3

在NumPy中，我有一个d x n数组A和一个L长度列表n，描述了我希望每列A在矩阵中结束的位置B。这个想法是i矩阵的列是对应的值B是sum的所有列中的列。ALi

我可以用一个for循环来做到这一点：

A = np.arange(15).reshape(3,5)
L = [0,1,2,1,1]
n_cols = 3
B = np.zeros((len(A), n_cols)) 
# assume I know the desired number of columns, 
# which is also one more than the maximum value of L
for i, a in enumerate(A.T):
    B[:, L[i]] += a

我想知道是否有办法通过切片数组A（或以其他矢量化方式）来做到这一点？

4

2 回答 2

3

您正在减少/折叠用于选择这些列的列A。L此外，您正在根据元素的唯一性更新输出数组的列L。

因此，您可以使用np.add.reduceat矢量化解决方案，如下所示 -

sidx = L.argsort()
col_idx, grp_start_idx = np.unique(L[sidx],return_index=True)
B_out = np.zeros((len(A), n_cols))
B_out[:,col_idx] = np.add.reduceat(A[:,sidx],grp_start_idx,axis=1)

运行时测试 -

In [129]: def org_app(A,n_cols):
     ...:     B = np.zeros((len(A), n_cols)) 
     ...:     for i, a in enumerate(A.T):
     ...:         B[:, L[i]] += a
     ...:     return B
     ...: 
     ...: def vectorized_app(A,n_cols):
     ...:     sidx = L.argsort()
     ...:     col_idx, grp_start_idx = np.unique(L[sidx],return_index=True)
     ...:     B_out = np.zeros((len(A), n_cols))
     ...:     B_out[:,col_idx] = np.add.reduceat(A[:,sidx],grp_start_idx,axis=1)
     ...:     return B_out
     ...: 

In [130]: # Setup inputs with an appreciable no. of cols & lesser rows
     ...: # so as that memory bandwidth to work with huge number of 
     ...: # row elems doesn't become the bottleneck
     ...: d,n_cols = 10,5000
     ...: A = np.random.rand(d,n_cols)
     ...: L = np.random.randint(0,n_cols,(n_cols,))
     ...: 

In [131]: np.allclose(org_app(A,n_cols),vectorized_app(A,n_cols))
Out[131]: True

In [132]: %timeit org_app(A,n_cols)
10 loops, best of 3: 33.3 ms per loop

In [133]: %timeit vectorized_app(A,n_cols)
100 loops, best of 3: 1.87 ms per loop

随着行数与 中的列数相当，A向量化方法的高内存带宽要求将抵消任何明显的加速。

于 2016-09-11T09:08:47.260 回答
1

`B` 上的迭代是否相同（未测试）？

 for I in range(B.shape[1]):
       B[:, I] = A[:, L==I].sum(axis=1)

数字循环会更少。但更重要的是，它可能会提供其他矢量化见解。

（编辑）在测试中，这可行，但速度要慢得多。

+========

scipy.sparse与矩阵乘积与一个矩阵的列求和。我们可以在这里做同样的事情吗？C在右列中使用 1制作数组

def my2(A,L):
    n_cols = L.shape[0]
    C = np.zeros((n_cols,n_cols),int)
    C[np.arange(n_cols), L] = 1
    return A.dot(C)

这比你的循环快 7 倍，比@Divakars reduceat代码快一点。

==========

In [126]: timeit vectorized_app(A,L)
The slowest run took 8.16 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 99.7 µs per loop
In [127]: timeit val2 = my2(A,L)
The slowest run took 10.46 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 81.6 µs per loop
In [128]: timeit org1(A,n_cols)
1000 loops, best of 3: 623 µs per loop
In [129]: d,n_cols
Out[129]: (10, 100)
于 2016-09-11T08:59:19.957 回答