python – 是否有更快的方法来添加两个2-d numpy数组

我们可以使用模数来模拟滚动/圈移的??循环行为,并使用广播指数覆盖所有行,我们将采用完全矢量化的方法,如此 –

n = b.shape[0]
idx = n-1 - np.mod(shift.cumsum()[:,None]-1 - np.arange(n),n)
a += b[idx].sum(0)

方法#2(Loopy one)

b_ext = np.row_stack((b,b[:-1] ))
start_idx = n-1 - np.mod(shift.cumsum()-1,n)
for j in range(start_idx.size):
    a += b_ext[start_idx[j]:start_idx[j]+n]

冒号表示法使用索引进行切片

一旦我们进入循环,这里的想法就是做最小的工作.我们在进入循环之前预先计算每次迭代的起始行索引.因此,我们在循环内部所需要做的就是使用冒号表示切片,这是一个数组视图并加起来.这应该比滚动需要计算所有那些导致副本昂贵的行索引要好得多.

在使用冒号和索引进行切片时,这里有更多关于视图和复制概念的内容 –

In [11]: a = np.random.randint(0,9,(10))

In [12]: a
Out[12]: array([8,1,7,5,6,0])

In [13]: a[3:8]
Out[13]: array([7,1])

In [14]: a[[3,4,7]]
Out[14]: array([7,1])

In [15]: np.may_share_memory(a,a[3:8])
Out[15]: True

In [16]: np.may_share_memory(a,a[[3,7]])
Out[16]: False

运行时测试

功能定义 –

def original_loopy_app(a,b):
    for j in range(shift.size):
        b=np.roll(b,axis=0)
        a += b

def vectorized_app(a,b):
    n = b.shape[0]
    idx = n-1 - np.mod(shift.cumsum()[:,n)
    a += b[idx].sum(0)

def modified_loopy_app(a,b):
    n = b.shape[0]
    b_ext = np.row_stack((b,b[:-1] ))
    start_idx = n-1 - np.mod(shift.cumsum()-1,n)
    for j in range(start_idx.size):
        a += b_ext[start_idx[j]:start_idx[j]+n]

情况1：

In [5]: # Setup input arrays
   ...: N = 200
   ...: M = 1000
   ...: a = np.random.randint(11,99,(N,N))
   ...: b = np.random.randint(11,N))
   ...: shift = np.random.randint(0,N,M)
   ...: 

In [6]: original_loopy_app(a1,b1)
   ...: vectorized_app(a2,b2)
   ...: modified_loopy_app(a3,b3)
   ...: 

In [7]: np.allclose(a1,a2) # Verify results
Out[7]: True

In [8]: np.allclose(a1,a3) # Verify results
Out[8]: True

In [9]: %timeit original_loopy_app(a1,b1)
   ...: %timeit vectorized_app(a2,b2)
   ...: %timeit modified_loopy_app(a3,b3)
   ...: 
10 loops,best of 3: 107 ms per loop
10 loops,best of 3: 137 ms per loop
10 loops,best of 3: 48.2 ms per loop

案例#2：

In [13]: # Setup input arrays (datasets are exactly 1/10th of original sizes)
    ...: N = 200
    ...: M = 10000
    ...: a = np.random.randint(11,N))
    ...: b = np.random.randint(11,N))
    ...: shift = np.random.randint(0,M)
    ...: 

In [14]: %timeit original_loopy_app(a1,b1)
    ...: %timeit modified_loopy_app(a3,b3)
    ...: 
1 loops,best of 3: 1.11 s per loop
1 loops,best of 3: 481 ms per loop

因此,我们正在考虑采用改进的循环方法进行2倍加速！