python - 使用 Python 和 Raspberry Pi 快速捕获和将 2D 图像堆叠为 3D numpy 数组

Question

我正在做一个 Raspberry Pi 项目，我需要每秒拍摄大约 30 张图像(没有电影)并使用 numpy 数组将每个 2D 图像堆叠到 3D 阵列，而不将每个 2D 捕获保存为文件(因为很慢)。

我发现这个 Python 代码可以尽可能快地拍摄图像，但我不知道如何将所有图像快速堆叠为 3D 图像堆栈。

import io
import time
import picamera
#from PIL import Image

def outputs():
    stream = io.BytesIO()
    for i in range(40):
        # This returns the stream for the camera to capture to
        yield stream
        # Once the capture is complete, the loop continues here
        # (read up on generator functions in Python to understand
        # the yield statement). Here you could do some processing
        # on the image...
        #stream.seek(0)
        #img = Image.open(stream)
        # Finally, reset the stream for the next capture
        stream.seek(0)
        stream.truncate()

with picamera.PiCamera() as camera:
    camera.resolution = (640, 480)
    camera.framerate = 80
    time.sleep(2)
    start = time.time()
    camera.capture_sequence(outputs(), 'jpeg', use_video_port=True)
    finish = time.time()
    print('Captured 40 images at %.2ffps' % (40 / (finish - start)))

你们中有人知道如何使用 Python 和 Raspberry Pi 相机模块将这段代码中拍摄的 2D 图像堆叠到 3D numpy 数组吗？无需将每个 2D 捕获保存为文件

最好的问候，奥古斯丁

Answer 1

这可能不能像复制粘贴一样直接工作，但应该演示如何预分配内存并在那里写入结果。我不熟悉 pycamera 但示例 here显示了内存中流的一些不同用法。

import numpy as np

def outputs(resolution, n_pics, clr_channels=3):
    # Note that the first dimension is height and second is width
    images = np.zeros((resolution[1], resolution[0], clr_channels, n_pics), dtype=np.uint8)
    stream = io.BytesIO()

    for i in range(n_pics):
        yield stream
        images[:,:,:,i] = Image.open(stream)

        stream.seek(0)
        stream.truncate()