Skip to main content

export - How to extract all images from avi when FrameRate is not Integer



Here is an avi movie (204*432 Pixels) which contains 22 different images with numbers from 1 to 22. I tested e.g. with VirtualDub and MATLAB that all extracted images are different.


Movie: https://drive.google.com/open?id=0B9wKP6yNcpyfUE1hb0UyNXhDRFk


(5.5MB)


When I extract the images with the mathematica code below image 20 is same as image 19. All the rest is correct.


The two same images are seen here:


Import[avifile, {"AVI", "ImageList", Range[19, 20, 1]}]

enter image description here



The error occurs due to the non integer FrameRate:



Import[avifile, {"FrameRate"}]

15.7143

VirtualDub and other software do not care about the frame rate. They simply extract sequentially image by image and the result is corrrect:


{Import["virtualdub_000019.png"], Import["virtualdub_000020.png"]}

enter image description here


Do you know a solution for mathematica?


My code for extracting grayscale images is:



avifile = "20170623_movie_for_testing_duplicate_images.avi";

numberImages = Length@Import[avifile, "Frames"]

22

Do[

image = ColorConvert[Import[avifile, {"Frames", i}], "Grayscale"];


fileNameCounter = ToString@PaddedForm[i, 6, NumberPadding -> {"0", ""}];

Export[StringJoin[fileNameCounter, ".png"], image];

, {i, 1, numberImages}

];

For comparison here are all extracted





Answer




Solution:


The error described below has nothing to do with mathematica (as you veryfied it) but with Quicktime. Mathematica used the Quicktime decoder which produced the mistake. Quicktime extracts image 20 as beeing the same as image 19.


After deinstalling Quicktime mathematica extracts the images correctly.


Thanks to everybody, especially for the hint of Theo Tiger who wrote: Can it be a codec issue?



See also these links:


https://github.com/SimonWoods/MathMF/


https://github.com/kmisiunas/ffmpeg-mathematica



"Mathematica's (v9 or v10) default methods uses QuickTime that produces artifact in uncompressed videos or duplicated frames."


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]