在 python 中生成定向光流 (HOOF) 直方图的最有效方法

Question

我有一个 498 帧长的图像序列，我使用 cv2.calcOpticalFlowFarneback 计算了光流。 因此，现在我有 497 个矢量图来表示我的运动矢量，这些矢量由幅度和方向描述。

我需要做的是生成一个直方图，在 x 轴上我有角度范围（以度为单位）。 更具体地说，我有 12 个 bin，其中第一个 bin 包含方向0 < angle < 30的所有向量，第二个 bin 包含方向30 < angle < 60等等。 相反，在 y 轴上，我需要得到每个 bin 中包含的那些向量的模的总和。

这里的问题是使用简单for循环和if语句来完成所有这些需要很长时间：

#magnitude and angle are two np.array of shape (497, 506, 1378)

bins = [1,2,3,4,5,6,7,8,9,10,11,12]
sum = np.zeros_like(bins)

for idx in range(np.array(magnitude).shape[0]): # for each flow map, i.e. for each image pair
    for mag, ang in zip(magnitude[idx].reshape(-1), angle[idx].reshape(-1)): 
        if ang >= 0 and ang <= 30:
            sum[0] += mag
        elif ang > 30 and ang <= 60:
            sum[1] += mag
        elif ang > 60 and ang <= 90:
            sum[2] += mag
        elif ang > 90 and ang <= 120:
            sum[3] += mag
        elif ang > 120 and ang <= 150:
            sum[4] += mag
        elif ang > 150 and ang <= 180:
            sum[5] += mag
        elif ang > 180 and ang <= 210:
            sum[6] += mag
        elif ang > 210 and ang <= 240:
            sum[7] += mag
        elif ang > 240 and ang <= 270:
            sum[8] += mag
        elif ang > 270 and ang <= 300:
            sum[9] += mag
        elif ang > 300 and ang <= 330:
            sum[10] += mag
        elif ang > 330 and ang <= 360:
            sum[11] += mag

计算大约需要 3 小时。 有人可以建议一种更好、更有效的方法来执行此计算吗？

提前致谢。

---

编辑

摆脱了条件句并使用 Numba 进一步加快了速度。 以下代码需要不到 10 秒的时间来计算：

import numpy as np
from numba import jit

@jit(nopython=True) # Set "nopython" mode for best performance, equivalent to @njit
def hoof(magnitude, angle):
    sum = np.zeros(13)

    for idx in range(magnitude.shape[0]): # for each flow map, i.e. for each image pair
        for mag, ang in zip(magnitude[idx].reshape(-1), angle[idx].reshape(-1)): 
            sum[int((ang)//30)] += mag
    
    sum[11] += sum[12]

    return sum[0:12]

Answer 1

条件很慢。 你应该尽可能避免它们。 此外， Numpy 矢量化和Numba JIT有助于大幅加快此类代码的速度。 这是一个未经测试的示例：

import numba as nb

@nb.jit
def compute(magnitude, angle):
    s = np.zeros(12)
    for idx in range(magnitude.shape[0]):
        for mag, ang in zip(magnitude[idx].reshape(-1), angle[idx].reshape(-1)):
            if ang == 0:
                s[0] += mag
            elif ang > 0 and ang <= 360: # The condition can be removed if always true
                s[(ang-1)//30] += mag
    return s

# Assume both are Numpy array and angle is of type int.
# Note that the first call will be slower unless you precise the types.
compute(magnitude, angle)