Skip to main content

export - High memory use while exporting a BMP


Something mysterious is going on when exporting images. I'm using Mathematica 9.0.1 on Windows 7 (64 bit), and find that the memory use is excessive when exporting to a bmp. But it doesn't happen with other image formats.


For example, running the following (with different values for k):


k = 1;  extensions = {"png", "tif", "jpg", "bmp"};  SeedRandom[42];
image = Image@RandomReal[{0, 1}, {1500, 1500, 3}];
base = MaxMemoryUsed[];

Export["C:\\mem." <> extensions[[k]], image];
{extensions[[k]], (MaxMemoryUsed[] - base)/1024.^2}
Quit[]

gives


{"png",  79.987}
{"tif", 28.617}
{"jpg", 60.453}
{"bmp", 927.602}


file sizes (added by s0rce):


png: 12.8 MB
tif: 12.8 MB
jpg: 1.29 MB
bmp: 6.43 MB

Why does exporting a bmp require over 10 times the memory of other image formats? And what can I do about it?


EDIT:


Here's the results of some more testing (code at the end).


format  texture  type    memory(MB)  time(s)  size(KB)


png random real 89 3.4 15 028
bit16 0 2.6 15 028
byte 0 1.5 7 515
smooth real 75 1.3 726
bit16 0 1.2 726
byte 0 0.5 85

tif random real 17 0.2 15 000
bit16 0 0.1 15 000

byte 0 0.1 7 500
smooth real 17 0.3 15 000
bit16 0 0.1 15 000
byte 0 0.2 7 500

jpg random real 68 0.4 1 500
bit16 0 0.8 1 503
byte 0 0.7 1 500
smooth real 68 0.3 50
bit16 0 0.3 50

byte 0 0.3 50

bmp random real 1069 25.3 7 500
bit16 967 22.8 7 500
byte 69 $Aborted 0
smooth real 1055 100.8 7 500
bit16 953 32.4 7 500
byte 71 1.4 7 500

Notes:




  • By default Mathematica will treat the image type as Real.

  • When exporting, a Real type image will need to be converted to an integer type. This requires extra memory and time.

    • tif and png support 16 bit pixels, so this will be used by default for Real images.

    • jpg and bmp only support up to 8 bits per pixel.



  • My computer was thrashing the memory disk cache for the bmp smooth real, so it took extra time.

  • The bmp random byte was using 100% for 5 minutes before I gave up.



Conclusions:



  • Use tif for speed.

  • Use png for saving disk space.

  • Never use jpg because it is lossy.

  • Never use bmp because it is broken.


Code:


k = 1; (* 1 .. 24 *)

{e, t, c} = Tuples[{Range[4], Range[3], Range[2]}][[k]];
extensions = {"png", "tif", "jpg", "bmp"};
types = {"real", "byte", "bit16"};
compress = {"random", "smooth"};
res = 1600;
SeedRandom[42];
data = Switch[compress[[c]],
"random", RandomReal[{0, 1}, {res, res, 3}],
"smooth", Table[{i, j, 1 - i j}, {i, 0., 1, 1/(res - 1)}, {j, 0., 1, 1/(res - 1)}]
];

image = Switch[types[[t]],
"real", Image@data,
"byte", Image[Round[(2^8 - 1) data], "Byte"],
"bit16", Image[Round[(2^16 - 1) data], "Bit16"]
];
filename = "C:\\_" <> types[[t]] <> "_" <> compress[[c]] <> "." <> extensions[[e]];
base = MaxMemoryUsed[];
time = AbsoluteTiming[Export[filename, image]][[1]];
CellPrint@TextCell[ToString@{
filename,(MaxMemoryUsed[] - base)/1024.^2, time, FileByteCount[filename]/1024.

}, "Text", CellAutoOverwrite -> False]
Quit[]

Answer



I believe this is a bug in Mathematica. It can be verified that the unreasonable memory usage is due to array unpacking by


On["Packing"]
ExportString[Image@RandomReal[1, {30, 30}], "BMP"];

So please report it to support@wolfram.com so this can be fixed for the benefit of all of us, and point them to this discussion.


Some further playing with Simon Woods's excellent spelunker and the (much underrated) built-in debugger revealed the culprit. It is an innocent looking line inside System`Convert`BitmapDump`ExportBMP24:


Module[{bitmap, ...},

...
bitmap = Floor[(255 bitmap)/max]; (* <-- this unpacks *)
...
]

Now bitmap appears to be a packed array of integer values and max is an integer as well. Did you already guess why this unpacks? Hover below for the spoiler.



255/max evaluates to a Rational and of course multiplying integers by Rationals will typically give a Rational and unpack the integer array. Using Floor[(255. bitmap)/max] would fix the problem.



This shows how counterintuitive Mathematica can be at times. I would have fallen into the same trap.





To verify that what I'm saying here is correct, one can do the following experiment:


MaxMemoryUsed[ExportString[Image@RandomReal[1, {1500, 1500}], "BMP"];]

(* ==> 1007548312 , i.e. 1 GB *)

Now evaluate DownValues[System`Convert`BitmapDump`ExportBMP24], search for Floor, change 255 to 255. and set the resulting expression back to DownValues[System`Convert`BitmapDump`ExportBMP24] = .... Let's measure the memory usage again:


MaxMemoryUsed[ExportString[Image@RandomReal[1, {1500, 1500}], "BMP"];]

(* ==> 289673368 i.e. 290 MB *)


Now it uses much less memory and ExportString finishes much faster.


The reason why it still uses quite a bit of memory is other flaws in the code which cause partial unpacking further down the pipeline.


Comments

Popular posts from this blog

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...

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...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...