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 计算了它的光流。因此现在我有 497 个矢量图表示我的运动矢量，这些矢量由大小和方向描述。

我需要做的是生成一个直方图，其中 x 轴上的角度范围以度为单位。更具体地说，我有 12 个箱子，其中第一个箱子包含方向为 0 < angle < 30 的所有矢量，第二个箱子包含方向为 30 < angle < 60 的所有矢量，依此类推。相反，在 y 轴上，我需要每个 bin 中包含的那些向量的模数之和。

这里的问题是使用简单的 for 循环和 if 语句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

这花了大约 3 个小时来计算。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

摆脱了条件并使用 Numba 进一步加速。以下代码的计算时间不到 10 秒：

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。您应该尽可能避免使用它们。 Numpy 向量化 和 Numba JIT 也有助于大大加快此类代码的速度。这是一个未经测试的例子：

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)

在 python 中生成定向光流 (HOOF) 直方图的最有效方法

Most efficient way to generate Histograms of Oriented Optical Flow (HOOF) in python

python

performance

histogram

opticalflow

computer-vision

编辑