ffmpeg - 从视频中提取准确的帧数 [英] ffmpeg - extract exact number of frames from video

问题描述

我想从视频中创建最多30张图像（并为他们绘制精灵图片）。

我尝试使用'select'与'mod'，但是如果总帧数不适合整合到所需数量的图像（30），那么我有时候会得到更多的图像，有时会更少。

例如，如果我的视频是72帧长，我的'mod'将是72/30，这是2.4。

我从一个python脚本运行，所以我''做一些类似如下的过滤器：

  select ='not（mod（n\，+ str mod）+））'
  

我认为mod必须是一个整数（？），所以我可以舍弃使用2，它给我36张图像或向上，给我24张图像

什么是最好的方式来准确地30？ - 显然间隔不一样，但是很好。

也许我可以使用for循环生成最接近所需间隔的帧的列表，然后传递作为选择过滤器？

例如要获得框架，我会做这样的事情：

  nframes = 72＃视频中的帧数
 outImages = 30＃我想要的图像数量
 mod = float（nframes）/ outImages＃2.4 
 
 frames = [] 
 
 idx = 1 
 
，而i < nframes：
 print str（idx）+：+ str（math.floor（i + 0.5））
 frames.append（int（math.floor（i + 0.5）））
 idx + = 1 
i + = mod 
  

然后我可以通过框架列表）到ffmpeg命令中？或者我可以告诉ffmpeg做类似的事情吗？

解决方案

如果你有一个框架列表，你可以运行

  ffmpeg -i in.mp4 -vf select ='eq（n，5）+ eq（n，11）+ eq（n， 15）..'-vsync 0 frames％d.jpg 
  

---

如果您可以将频率分解为有理数字（并进行一些数学计算），则可以直接使用选择过滤器执行此操作。

f = 2.4 与 12/5 相同。所以这意味着你需要从每12个5帧。你可以将它分解为12中的3 + 1中的1。由于在这种情况下，后者将与前一个选择一致，所以我们可以选择一个帧，即第11个每个第12帧。

  ffmpeg -i in.mp4 -vf select ='not（mod（n，3））+不（mod（n + 1,12））'-vsync 0 frames％d.jpg 
  

想法是将您的频率表示为相互之和， 1 / f = 1 / m + 1 / n + 1 / p 。您可以使用偏移设备 n + c 或 nc 如果其中一个分母是其他分母的倍数。 / p>

---

这样做的一个主要方法是

  ffmpeg -i in.mp4 -vf select ='not（mod（n，12））+ not（mod（n + 3,12））+ not（mod（n + 5,12）） +不（mod（n + 7,12））+ not（mod（n + 11,12））`-vsync 0 frames％d.jpg 
  

其中select子句的总数等于每组X帧所需的帧数。

---

无论如何，您可以通过添加 -vframes 30

来粗略地限制帧总数

I want to create a maximum of 30 images from a video (and tile them for a sprite sheet).

I've tried using the 'select' with 'mod' but if the total number of frames does not fit neatly into the desired number of images (30) then I sometimes end up with more images, sometimes less.

For example if my video is 72 frames long, my 'mod' would be 72 / 30, which is 2.4.

I'm running this from a python script so i'm doing something like the following for the filter:

select='not(mod(n\," + str(mod) + "))'

I think the mod has to be an integer (?) so I could either round down and use 2 which gives me 36 images or round up which gives me 24 images

Whats the best way to get exactly 30? - obviously the interval wouldn't be identical but thats fine.

Maybe I could use a for loop to generate a list of the frames closest to the desired interval and then pass that in as the select filter?

e.g. to get the frames I would do something like this:

nframes = 72 # number of frames in video
outImages = 30 # number of images I want
mod = float(nframes) / outImages # 2.4

frames = []

idx = 1

while i < nframes:
    print str(idx) + ": " + str(math.floor(i+0.5)) 
    frames.append(int(math.floor(i+0.5)))
    idx += 1
    i += mod

Then am I able to pass that (the frames list) into the ffmpeg command? Or can I tell ffmpeg to do something similar?

解决方案

If you have a list of frames, you can just run

ffmpeg -i in.mp4 -vf select='eq(n,5)+eq(n,11)+eq(n,15)..' -vsync 0 frames%d.jpg

---

There is a way to do this directly with select filter if you can decompose the frequency into a rational number (and do a bit of maths).

Let's take f = 2.4 which is the same as 12/5. So that means you need 5 frames from every 12. You can decompose that as 1 out of 3 + 1 out of 12. Since, in this case, the latter will coincide with the former selection, we can pick one frame earlier i.e. the 11th of every 12th frame.

ffmpeg -i in.mp4 -vf select='not(mod(n,3))+not(mod(n+1,12))' -vsync 0 frames%d.jpg

The idea is to represent your frequency as a sum of reciprocals, 1/f = 1/m + 1/n + 1/p. You can use the offset device n+c or n-c if one of the denominators is a multiple of the others.

---

A cruder way to do this is

ffmpeg -i in.mp4 -vf select='not(mod(n,12))+not(mod(n+3,12))+not(mod(n+5,12))+not(mod(n+7,12))+not(mod(n+11,12))` -vsync 0 frames%d.jpg

where total number of select clauses is equal to the number of frames needed from each set of X frames.

---

In any case, you can crudely limit the total number of frames by adding -vframes 30

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