直接在 Scipy 稀疏矩阵上使用 Intel mkl 库以更少的内存计算 A 点 A.T [英] Directly use Intel mkl library on Scipy sparse matrix to calculate A dot A.T with less memory
问题描述
我想从 python 调用 mkl.mkl_scsrmultcsr.目标是计算 压缩稀疏行 格式.稀疏矩阵C是A与A的转置的矩阵乘积，其中A也是csr格式的稀疏矩阵.当用 scipy 计算 C = A dot (AT) 时，scipy 似乎 (?)矩阵).所以，我想尝试直接使用 mkl c 函数来减少内存使用.
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.
这里适用于另一个 mkl 函数的答案.在那个答案中，mkl 函数快了 4 倍.
Here is an answer that works for another mkl function. In that answer, mkl function was faster by 4 times.
经过4天的努力，下面的版本终于可以运行了.我想知道我是否浪费了任何记忆.ctype 会复制 numpy 数组吗?测试函数中的 csr->csc 转换是必要的吗?intel c函数可以计算(A.T)点A，或者A点A，但不能计算A点(A.T).
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).
再次感谢.
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]+':
%s'%mat[i]
str_print = str_print.replace('
', '
'+' '*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()
结果:
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]]
学到的东西:
- 对于矩阵 A、B 和 C，索引从 1 开始.
- 我在原始代码中混淆了 c_indptr 和 c_indices.应该是= scipy csr_matrix 的 indptr.ja = scipy csr_matrix 的索引.
来自代码此处.一切都作为指针传递给 mkl_?csrmultcsr.
mkl_scsrmultcsr(&ta, &r[1], &sort, &m, &m, &m, a, ja, ia, a, ja, ia, c, jc, ic,&nzmax, &info);
- index starts from 1 for matrices A, B, and C.
- 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.
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);
我想要一个可以使用从零开始的索引的 mkl 函数.mkl.mkl_scsrmultcsr 函数只能使用基于一的索引.(或者我可以对所有内容进行基于一个的索引.这意味着对大多数线性代数步骤使用 intel c 函数而不是 scipy/numpy.)
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.)
推荐答案
查看 scipy 稀疏乘积的 Python 代码.请注意，它分 2 次调用编译后的代码.
Look at the Python code for the scipy sparse product. Notice that it calls the compiled code in 2 passes.
看起来 mkl 代码做了同样的事情
It looks like the mkl code does the same thing
https://software.intel.com/en-us/node/468640
如果request=1，例程只计算长度为m+1的数组ic的值，这个数组的内存必须预先分配.退出时，值 ic(m+1) - 1 是数组 c 和 jc 中元素的实际数量.
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.
如果 request=2，则该例程之前已使用参数 request=1 调用，输出数组 jc 和 c 在调用程序中分配，并且它们的长度至少为 ic(m+1) - 1.
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.
你首先根据C的行数分配ic(你从输入中知道)，并用request=1调用mkl代码.
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.
对于 request=2 你必须根据 ic(m+1) - 1.这与输入数组中 nnz 的数量不同.
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.
您正在使用 request1 = c_int(0)，这要求 c 数组的大小正确 - 如果不实际执行 request 1).
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 分配indptr(注意大小)，并传递指针(数据在此pass不重要)
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 分配indptr(不同大小)，并基于nnzindices和data.>
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)
最后使用这些数组创建一个新数组.
Last make a new array using these arrays.
return self.__class__((data,indices,indptr),shape=(M,N))
mkl 库应该以同样的方式使用.
The mkl library should be used in the same way.
====================
===================
https://github.com/scipy/scipy/blob/master/scipy/sparse/sparsetools/csr.h
有 csr_matmat_pass1 和 csr_matmat_pass2
====================
====================
如果有帮助，这里是这些传递的纯 Python 实现.不尝试利用任何数组操作的字面翻译.
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
A 和 B 都必须是 csr.它使用转置它需要转换，例如B = A.T.tocsr().
Both A and B have to be csr. It using a transpose it needs conversion, e.g. B = A.T.tocsr().
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