我需要一个快速的逐元素最大值，它将 n×m scipy 稀疏矩阵元素的每一行与稀疏的 1×m 矩阵进行比较。这在 Numpy 中使用 np.maximum(mat, vec) 完美运行通过 Numpy 的广播。
然而，Scipy 的 .maximum()没有广播。我的矩阵很大，所以我不能将它转换为一个 numpy 数组。
我目前的解决方法是使用 mat[row,:].maximum(vec) 遍历多行垫子.这个大循环是破坏我的代码效率(必须多次完成)。我的缓慢解决方案在下面的第二个代码片段中 - 有更好的解决方案吗？
# Example
import numpy as np
from scipy import sparse
mat = sparse.csc_matrix(np.arange(12).reshape((4,3)))
vec = sparse.csc_matrix([-1, 5, 100])
# Numpy's np.maximum() gives the **desired result** using broadcasting (but it can't handle sparse matrices):
numpy_result = np.maximum( mat.toarray(), vec.toarray() )
print( numpy_result )
# [[ 0 5 100]
# [ 3 5 100]
# [ 6 7 100]
# [ 9 10 100]]
# Scipy only compares the top row of mat to vec (no broadcasting!):
scipy_result = mat.maximum(vec)
print( scipy_result.toarray() )
# [[ 0 5 100]
# [ 3 4 5]
# [ 6 7 8]
# [ 9 10 11]]
#Reversing the order of mat and vec in the call to vec.maximum(mat) results in a single row output, and also frequently seg faults (!):
import numpy as np
from scipy import sparse
import timeit
mat = sparse.csc_matrix( sparse.random(20000, 4000, density=.01, data_rvs=lambda s: np.random.randint(0, 5000, size=s)) )
vec = sparse.csc_matrix( sparse.random(1, 4000, density=.01, data_rvs=lambda s: np.random.randint(0, 5000, size=s)) )
def sparse_elementwise_maximum(mat, vec):
output = sparse.lil_matrix(mat.shape)
for row_idx in range( mat.shape[0] ):
output[row_idx] = mat[row_idx,:].maximum(vec)
return output
# Time it
num_timing_loops = 3.0
starttime = timeit.default_timer()
for _ in range(int(num_timing_loops)):
sparse_elementwise_maximum(mat, vec)
print('time per call is:', (timeit.default_timer() - starttime)/num_timing_loops, 'seconds')
# 15 seconds per call (way too slow!)
我接受 Max 的回答，因为该问题专门针对高性能解决方案，而 Max 的解决方案在我尝试的各种输入上提供了 1000x-2500x 的巨大加速，但代价是增加了更多代码行和 Numba 编译。但是，对于一般用途，Daniel F 的 one-liner 是一个很好的解决方案，在我尝试过的示例上提供了 10 到 50 倍的加速——我可能会用于许多其他事情。
最佳答案
低级方法
与往常一样，您可以考虑如何为该操作构建适当的稀疏矩阵格式，对于 csr 矩阵，主要组件是 shape、data_arr、indices 和 ind_ptr。
使用 scipy.sparse.csr 对象的这些部分，使用编译语言(C、C++、Cython、Python-Numba)实现高效算法非常简单，但可能有点耗时。 Int 他的实现我使用了 Numba，但是将它移植到 C++ 应该很容易(语法更改)并且可能避免切片。
实现(初试)
import numpy as np
import numba as nb
# get all needed components of the csr object and create a resulting csr object at the end
def sparse_elementwise_maximum_wrap(mat,vec):
mat_csr=mat.tocsr()
vec_csr=vec.tocsr()
shape_mat=mat_csr.shape
indices_mat=mat_csr.indices
indptr_mat=mat_csr.indptr
data_mat=mat_csr.data
indices_vec=vec_csr.indices
data_vec=vec_csr.data
res=sparse_elementwise_maximum_nb(indices_mat,indptr_mat,data_mat,shape_mat,indices_vec,data_vec)
res=sparse.csr_matrix(res, shape=shape_mat)
return res
@nb.njit(cache=True)
def sparse_elementwise_maximum_nb(indices_mat,indptr_mat,data_mat,shape_mat,vec_row_ind,vec_row_data):
data_res=[]
indices_res=[]
indptr_mat_res=[]
indptr_mat_=0
indptr_mat_res.append(indptr_mat_)
for row_idx in range(shape_mat[0]):
mat_row_ind=indices_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
mat_row_data=data_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
mat_ptr=0
vec_ptr=0
while mat_ptr<mat_row_ind.shape[0] and vec_ptr<vec_row_ind.shape[0]:
ind_mat=mat_row_ind[mat_ptr]
ind_vec=vec_row_ind[vec_ptr]
#value for both matrix and vector is present
if ind_mat==ind_vec:
data_res.append(max(mat_row_data[mat_ptr],vec_row_data[vec_ptr]))
indices_res.append(ind_mat)
mat_ptr+=1
vec_ptr+=1
indptr_mat_+=1
#only value for the matrix is present vector is assumed 0
elif ind_mat<ind_vec:
if mat_row_data[mat_ptr] >0:
data_res.append(mat_row_data[mat_ptr])
indices_res.append(ind_mat)
indptr_mat_+=1
mat_ptr+=1
#only value for the vector is present matrix is assumed 0
else:
if vec_row_data[vec_ptr] >0:
data_res.append(vec_row_data[vec_ptr])
