Numba和多维添加项-不适用于numpy.newaxis？

Question

[尝试在python上加速DP算法，numba似乎是合适的候选人。

我正在用提供3D数组的1D数组减去2D数组。然后，我沿第三维使用.argmin()获得2D数组。这对numpy正常工作，但对numba无效。

玩具代码重现问题：

from numba import jit
import numpy as np

inflow      = np.arange(1,0,-0.01)                  # Dim [T]
actions     = np.arange(0,1,0.05)                   # Dim [M]
start_lvl   = np.random.rand(500).reshape(-1,1)*49  # Dim [Nx1]
disc_lvl    = np.arange(0,1000)                     # Dim [O]

@jit(nopython=True)
def my_func(disc_lvl, actions, start_lvl, inflow):
    for i in range(0,100):
        # Calculate new level at time i
        new_lvl = start_lvl + inflow[i] + actions       # Dim [N x M]

        # For each new_level element, find closest discretized level
        diff    = (disc_lvl-new_lvl[:,:,np.newaxis])    # Dim [N x M x O]
        idx_lvl = abs(diff).argmin(axis=2)              # Dim [N x M]

        return True

# function works fine without numba
success = my_func(disc_lvl, actions, start_lvl, inflow)

为什么上面的代码无法运行？取出@jit(nopython=True)时会这样做。是否有一个工作回合可以使用numba进行以下计算？

我尝试了具有numpy重复和expand_dims的变体，并且明确定义了jit函数的输入类型，但没有成功。

Answer 1

您需要进行一些更改才能使其正常工作：

1. 使用arr[:, :, None]添加尺寸：对于Numba，它看起来像getitem，所以更喜欢使用reshape
2. 使用np.abs代替内置的abs
3. 带有argmin关键字参数的axis为not implemented。最好使用Numba旨在优化的循环。

已修复所有这些，您可以运行jitted函数：

from numba import jit
import numpy as np

inflow = np.arange(1,0,-0.01)  # Dim [T]
actions = np.arange(0,1,0.05)  # Dim [M]
start_lvl = np.random.rand(500).reshape(-1,1)*49  # Dim [Nx1]
disc_lvl = np.arange(0,1000)  # Dim [O]

@jit(nopython=True)
def my_func(disc_lvl, actions, start_lvl, inflow):
    for i in range(0,100):
        # Calculate new level at time i
        new_lvl = start_lvl + inflow[i] + actions  # Dim [N x M]

        # For each new_level element, find closest discretized level
        new_lvl_3d = new_lvl.reshape(*new_lvl.shape, 1)
        diff = np.abs(disc_lvl - new_lvl_3d)  # Dim [N x M x O]

        idx_lvl = np.empty(new_lvl.shape)
        for i in range(diff.shape[0]):
            for j in range(diff.shape[1]):
                idx_lvl[i, j] = diff[i, j, :].argmin()

        return True

# function works fine without numba
success = my_func(disc_lvl, actions, start_lvl, inflow)

Answer 2

找到我的第一篇文章的更正代码，您可以在有和没有numba库的固定模式下执行。在本示例中，我个人观察到速度提高了2倍。

from numba import jit
import numpy as np
import datetime as dt

inflow = np.arange(1,0,-0.01)                       # Dim [T]
nbTime = np.shape(inflow)[0]
actions = np.arange(0,1,0.01)                       # Dim [M]
start_lvl = np.random.rand(500).reshape(-1,1)*49    # Dim [Nx1]
disc_lvl = np.arange(0,1000)                        # Dim [O]

@jit(nopython=True)
def my_func(nbTime, disc_lvl, actions, start_lvl, inflow):
    # Initialize result 
    res = np.empty((nbTime,np.shape(start_lvl)[0],np.shape(actions)[0]))

    for t in range(0,nbTime):
        # Calculate new level at time t
        new_lvl = start_lvl + inflow[t] + actions  # Dim [N x M]      
        print(t)

        # For each new_level element, find closest discretized level
        new_lvl_3d = new_lvl.reshape(*new_lvl.shape, 1)
        diff = np.abs(disc_lvl - new_lvl_3d)  # Dim [N x M x O]

        idx_lvl = np.empty(new_lvl.shape)
        for i in range(diff.shape[0]):
            for j in range(diff.shape[1]):
                idx_lvl[i, j] = diff[i, j, :].argmin()

        res[t,:,:] = idx_lvl

    return res

# Call function and print running time
start_time = dt.datetime.now()
result = my_func(nbTime, disc_lvl, actions, start_lvl, inflow)
print('Execution time :',(dt.datetime.now() - start_time))