向量化 NumPy 数组中的迭代加法

Question

对于 2D 索引的随机数组中的每个元素（可能有重复项），我想“+=1”到 2D 零数组中的相应网格。但是，我不知道如何优化计算。使用标准的 for 循环，如下所示，

def interadd():
    U = 100 
    input = np.random.random(size=(5000,2)) * U
    idx = np.floor(input).astype(np.int) 

    grids = np.zeros((U,U))      
    for i in range(len(input)):
        grids[idx[i,0],idx[i,1]] += 1
    return grids

运行时间可能非常重要：

>> timeit(interadd, number=5000)
43.69953393936157

有没有办法向量化这个迭代过程？

Answer 1

Divakar 的回答让我尝试以下方法，这似乎是迄今为止最快的方法：

lin_idx = idx[:,0]*U + idx[:,1]
grids = np.bincount(lin_idx, minlength=U**2).reshape(U, U)

时间安排：

In [184]: U = 100 
     ...: input = np.random.random(size=(5000,2)) * U
     ...: idx = np.floor(input).astype(np.int)

In [185]: %timeit interadd3(U, idx)  # By DSM / XYD
1000 loops, best of 3: 1.68 ms per loop

In [186]: %timeit unique_counts(U, idx)  # By Divakar
1000 loops, best of 3: 676 µs per loop

In [187]: %%timeit
     ...: lin_idx = idx[:,0]*U + idx[:,1]
     ...: grids = np.bincount(lin_idx, minlength=U*U).reshape(U, U)
     ...: 
10000 loops, best of 3: 97.5 µs per loop

Answer 2

您可以将R,C 索引从idx 转换为线性索引，然后找出唯一的索引及其计数，最后将它们存储在输出grids 作为最终输出。这是实现相同的实现 -

# Calculate linear indices corressponding to idx
lin_idx = idx[:,0]*U + idx[:,1]

# Get unique linear indices and their counts
unq_lin_idx,idx_counts = np.unique(lin_idx,return_counts=True)

# Setup output array and store index counts in raveled/flattened version
grids = np.zeros((U,U))  
grids.ravel()[unq_lin_idx] = idx_counts

运行时测试 -

这里是涵盖所有方法（包括 @DSM's approaches）并使用与该解决方案中列出的相同定义的运行时测试 -

In [63]: U = 100
    ...: idx = np.floor(np.random.random(size=(5000,2))*U).astype(np.int)
    ...: 

In [64]: %timeit interadd(U, idx)
100 loops, best of 3: 7.57 ms per loop

In [65]: %timeit interadd2(U, idx)
100 loops, best of 3: 2.59 ms per loop

In [66]: %timeit interadd3(U, idx)
1000 loops, best of 3: 1.24 ms per loop

In [67]: def unique_counts(U, idx):
    ...:     lin_idx = idx[:,0]*U + idx[:,1]
    ...:     unq_lin_idx,idx_counts = np.unique(lin_idx,return_counts=True)
    ...:     grids = np.zeros((U,U))  
    ...:     grids.ravel()[unq_lin_idx] = idx_counts
    ...:     return grids
    ...: 

In [68]: %timeit unique_counts(U, idx)
1000 loops, best of 3: 595 µs per loop

运行时表明，提议的基于 np.unique 的方法比第二快的方法快 50% 以上。

Answer 3

您可以使用np.add.at 加快速度，它可以正确处理重复索引的情况：

def interadd(U, idx):
    grids = np.zeros((U,U))      
    for i in range(len(idx)):
        grids[idx[i,0],idx[i,1]] += 1
    return grids

def interadd2(U, idx):
    grids = np.zeros((U,U))
    np.add.at(grids, idx.T.tolist(), 1)
    return grids

def interadd3(U, idx):
    # YXD suggestion
    grids = np.zeros((U,U))
    np.add.at(grids, (idx[:,0], idx[:,1]), 1)
    return grids

给了

>>> U = 100
>>> idx = np.floor(np.random.random(size=(5000,2))*U).astype(np.int)
>>> (interadd(U, idx) == interadd2(U, idx)).all()
True
>>> %timeit interadd(U, idx)
100 loops, best of 3: 8.48 ms per loop
>>> %timeit interadd2(U, idx)
100 loops, best of 3: 2.62 ms per loop

---

还有YXD的建议：

>>> (interadd(U, idx) == interadd3(U, idx)).all()
True
>>> %timeit interadd3(U, idx)
1000 loops, best of 3: 1.09 ms per loop