Skip to main content

performance tuning - How can I generate a $2n$-point moving maximum of a list as efficiently as MaxFilter?


This is based on the prior question, Moving maximum function?, where @rasher provided two winning solutions (i.e. clearly the most efficient):



MaxFilter[list, 1]

and


Max /@ Transpose[{Rest[Append[#, 0]], #, Most[Prepend[#, 0]]}] &[list]

Taking the same example list,


{5, 6, 9, 3, 2, 6, 7, 8, 1, 1, 4, 7}

I'd like to be able to specify a window $w$ such that a rolling maximum starts at index $w$, generating, for example for $w=5$ (lined up on purpose):


            {9, 9, 9, 8, 8, 8, 8, 8}


That's not too hard; as @TylerDurden hinted, we can simply use MaxFilter with range $(w-1)/2=2$ and drop each end ("shifting," as he imagined):


w=5;
Drop[Drop[MaxFilter[list, (w - 1)/2], (w - 1)/2], -(w - 1)/2]
(* {9, 9, 9, 8, 8, 8, 8, 8} *)

But of course, this can't be done with an even number $w$.


So, giving up, I decided to try to generalize @rasher's second solution—i.e. staggering $w$ copies of the list (in his case, essentially hardcoded to $w=3$) and getting maximums across the transpose:


AnotherMaxFilter[list_, w_] :=
Max /@ (

Array[
ConstantArray[0, w - #]~Join~list &,
w
]~Flatten~{2}
)[[w~Range~Length@list, Range@w]]

Note: I'm using Flatten to do a jagged transpose.


However, my attempt wasn't efficient at all:


list = RandomInteger[10, 10^7];


AbsoluteTiming[AnotherMaxFilter[list, 3]][[1]]
(* 5.257301 *)

AbsoluteTiming[AnotherMaxFilter[list, 5]][[1]]
(* 7.153409 *)

AbsoluteTiming[AnotherMaxFilter[list, 7]][[1]]
(* 14.786846 *)

Is there a way to improve my generalization to reach @rasher's original efficiency? Or, is there a clever way to get MaxFilter to max over even-numbered windows?



I'd also very much appreciate if someone could explain what about my code could have introduced such inefficiency.



Answer



I found a clever way to make MaxFilter work with even windows, and used conditionals to combine the odd- and even- cases. This solution is, surprisingly, more efficient than even @MrWizard's compiled function, and increasingly so as the window size increases.



MovingMax[list_, w_] := If[w > Length@list, {}, Module[{r, tmp},
r = Floor[w/2];
If[
OddQ@w
,
MaxFilter[list, r][[r + 1 ;; -(r + 1)]]

,
(* MaxFilter only supports odd windows; here's a hack for even windows. *)
tmp = MaxFilter[Riffle[list, Min@list, w], r][[r + 1 ;; -(r + 1)]];
If[
w > Length@tmp + 1
,
(* If window is greater than number of resulting elements, drop nothing. *)
tmp
,
Drop[tmp, {w, Length@tmp, w}]]

]
]
];

To create MovingMin, it will work to just swap all references to Max and Min with the opposites.



If the window size $w$ is odd, then we can just use MaxFilter with a radius of $(w-1)/2$ and drop the same amount of elements (i.e. as the radius) from each end of the resulting list.


If the window size $w$ is even, my idea was to insert a $-\infty$ for every $w$th element, e.g. turning


{1, 2, 3, 4, 5, 6, 7}


into


{1, 2, 3, -∞, 4, 5, 6, -∞, 7}

if $w$ were $4$. Because then, MaxFilter would essentially be finding the maximums of...


         1,  2,  3    = 3     // incomplete list; drop
1, 2, 3, -∞ = 3 // incomplete list; drop
1, 2, 3, -∞, 4 = 4
2, 3, -∞, 4, 5 = 5
3, -∞, 4, 5, 6 = 6
-∞, 4, 5, 6, -∞ = 6 // incomplete list; drop

4, 5, 6, -∞, 7 = 7
5, 6, -∞, 7 = 7 // incomplete list; drop
6, -∞, 7 = 7 // incomplete list; drop

...so after dropping the first and last two elements (same as we do for the odd case), we just have to additionally drop every $w$th element, and we're left with the maximums we're looking for.


Later, I replaced $-\infty$ with Min@list as the former method was adding an inefficiency (see @MrWizard's comment).



Testing on a list of 20,000,000 random integers and varying window sizes, here were the results:


        enter image description here





  • @MrWizard's cf performs well, but increases in time linearly with the window size.




