
print "C after calling mkl:"
show_csr_internal(C)

C_fix = fix_for_scipy(C,A)
print "C_fix for scipy:"
show_csr_internal(C_fix)
print C_fix.todense()

print "Original C after fixing:"
show_csr_internal(C)

print "scipy's (A).dot(A.T)"
scipy_ans = (A_original).dot(A_original.T)
#scipy_ans = spsp.csr_matrix(scipy_ans)
show_csr_internal(scipy_ans)
print scipy_ans.todense()

if __name__ == "__main__":
test()


Result:

A:
data:
[ 1. 1. 1. ..., 1. 1. 1.]
indptr:
[ 0 0 0 ..., 22673 22673 22673]
indices:
[1 0 2 ..., 2 1 2]
[[ 0. 0. 0.]
[ 0. 0. 0.]
[ 0. 0. 0.]
...,
[ 0. 0. 0.]
[ 0. 0. 0.]
[ 0. 0. 0.]]
C:
data:
[ 0. 0. 0. 0. 0. 0. 0. 0. 0.]
indptr:
[0 0 0 0]
indices:
[0 0 0 0 0 0 0 0 0]
=call mkl function=
(ret, info): [(2, 0)]
C after calling mkl:
data:
[ 7576. 77. 83. 77. 7607. 104. 83. 104. 7490.]
indptr:
[ 1 4 7 10]
indices:
[1 2 3 1 2 3 1 2 3]
fix n 3
fix nz 9
C_fix for scipy:
data:
[ 7576. 77. 83. 77. 7607. 104. 83. 104. 7490.]
indptr:
[0 3 6 9]
indices:
[0 1 2 0 1 2 0 1 2]
[[ 7576. 77. 83.]
[ 77. 7607. 104.]
[ 83. 104. 7490.]]
Original C after fixing:
data:
[ 7576. 77. 83. 77. 7607. 104. 83. 104. 7490.]
indptr:
[0 3 6 9]
indices:
[0 1 2 0 1 2 0 1 2]
scipy's (A.T).dot(A)
data:
[ 83 77 7576 104 77 7607 83 104 7490]
indptr:
[0 3 6 9]
indices:
[2 1 0 2 0 1 0 1 2]
[[7576 77 83]
[ 77 7607 104]
[ 83 104 7490]]


Things learned:


  1. index starts from 1 for matrices A, B, and C.

  2. I mixed up c_indptr and c_indices in the original code. Should be ia
    = indptr of scipy csr_matrix. ja = indices of scipy csr_matrix.

  3. from the code here. Everything is passed to the mkl_?csrmultcsr as a pointer.
    mkl_scsrmultcsr(&ta, &r[1], &sort, &m, &m, &m, a, ja, ia, a, ja, ia, c, jc, ic, &nzmax, &info);

  4. I want a mkl function that can work with zero-based index. The mkl.mkl_scsrmultcsr function can only work with one-based indexing. (Or I can one-based indexing for everything. This means using intel c function instead of scipy/numpy for most linear algebra steps.)


Answer

Look at the Python code for the scipy sparse product. Notice that it calls the compiled code in 2 passes.

It looks like the mkl code does the same thing

https://software.intel.com/en-us/node/468640

If request=1, the routine computes only values of the array ic of length m + 1, the memory for this array must be allocated beforehand. On exit the value ic(m+1) - 1 is the actual number of the elements in the arrays c and jc.

If request=2, the routine has been called previously with the parameter request=1, the output arrays jc and c are allocated in the calling program and they are of the length ic(m+1) - 1 at least.

You first allocated ic based on the number of rows of C (you know that from the inputs), and call the mkl code with request=1.

For request=2 you have to allocate c and jc arrays, based on the size in ic(m+1) - 1. This is not the same as the number of nnz in the input arrays.

You are using request1 = c_int(0), which requires that the c arrays be the correct size - which you don't know without actually performing the dot (or a request 1).

