Skip to main content

import - When importing GIF animation, how to find the correct list of "DisplayDurations"?



When importing GIF animations with variable frame durations, the settings for "DisplayDurations" are not recovered correctly. As an example, I'm using the movie from this earlier question:


slowDown


Here I'm repeating the code to generate it, putting the gradually increasing frame durations in a variable durations:


frames = 
Table[Graphics[
Text[Style["Slow Down", FontFamily -> "Futura", FontColor -> Blue,
FontSize -> 48], {0, y}], Background -> Cyan,
PlotRange -> {{-1, 1}, {.1, 1.5}}], {y, 1.2, .2, -.1}];
durations = Append[Range[Length[frames] - 1]/20, 2];
Export["slowDownMovie.gif", frames, "DisplayDurations" -> durations,

"AnimationRepetitions" -> Infinity];

durations//N


{0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 2.}



Now compare to the durations that we get when re-importing the same animated GIF:


newDurations = Import["slowDownMovie.gif", "DisplayDurations"]



{20., 20., 20., 20., 20., 20., 20., 20., 20., 20., 20.}



newDurations == durations


False



As expected from this negative result, ListAnimate also plays the animation at an incorrect, constant frame delay. It seems clear that this is a bug.


Before I start looking for a platform-specific hack to fix this: is there any way to get the correct display durations from a GIF with the Import function? I looked into the "Frames" element of the GIF import, but that also seems to have the wrong display duration in it.



I haven't looked at the raw data, but if one has to do that it may be better to find a non-Mathematica hack instead.


Edit


Wolfram support has confirmed it's a bug [TS 19773].



Answer



I don't know if you can persuade Import to return the correct durations, but luckily the format of .gif files is pretty straight forward so it's not that hard to extract the correct durations manually from the raw data.


In an animated gif the frame durations are stored in a so called Graphic Control Extension or GCE preceding each frame. A GCE starts with the byte sequence 21 F9 04 followed by 4 data bytes and is closed by 00. Of these data bytes, the second and third byte are the frame duration in hundreds of seconds ordered least significant byte first.


So a crude way to extract the frame durations is to read in the file, find the positions of the byte sequence 21 F9 04 and extract the fourth and fifth byte after each of these positions, e.g.


lst = BinaryReadList["~/test.gif"];
seq = FromDigits[#, 16] & /@ StringSplit["21 f9 04"];
pos = Position[Partition[lst, 3, 1], seq];

durations2 = (Extract[lst, pos + 4] + 256 Extract[lst, pos + 5])/100

(* ==> {1/20, 1/10, 3/20, 1/5, 1/4, 3/10, 7/20, 2/5, 9/20, 1/2, 2} *)

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...