  • @rasher's Partition-based method ran my machine out of memory for larger window sizes. (But @rasher meant to improve on my inefficient AnotherMaxFilter, not submit a serious competitor, so it's unfair to compare his solution; I only included it out of curiosity.)




  • The MaxFilter functions (split between odd and even since they're essentially two unrelated functions), in contrast, decrease in time consumption as window size increases! I imagine this is because MaxFilter probably optimizes by caching the index of its latest maximum as it runs, e.g.


     6 8 12 14 6 9 11 7 13 17 3 9 20 20 12 18 18 1 3 16
    | |

    +-------+

    where the first maximum is 14 and the cached index is 4. This way, as it moves forward, it only needs to compare one number, e.g.


     6 8 12 14 6 9 11 7 13 17 3 9 20 20 12 18 18 1 3 16
    | |
    +-------+

    where since 6 is not greater than the current maximum of 14, there's no need to compare the new set of 4 elements. This can continue until the cached index "expires," e.g.


     6 8 12 14 6 9 11 7 13 17 3 9 20 20 12 18 18 1 3 16
    | |

    +-------+

    when a new maximum must be calculated (e.g. 11, now caching index 7.)


    With such an algorithm, greater window sizes would mean greater savings.





Original Code


MovingMax[list_, w_, lowerBound_: - Infinity] := Module[{r, tmp},
r = Floor[w/2];

If[
OddQ@w
,
Drop[Drop[MaxFilter[list, r], r], -r]
,
tmp = Drop[Drop[MaxFilter[Riffle[list, lowerBound, w], r], r], -r];
Drop[tmp, {w, Length@tmp, w}]
]
];


This code is much more efficient if given an actual lower bound, e.g. -10^-6, rather than using -∞. @MrWizard explains the reason in the comments.


Original Test Code


Here's the (unedited, messy) test code I used.


FCompiled = 
Compile[{{x, _Integer, 1}, {n, _Integer}},
Module[{i = n,
a = Take[x, n]}, (a[[Mod[i, n, 1]]] = #; i++; Max[a]) & /@
Drop[x, n - 1]]];
FPartition = (Max /@ Partition[#1, #2, 1]) &;
FMaxFilterOdd =

If[OddQ@#2,
Drop[Drop[MaxFilter[#, (#2 - 1)/2], (#2 - 1)/2], -(#2 - 1)/2]] &;
FMaxFilterEven =
If[EvenQ@#2,
Module[{tmp},
Drop[tmp =
Drop[Drop[
MaxFilter[Riffle[#, -Infinity, #2], #2/2], #2/2], -#2/
2], {#2, Length@tmp, #2}]]] &;


list = RandomInteger[{-1*^6, 1*^6}, 100000];
timings = results = ConstantArray[Null, {4, 8}];

test[fn_, row_] := (
timings[[row, 1]] = ((results[[row, 1]] = fn[list, 3]); // AbsoluteTiming // First);
timings[[row, 2]] = ((results[[row, 2]] = fn[list, 6]); // AbsoluteTiming // First);
timings[[row, 3]] = ((results[[row, 3]] = fn[list, 9]); // AbsoluteTiming // First);
timings[[row, 4]] = ((results[[row, 4]] = fn[list, 12]); // AbsoluteTiming // First);
If[
! SameQ[fn, FPartition]

,
timings[[row, 5]] = ((results[[row, 5]] = fn[list, 301]); // AbsoluteTiming // First);
timings[[row, 6]] = ((results[[row, 6]] = fn[list, 602]); // AbsoluteTiming // First);
timings[[row, 7]] = ((results[[row, 7]] = fn[list, 903]); // AbsoluteTiming // First);
timings[[row, 8]] = ((results[[row, 8]] = fn[list, 1204]); // AbsoluteTiming // First);
]
);

test[FCompiled, 1];
test[FPartition, 2];

test[FMaxFilterOdd, 3];
test[FMaxFilterEven, 4];

TableForm[timings // Transpose,
TableHeadings -> {{3, 6, 9, 12, 301, 602, 903, 1204}, {"@MrWizard's",
"@rasher's", "MF-based (odd)", "MF-based (even)"}}]

Analysis of New Version


@MrWizard suggested two improvements: replacing Drop[Drop[..., r, -r]] with [[r+1;;-r-1]], and using Min@list as the lower bound. I ran some new tests:


Test results.



The first improvement's impact was clear. The second improvement would of course slow it down (since computing a Min can't be faster than a pre-supplied lower bound), but it was worth eliminating a parameter. Therefore both improvements have been incorporated into the new "tl;dr."


Comments

Popular posts from this blog

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}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....