用Glumpy将NumPy数组显示为连续更新图像

Question

我已经使用NumPy和SciPy在Python中运行了一个仿真模型，它生成一个2D NumPy数组作为每次迭代的输出。我一直使用matplotlib和imshow函数将此输出显示为图像。但是，我发现了Glumpy，在它的文档页上写着：

由于有了IPython shell，可以在交互模式下运行glumpy，当它们的内容发生更改时，您可以在显示的数组中体验实时更新。

然而，我似乎想不出如何用他们给出的例子来解决这个问题。基本上，我的模型作为一个函数运行，其中包含一个大的for循环，用于循环我正在运行的迭代次数。在for循环每次迭代结束时，我想显示数组。目前，我正在使用matplotlib将图像保存到png文件中，因为通过matplotlib在屏幕上显示它似乎会冻结python进程。

我肯定有办法用Glumpy来做这件事，我只是不知道怎么做，我找不到任何有用的教程。

Answer 1

Glumpy文档相当不存在！下面是一个简单的模拟示例，将数组可视化与glumpy与matplotlib进行比较

import numpy as np
import glumpy
from OpenGL import GLUT as glut
from time import time
from matplotlib.pyplot import subplots,close
from matplotlib import cm

def randomwalk(dims=(256,256),n=3,sigma=10,alpha=0.95,seed=1):
    """ A simple random walk with memory """
    M = np.zeros(dims,dtype=np.float32)
    r,c = dims
    gen = np.random.RandomState(seed)
    pos = gen.rand(2,n)*((r,),(c,))
    old_delta = gen.randn(2,n)*sigma
    while 1:
        delta = (1.-alpha)*gen.randn(2,n)*sigma + alpha*old_delta
        pos += delta
        for ri,ci in pos.T:
            if not (0. <= ri < r) : ri = abs(ri % r)
            if not (0. <= ci < c) : ci = abs(ci % c)
            M[ri,ci] += 1
        old_delta = delta
        yield M

def mplrun(niter=1000):
    """ Visualise the simulation using matplotlib, using blit for 
    improved speed"""
    fig,ax = subplots(1,1)
    rw = randomwalk()
    im = ax.imshow(rw.next(),interpolation='nearest',cmap=cm.hot,animated=True)
    fig.canvas.draw()
    background = fig.canvas.copy_from_bbox(ax.bbox) # cache the background

    tic = time()
    for ii in xrange(niter):
        im.set_data(rw.next())          # update the image data
        fig.canvas.restore_region(background)   # restore background
        ax.draw_artist(im)          # redraw the image
        fig.canvas.blit(ax.bbox)        # redraw the axes rectangle

    close(fig)
    print "Matplotlib average FPS: %.2f" %(niter/(time()-tic))

def gprun(niter=1000):
    """ Visualise the same simulation using Glumpy """
    rw = randomwalk()
    M = rw.next()

    # create a glumpy figure
    fig = glumpy.figure((512,512))

    # the Image.data attribute is a referenced copy of M - when M
    # changes, the image data also gets updated
    im = glumpy.image.Image(M,colormap=glumpy.colormap.Hot)

    @fig.event
    def on_draw():
        """ called in the simulation loop, and also when the
        figure is resized """
        fig.clear()
        im.update()
        im.draw( x=0, y=0, z=0, width=fig.width, height=fig.height )

    tic = time()
    for ii in xrange(niter):
        M = rw.next()           # update the array          
        glut.glutMainLoopEvent()    # dispatch queued window events
        on_draw()           # update the image in the back buffer
        glut.glutSwapBuffers()      # swap the buffers so image is displayed

    fig.window.hide()
    print "Glumpy average FPS: %.2f" %(niter/(time()-tic))

if __name__ == "__main__":
    mplrun()
    gprun()

使用matplotlib和GTKAgg作为后端，并使用blit避免每次绘制背景，我可以达到大约95个FPS。使用Glumpy，我可以得到大约250-300FPS，尽管我目前在笔记本电脑上的图形设置相当糟糕。话虽如此，Glumpy在工作上还是比较谨慎的，除非您正在处理巨大的矩阵，或者您需要一个非常高的框架，不管出于什么原因，我还是坚持使用matplotlib和blit。

Answer 2

使用公式0.2.8，您可以使用pf.screen创建一个非阻塞屏幕：

import pyformulas as pf
import numpy as np

canvas = np.floor(np.random.normal(scale=50, size=(480,640,3)) % 256).astype(np.uint8)
screen = pf.screen(canvas)

while screen.exists():
    canvas = np.floor(np.random.normal(scale=50, size=(480,640,3)) % 256).astype(np.uint8)
    screen.update(canvas)

#screen.close()

免责声明:我是肾盂配方的维护者。