python - 什么是创建 NumPy 数组的多个幂的矢量化方法？

Question

我有一个 NumPy 数组：

arr = [[1, 2],
       [3, 4]]

我想创建一个新数组，其中包含arr高达 a 的幂order：

# arr_new = [arr^0, arr^1, arr^2, arr^3,...arr^order]
arr_new = [[1, 1, 1, 2, 1, 4, 1, 8],
           [1, 1, 3, 4, 9, 16, 27, 64]]

我目前的方法使用for循环：

# Pre-allocate an array for powers
arr = np.array([[1, 2],[3,4]])
order = 3
rows, cols = arr.shape
arr_new = np.zeros((rows, (order+1) * cols))

# Iterate over each exponent
for i in range(order + 1):
    arr_new[:, (i * cols) : (i + 1) * cols] = arr**i
print(arr_new)

是否有更快（即矢量化）的方法来创建数组的幂？

---

基准测试

感谢@hpaulj 和@Divakar 和@Paul Panzer 的回答。我在以下测试阵列上对基于循环和基于广播的操作进行了基准测试。

arr = np.array([[1, 2],
                [3,4]])
order = 3

arrLarge = np.random.randint(0, 10, (100, 100))  # 100 x 100 array
orderLarge = 10

loop_based功能是：

def loop_based(arr, order):
    # pre-allocate an array for powers
    rows, cols = arr.shape
    arr_new = np.zeros((rows, (order+1) * cols))
    # iterate over each exponent
    for i in range(order + 1):
        arr_new[:, (i * cols) : (i + 1) * cols] = arr**i
    return arr_new

使用的broadcast_based功能hstack是：

def broadcast_based_hstack(arr, order):
    # Create a 3D exponent array for a 2D input array to force broadcasting
    powers = np.arange(order + 1)[:, None, None]
    # Generate values (third axis contains array at various powers)
    exponentiated = arr ** powers
    # Reshape and return array
    return np.hstack(exponentiated)   # <== using hstack function

使用的broadcast_based功能reshape是：

def broadcast_based_reshape(arr, order):
    # Create a 3D exponent array for a 2D input array to force broadcasting
    powers = np.arange(order + 1)[:, None]
    # Generate values (3-rd axis contains array at various powers)
    exponentiated = arr[:, None] ** powers
    # reshape and return array
    return exponentiated.reshape(arr.shape[0], -1)  # <== using reshape function

broadcast_based使用累积乘积cumprod和的函数reshape：

def broadcast_cumprod_reshape(arr, order):
    rows, cols = arr.shape
    # Create 3D empty array where the middle dimension is
    # the array at powers 0 through order
    out = np.empty((rows, order + 1, cols), dtype=arr.dtype)
    out[:, 0, :] = 1   # 0th power is always 1
    a = np.broadcast_to(arr[:, None], (rows, order, cols))
    # Cumulatively multiply arrays so each multiplication produces the next order
    np.cumprod(a, axis=1, out=out[:,1:,:])
    return out.reshape(rows, -1)

在 Jupyter notebook 上，我使用了timeit命令并得到了以下结果：

小阵列（2x2）：

%timeit -n 100000 loop_based(arr, order)
7.41 µs ± 174 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit -n 100000 broadcast_based_hstack(arr, order)
10.1 µs ± 137 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit -n 100000 broadcast_based_reshape(arr, order)
3.31 µs ± 61.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit -n 100000 broadcast_cumprod_reshape(arr, order)
11 µs ± 102 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

大型阵列（100x100）：

%timeit -n 1000 loop_based(arrLarge, orderLarge)
261 µs ± 5.82 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit -n 1000 broadcast_based_hstack(arrLarge, orderLarge)
225 µs ± 4.15 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit -n 1000 broadcast_based_reshape(arrLarge, orderLarge)
223 µs ± 2.16 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit -n 1000 broadcast_cumprod_reshape(arrLarge, orderLarge)
157 µs ± 1.02 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

结论：

reshape对于较小的阵列，使用基于广播的方法似乎更快。但是，对于大型阵列，该cumprod方法可扩展性更好且速度更快。

标签： pythonarraysnumpyvectorization

Answer 1

将阵列扩展到更高的暗度，并在以下方面的帮助下broadcasting发挥它的魔力reshaping-

In [16]: arr = np.array([[1, 2],[3,4]])

In [17]: order = 3

In [18]: (arr[:,None]**np.arange(order+1)[:,None]).reshape(arr.shape[0],-1)
Out[18]: 
array([[ 1,  1,  1,  2,  1,  4,  1,  8],
       [ 1,  1,  3,  4,  9, 16, 27, 64]])

请注意，arr[:,None]本质上是arr[:,None,:]，但为简洁起见，我们可以跳过尾随省略号。

更大数据集的计时 -

In [40]: np.random.seed(0)
    ...: arr = np.random.randint(0,9,(100,100))
    ...: order = 10

# @hpaulj's soln with broadcasting and stacking
In [41]: %timeit np.hstack(arr **np.arange(order+1)[:,None,None])
1000 loops, best of 3: 734 µs per loop

In [42]: %timeit (arr[:,None]**np.arange(order+1)[:,None]).reshape(arr.shape[0],-1)
1000 loops, best of 3: 401 µs per loop

重塑部分实际上是免费的，这就是我们在这里获得性能的地方，当然还有广播部分，如下面的细分所示 -

In [52]: %timeit (arr[:,None]**np.arange(order+1)[:,None])
1000 loops, best of 3: 390 µs per loop

In [53]: %timeit (arr[:,None]**np.arange(order+1)[:,None]).reshape(arr.shape[0],-1)
1000 loops, best of 3: 401 µs per loop