indices_res.append(ind_vec)
indptr_mat_+=1
vec_ptr+=1
for i in range(mat_ptr,mat_row_ind.shape[0]):
if mat_row_data[i] >0:
data_res.append(mat_row_data[i])
indices_res.append(mat_row_ind[i])
indptr_mat_+=1
for i in range(vec_ptr,vec_row_ind.shape[0]):
if vec_row_data[i] >0:
data_res.append(vec_row_data[i])
indices_res.append(vec_row_ind[i])
indptr_mat_+=1
indptr_mat_res.append(indptr_mat_)
return np.array(data_res),np.array(indices_res),np.array(indptr_mat_res)
在这种方法中，列表被动态调整大小的数组替换。我以 60 MB 的步长增加了输出的大小。在创建 csr 对象时，也没有复制的数据，只有引用。如果你想避免内存开销，你必须最后复制数组。
@nb.njit(cache=True)
def sparse_elementwise_maximum_nb(indices_mat,indptr_mat,data_mat,shape_mat,vec_row_ind,vec_row_data):
mem_step=5_000_000
#preallocate memory for 5M non-zero elements (60 MB in this example)
data_res=np.empty(mem_step,dtype=data_mat.dtype)
indices_res=np.empty(mem_step,dtype=np.int32)
data_res_p=0
indptr_mat_res=np.empty((shape_mat[0]+1),dtype=np.int32)
indptr_mat_res[0]=0
indptr_mat_res_p=1
indptr_mat_=0
for row_idx in range(shape_mat[0]):
mat_row_ind=indices_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
mat_row_data=data_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
#check if resizing is necessary
if data_res.shape[0]<data_res_p+shape_mat[1]:
#add at least memory for another mem_step elements
size_to_add=mem_step
if shape_mat[1] >size_to_add:
size_to_add=shape_mat[1]
data_res_2 =np.empty(data_res.shape[0] +size_to_add,data_res.dtype)
indices_res_2=np.empty(indices_res.shape[0]+size_to_add,indices_res.dtype)
for i in range(data_res_p):
data_res_2[i]=data_res[i]
indices_res_2[i]=indices_res[i]
data_res=data_res_2
indices_res=indices_res_2
mat_ptr=0
vec_ptr=0
while mat_ptr<mat_row_ind.shape[0] and vec_ptr<vec_row_ind.shape[0]:
ind_mat=mat_row_ind[mat_ptr]
ind_vec=vec_row_ind[vec_ptr]
#value for both matrix and vector is present
if ind_mat==ind_vec:
data_res[data_res_p]=max(mat_row_data[mat_ptr],vec_row_data[vec_ptr])
indices_res[data_res_p]=ind_mat
data_res_p+=1
mat_ptr+=1
vec_ptr+=1
indptr_mat_+=1
#only value for the matrix is present vector is assumed 0
elif ind_mat<ind_vec:
if mat_row_data[mat_ptr] >0:
data_res[data_res_p]=mat_row_data[mat_ptr]
indices_res[data_res_p]=ind_mat
data_res_p+=1
indptr_mat_+=1
mat_ptr+=1
#only value for the vector is present matrix is assumed 0
else:
if vec_row_data[vec_ptr] >0:
data_res[data_res_p]=vec_row_data[vec_ptr]
indices_res[data_res_p]=ind_vec
data_res_p+=1
indptr_mat_+=1
vec_ptr+=1
for i in range(mat_ptr,mat_row_ind.shape[0]):
if mat_row_data[i] >0:
data_res[data_res_p]=mat_row_data[i]
indices_res[data_res_p]=mat_row_ind[i]
data_res_p+=1
indptr_mat_+=1
for i in range(vec_ptr,vec_row_ind.shape[0]):
if vec_row_data[i] >0:
data_res[data_res_p]=vec_row_data[i]
indices_res[data_res_p]=vec_row_ind[i]
data_res_p+=1
indptr_mat_+=1
indptr_mat_res[indptr_mat_res_p]=indptr_mat_
indptr_mat_res_p+=1
return data_res[:data_res_p],indices_res[:data_res_p],indptr_mat_res
这种方法的性能和可用性在很大程度上取决于输入。在这种方法中分配了最大内存(这很容易导致内存不足错误)。
@nb.njit(cache=True)
def sparse_elementwise_maximum_nb(indices_mat,indptr_mat,data_mat,shape_mat,vec_row_ind,vec_row_data,shrink_to_fit):
max_non_zero=shape_mat[0]*vec_row_data.shape[0]+data_mat.shape[0]
data_res=np.empty(max_non_zero,dtype=data_mat.dtype)
indices_res=np.empty(max_non_zero,dtype=np.int32)
data_res_p=0
indptr_mat_res=np.empty((shape_mat[0]+1),dtype=np.int32)
indptr_mat_res[0]=0
indptr_mat_res_p=1
indptr_mat_=0
for row_idx in range(shape_mat[0]):
mat_row_ind=indices_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
mat_row_data=data_mat[indptr_mat[row_idx]:indptr_mat[row_idx+1]]
mat_ptr=0
vec_ptr=0
while mat_ptr<mat_row_ind.shape[0] and vec_ptr<vec_row_ind.shape[0]:
ind_mat=mat_row_ind[mat_ptr]
