有没有更快的方法来添加两个2-d numpy数组 [英] Is there a faster way to add two 2-d numpy array

问题描述

比方说，我有两个尺寸相同的大型二维numpy数组(例如2000x2000).我想对它们进行元素明智的总结.我想知道是否有比np.add()更快的方法

Let say I have two large 2-d numpy array of same dimensions (say 2000x2000). I want to sum them element wise. I was wondering if there is a faster way than np.add()

编辑:我正在添加一个类似的示例，说明我现在正在使用的内容.有什么方法可以加快速度吗?

I am adding a similar example of what I am using now. Is there a way to speed up this?

#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)

推荐答案

方法1(矢量化)

我们可以使用modulus来模拟roll/circshift的循环行为，并使用广播索引覆盖所有行，我们将拥有完全矢量化的方法，就像这样-

Approach #1 (Vectorized)

We can use modulus to simulate the circulating behavior of roll/circshift and with broadcasted indices to cover all rows, we would have a fully vectorized approach, like so -

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(循环1)

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

一旦进入循环，这里的想法就是做最少的工作.在进入循环之前，我们正在预先计算每个迭代的起始行索引.因此，我们在循环内需要做的所有事情就是使用冒号表示法进行切片，这是对数组的视图并累加.这应该比rolling更好，因为rolling需要计算所有这些行索引，从而导致复制成本很高.

The idea here to do minimal work once we are inside the loop. We are pre-computing the start row index of each iteration before going into the loop. So, all we need to do once inside the loop is slicing using colon notation, which is a view into the array and adding up. This should be much better than rolling that needs to compute all of those row indices that results in a copy that is expensive.

在对冒号和索引进行切片时，视图和复制的概念会更多一些-

Here's a bit more into the view and copy concepts when slicing with colon and indices -

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-

---

Runtime test

Function defintions -

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

因此，我们正在研究采用改进的循环方法来提高 2x+ 的速度！

So, we are looking at 2x+ speedup there with the modified loopy approach!

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