是否有更快的方法来添加两个2-d numpy数组

时间:2016-12-18 05:29:04

标签: python performance python-2.7 numpy optimization

假设我有两个相同尺寸的大型2-d numpy数组(比如2000x2000)。我想要明智地总结它们。我想知道是否有比np.add()

更快的方法

编辑:我正在添加一个类似于我现在使用的示例。有没有办法加快这个?

#a and b are the two matrices I already have.Dimension is 2000x2000
#shift is also a list that is previously known
for j in range(100000):
    b=np.roll(b, shift[j] , axis=0)
    a=np.add(a,b)

1 个答案:

答案 0 :(得分:7)

方法#1(矢量化)

我们可以使用modulus模拟roll/circshift的循环行为,并使用广播索引覆盖所有行,我们将采用完全向量化的方法,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]

冒号表示法与使用索引进行切片

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。这应该比需要计算所有那些导致副本昂贵的行索引的rolling好得多。

在使用冒号和索引进行切片时,在视图和复制概念中更多一点 -

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

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

In [13]: a[3:8]
Out[13]: array([7, 5, 0, 6, 1])

In [14]: a[[3,4,5,6,7]]
Out[14]: array([7, 5, 0, 6, 1])

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

In [16]: np.may_share_memory(a, a[[3,4,5,6,7]])
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, shift[j] , axis=0)
        a += b

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

案例#1:

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

所以,我们正在使用修改后的循环方法来查看 2x+ 加速!