ind_vec=vec_row_ind[vec_ptr]
#value for both matrix and vector is present
if ind_mat==ind_vec:
data_res[data_res_p]=max(mat_row_data[mat_ptr],vec_row_data[vec_ptr])
indices_res[data_res_p]=ind_mat
data_res_p+=1
mat_ptr+=1
vec_ptr+=1
indptr_mat_+=1
#only value for the matrix is present vector is assumed 0
elif ind_mat<ind_vec:
if mat_row_data[mat_ptr] >0:
data_res[data_res_p]=mat_row_data[mat_ptr]
indices_res[data_res_p]=ind_mat
data_res_p+=1
indptr_mat_+=1
mat_ptr+=1
#only value for the vector is present matrix is assumed 0
else:
if vec_row_data[vec_ptr] >0:
data_res[data_res_p]=vec_row_data[vec_ptr]
indices_res[data_res_p]=ind_vec
data_res_p+=1
indptr_mat_+=1
vec_ptr+=1
for i in range(mat_ptr,mat_row_ind.shape[0]):
if mat_row_data[i] >0:
data_res[data_res_p]=mat_row_data[i]
indices_res[data_res_p]=mat_row_ind[i]
data_res_p+=1
indptr_mat_+=1
for i in range(vec_ptr,vec_row_ind.shape[0]):
if vec_row_data[i] >0:
data_res[data_res_p]=vec_row_data[i]
indices_res[data_res_p]=vec_row_ind[i]
data_res_p+=1
indptr_mat_+=1
indptr_mat_res[indptr_mat_res_p]=indptr_mat_
indptr_mat_res_p+=1
if shrink_to_fit==True:
data_res=np.copy(data_res[:data_res_p])
indices_res=np.copy(indices_res[:data_res_p])
else:
data_res=data_res[:data_res_p]
indices_res=indices_res[:data_res_p]
return data_res,indices_res,indptr_mat_res
# get all needed components of the csr object and create a resulting csr object at the end
def sparse_elementwise_maximum_wrap(mat,vec,shrink_to_fit=True):
mat_csr=mat.tocsr()
vec_csr=vec.tocsr()
shape_mat=mat_csr.shape
indices_mat=mat_csr.indices
indptr_mat=mat_csr.indptr
data_mat=mat_csr.data
indices_vec=vec_csr.indices
data_vec=vec_csr.data
res=sparse_elementwise_maximum_nb(indices_mat,indptr_mat,data_mat,shape_mat,indices_vec,data_vec,shrink_to_fit)
res=sparse.csr_matrix(res, shape=shape_mat)
return res
Numba 有编译开销或一些开销来从缓存加载函数。如果要获取运行时而不是编译+运行时，请不要考虑第一次调用。
import numpy as np
from scipy import sparse
mat = sparse.csr_matrix( sparse.random(20000, 4000, density=.01, data_rvs=lambda s: np.random.randint(0, 5000, size=s)) )
vec = sparse.csr_matrix( sparse.random(1, 4000, density=.01, data_rvs=lambda s: np.random.randint(0, 5000, size=s)) )
%timeit output=sparse_elementwise_maximum(mat, vec)
#for csc input
37.9 s ± 224 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#for csr input
10.7 s ± 90.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#Daniel F
%timeit sparse_maximum(mat, vec)
164 ms ± 1.74 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
#low level implementation (first try)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec)
89.7 ms ± 2.51 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
#low level implementation (optimized, csr)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec)
16.5 ms ± 122 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#low level implementation (preallocation, without copying at the end)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec)
16.5 ms ± 122 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#low level implementation (preallocation, with copying at the end)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec)
16.5 ms ± 122 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec,shrink_to_fit=False)
14.9 ms ± 110 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit res=sparse_elementwise_maximum_wrap(mat,vec,shrink_to_fit=True)
21.7 ms ± 399 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#For comparison, copying the result takes
%%timeit
np.copy(res.data)
np.copy(res.indices)
np.copy(res.indptr)
7.8 ms ± 47.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
关于python - 具有广播的稀疏 Scipy 矩阵和向量的元素最大值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/64872560/