Threshold calculation without "if else"
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Given following logic
T=1, if x < m1
T=2, if x >= m1 && x < m2
T=3, if x >= m2 && x < m3
T=4, if x >= m3
x is a parameter; m1, m2, m3 are thresholds to determine the conversion from x to T. Is there a way to calculate T without using "if"? Say, "rem" or "mod" or something without logic operation?
4 个评论
Kyle Wang
2015-3-10
Star Strider
2015-3-10
编辑：Star Strider
2015-3-10
How could you possibly solve it without logic operations? That’s how you defined it!
If you believe it could be solved without logic operations, go for it ! I’ll be interested in seeing if your solution produces results equivalent to Guillaume’s or mine.
Kyle Wang
2015-3-10
Never mind that your solution only works with scalars whereas ours work on matrices of any shape, you're using floating point division, one of the slowest operation for a processor instead of comparison, one of the fastest operation.
Kind of pointless really. Particularly, that div./(div+eps) which is just a slow and roundabout way of saying div >= 1.
采纳的回答
Guillaume
2015-3-10
T = 1 + (x >= m1) + (x >= m2) + (x >= m3)
would work. May not be more efficient that if though.
5 个评论
Kyle Wang
2015-3-10
Guillaume
2015-3-10
Logic operations are the simplest operations that a processor can do. I don't see why you'd want to avoid them. Furthermore, they're definitively part of pure calculation.
If you really want to go the roundabout and slow way, you could do:
[~, T] = histc(x, [-Inf m1 m2 m3 Inf]);
Of course under the hood histc uses logic operations.
Your problem is one of logic operations.
Jan
2015-3-11
@Guillaume: histc is the most efficient solution.
Guillaume
2015-3-11
Timing comparison:
- With scalar:
x = rand;
m1 = 0.25; m2 = 0.5; m3 = 0.75;
%method 1: my solution
timeit(@() 1 + (x >= m1) + (x >= m2) + (x >= m3), 1)
%method 2: Star's solution
timeit(@() 1*(x < m1) + 2*((x >= m1) & (x < m2)) + 3*((x >= m2) & (x < m3)) + 4*(x >= m3), 1)
%method 3: Kyle's solution
timeit(@() sum([1 floor(x./[m1 m2 m3]) ./ (floor(x./[m1 m2 m3]) + eps)]), 1)
%method 4: histc
timeit(@() histc(x, [-Inf m1 m2 m3 Inf]), 2)
%method 5: Image's solution
timeit(@() imquantize(x, [m1 m2 m3]), 1)
Timings:
- method 1 (mine): 6.6384955612319e-06
- method 2 (star): 8.68228864632548e-06
- method 3 (kyle): 9.01446251150368e-06
- method 4 (histc): 1.3435760719302e-05
- method 5 (image): 2.7342570776026e-05
sum of logical is fastest, histc and imquantize are ten times slower. Kyle's method is surprisingly not too bad but still slower.
- with matrices
x = rand(2000);
m1 = 0.25; m2 = 0.5; m3 = 0.75;
%method 1: my solution
timeit(@() 1 + (x >= m1) + (x >= m2) + (x >= m3), 1)
%method 2: Star's solution
timeit(@() 1*(x < m1) + 2*((x >= m1) & (x < m2)) + 3*((x >= m2) & (x < m3)) + 4*(x >= m3), 1)
%method 3: Kyle's solution
timeit(@() sum([1 floor(x./[m1 m2 m3]) ./ (floor(x./[m1 m2 m3]) + eps)]), 1)
%method 4: histc
timeit(@() histc(x, [-Inf m1 m2 m3 Inf]), 2)
%method 5: Image's solution
timeit(@() imquantize(x, [m1 m2 m3]), 1)
- method 1 (mine): 0.0722110642432628
- method 2 (star): 0.173027548846752
- method 3 (kyle): Does not work!
- method 4 (histc): 0.0570872784882915
- method 5 (image): 0.05592453728061645
histc and imquantize are slightly faster (probably because they only loop once over the matrix). Kyle's method does not work on matrices / vectors
更多回答（1 个）
Image Analyst
2015-3-11
There sure is, if you have the Image Processing Toolbox . The function is called imquantize() . Here's a full demo:
% T=1, if x < m1
% T=2, if x >= m1 && x < m2
% T=3, if x >= m2 && x < m3
% T=4, if x >= m3
x = imread('cameraman.tif');
subplot(1,2,1);
imshow(x);
levels = [50, 150, 230]; % The "m" threshold levels.
classifiedImage = uint8(imquantize(x, levels));
subplot(1, 2, 2);
imshow(classifiedImage, []);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
There is logic, but it's hidden - internal to the imquantize() function so you don't see it and you don't have to explicitly do logical operations yourself, you just call the function.
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