Question

我正在开发一个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数组，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到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)))

您是否有人知道如何使用Python和Raspberry Pi相机模块将此代码中拍摄的2D图像堆叠到3D numpy数组中？不将每个2D捕获保存为文件

最好的问候，Agustín

Answer 1

这可能无法直接作为copy-n-paste工作，但应该演示如何预先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()

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

1 个答案: