查找progressb的下载速度

Question

我正在写一个脚本从一个网站下载视频。我添加了一个报告钩子来获取下载进度。到目前为止，它显示了下载数据的百分比和大小。我觉得增加下载速度和预计到达时间会很有趣。
问题是，如果我用一个简单的speed = chunk_size/time显示的速度足够精确，但是像疯了一样跳来跳去。所以，我使用了下载单个块的时间历史。比如，speed = chunk_size*n/sum(n_time_history)。
现在它显示了一个稳定的下载速度，但它肯定是错误的，因为它的值以几位/秒为单位，而下载的文件显然以更快的速度增长。
有人能告诉我我哪里出错了吗？在

这是我的密码。在

def dlProgress(count, blockSize, totalSize):
    global init_count
    global time_history
    try:
        time_history.append(time.monotonic())
    except NameError:
        time_history = [time.monotonic()]
    try:
        init_count
    except NameError:
        init_count = count
    percent = count*blockSize*100/totalSize
    dl, dlu = unitsize(count*blockSize)             #returns size in kB, MB, GB, etc.
    tdl, tdlu = unitsize(totalSize)
    count -= init_count                             #because continuation of partial downloads is supported
    if count > 0:
        n = 5                                       #length of time history to consider
        _count = n if count > n else count
        time_history = time_history[-_count:]
        time_diff = [i-j for i,j in zip(time_history[1:],time_history[:-1])]
        speed = blockSize*_count / sum(time_diff)
    else: speed = 0
    n = int(percent//4)
    try:
        eta = format_time((totalSize-blockSize*(count+1))//speed)
    except:
        eta = '>1 day'
    speed, speedu = unitsize(speed, True)           #returns speed in B/s, kB/s, MB/s, etc.
    sys.stdout.write("\r" + percent + "% |" + "#"*n + " "*(25-n) + "| " + dl + dlu  + "/" + tdl + tdlu + speed + speedu + eta)
    sys.stdout.flush()

Edit:
Corrected the logic. Download speed shown is now much better.
As I increase the length of history used to calculate the speed, the stability increases but sudden changes in speed (if download stops, etc.) aren't shown.
How do I make it stable, yet sensitive to large changes?

我意识到现在的问题更倾向于数学，但如果有人能帮我解决问题或给我指出正确的方向，那就太好了。
另外，请告诉我是否有更有效的方法来实现这一点。在

Answer 1

_count = n if count > n else count
time_history = time_history[-_count:]
time_weights = list(range(1,len(time_history))) #just a simple linear weights
time_diff = [(i-j)*k for i,j in zip(time_history[1:], time_history[:-1],time_weights)]
speed = blockSize*(sum(time_weights)) / sum(time_diff)

为了使其更稳定，并且在下载峰值上升或下降时不会发生反应，您还可以添加以下内容：

^{pr2}$

这将删除time_history中的最高和最低峰值，这将使显示的数字更加稳定。如果你想挑肥拣瘦，你可以在移除之前生成权重，然后使用time_diff.index(min(time_diff))过滤映射值。在

同样，使用非线性函数（如sqrt()）来生成权重将给您带来更好的结果。哦，正如我在评论中所说的：添加统计方法来过滤时间应该稍微好一点，但是我怀疑它不值得增加额外的开销。在