==================

File:        /usr/lib/python3/dist-packages/scipy/sparse/compressed.py
Definition:  A._mul_sparse_matrix(self, other)

pass 1 allocates indptr (note size), and passes the pointers (data doesn't matter at this pass)

    indptr = np.empty(major_axis + 1, dtype=idx_dtype)

    fn = getattr(_sparsetools, self.format + '_matmat_pass1')
    fn(M, N,
       np.asarray(self.indptr, dtype=idx_dtype),
       np.asarray(self.indices, dtype=idx_dtype),
       np.asarray(other.indptr, dtype=idx_dtype),
       np.asarray(other.indices, dtype=idx_dtype),
       indptr)

    nnz = indptr[-1]

pass 2 allocates indptr (different size), and based on nnz indices and data.

    indptr = np.asarray(indptr, dtype=idx_dtype)
    indices = np.empty(nnz, dtype=idx_dtype)
    data = np.empty(nnz, dtype=upcast(self.dtype, other.dtype))

    fn = getattr(_sparsetools, self.format + '_matmat_pass2')
    fn(M, N, np.asarray(self.indptr, dtype=idx_dtype),
       np.asarray(self.indices, dtype=idx_dtype),
       self.data,
       np.asarray(other.indptr, dtype=idx_dtype),
       np.asarray(other.indices, dtype=idx_dtype),
       other.data,
       indptr, indices, data)

Last make a new array using these arrays.

    return self.__class__((data,indices,indptr),shape=(M,N))

The mkl library should be used in the same way.

===================

https://github.com/scipy/scipy/blob/master/scipy/sparse/sparsetools/csr.h

has c code for csr_matmat_pass1 and csr_matmat_pass2

====================

In case it helps, here's a pure Python implementation of these passes. A literal translation without any attempt to take advantage of any array operations.

def pass1(A, B):
    nrow,ncol=A.shape
    Aptr=A.indptr
    Bptr=B.indptr
    Cp=np.zeros(nrow+1,int)
    mask=np.zeros(ncol,int)-1
    nnz=0
    for i in range(nrow):
        row_nnz=0
        for jj in range(Aptr[i],Aptr[i+1]):
            j=A.indices[jj]
            for kk in range(Bptr[j],Bptr[j+1]):
                k=B.indices[kk]
                if mask[k]!=i:
                    mask[k]=i
                    row_nnz += 1
        nnz += row_nnz
        Cp[i+1]=nnz
    return Cp

def pass2(A,B,Cnnz):
    nrow,ncol=A.shape
    Ap,Aj,Ax=A.indptr,A.indices,A.data
    Bp,Bj,Bx=B.indptr,B.indices,B.data

    next = np.zeros(ncol,int)-1
    sums = np.zeros(ncol,A.dtype)

    Cp=np.zeros(nrow+1,int)
    Cj=np.zeros(Cnnz,int)
    Cx=np.zeros(Cnnz,A.dtype)
    nnz = 0
    for i in range(nrow):
        head = -2
        length = 0
        for jj in range(Ap[i], Ap[i+1]):
            j, v = Aj[jj], Ax[jj]
            for kk in range(Bp[j], Bp[j+1]):
                k = Bj[kk]
                sums[k] += v*Bx[kk]
                if (next[k]==-1):
                    next[k], head=head, k
                    length += 1
        print(i,sums, next)
        for _ in range(length):
            if sums[head] !=0:
                Cj[nnz], Cx[nnz] = head, sums[head]
                nnz += 1
            temp = head
            head = next[head]
            next[temp], sums[temp] = -1, 0
        Cp[i+1] = nnz            
    return Cp, Cj, Cx                  

def pass0(A,B):
    Cp = pass1(A,B)
    nnz = Cp[-1]
    Cp,Cj,Cx=pass2(A,B,nnz)
    N,M=A.shape[0], B.shape[1]
    C=sparse.csr_matrix((Cx, Cj, Cp), shape=(N,M))
    return C

Both A and B have to be csr. It using a transpose it needs conversion, e.g. B = A.T.tocsr().