필터 지우기
필터 지우기

Boxcount in a DEM ALOS image

조회 수: 1 (최근 30일)
Arlete Conde
Arlete Conde 2022년 7월 26일
Hello. I am trying to make the boxcount in a satellite image (DEM ALOS) and this was the result:
Error using logical
NaN values cannot be converted to logicals.
Error in boxcount (line 64)
c = logical(squeeze(c));
Error in Alos_clip01 (line 6)
boxcount(c)
c = imread('Alos_clip01.tif');
imagesc(c)
colormap gray
axis square
%%
boxcount(c)
%%
[n, r] = boxcount(c);
loglog(r, n,'bo-', r, (r/r(end)).^(-2), 'r--')
xlabel('r')
ylabel('n(r)')
legend('actual box-count','space-filling box-count');
%%
boxcount(c, 'slope')
Boxcount code:
function [n,r] = boxcount(c,varargin)
%BOXCOUNT Box-Counting of a D-dimensional array (with D=1,2,3).
% [N, R] = BOXCOUNT(C), where C is a D-dimensional array (with D=1,2,3),
% counts the number N of D-dimensional boxes of size R needed to cover
% the nonzero elements of C. The box sizes are powers of two, i.e.,
% R = 1, 2, 4 ... 2^P, where P is the smallest integer such that
% MAX(SIZE(C)) <= 2^P. If the sizes of C over each dimension are smaller
% than 2^P, C is padded with zeros to size 2^P over each dimension (e.g.,
% a 320-by-200 image is padded to 512-by-512). The output vectors N and R
% are of size P+1. For a RGB color image (m-by-n-by-3 array), a summation
% over the 3 RGB planes is done first.
%
% The Box-counting method is useful to determine fractal properties of a
% 1D segment, a 2D image or a 3D array. If C is a fractal set, with
% fractal dimension DF < D, then N scales as R^(-DF). DF is known as the
% Minkowski-Bouligand dimension, or Kolmogorov capacity, or Kolmogorov
% dimension, or simply box-counting dimension.
%
% BOXCOUNT(C,'plot') also shows the log-log plot of N as a function of R
% (if no output argument, this option is selected by default).
%
% BOXCOUNT(C,'slope') also shows the semi-log plot of the local slope
% DF = - dlnN/dlnR as a function of R. If DF is contant in a certain
% range of R, then DF is the fractal dimension of the set C. The
% derivative is computed as a 2nd order finite difference (see GRADIENT).
%
% The execution time depends on the sizes of C. It is fastest for powers
% of two over each dimension.
%
% Examples:
%
% % Plots the box-count of a vector containing randomly-distributed
% % 0 and 1. This set is not fractal: one has N = R^-2 at large R,
% % and N = cste at small R.
% c = (rand(1,2048)<0.2);
% boxcount(c);
%
% % Plots the box-count and the fractal dimension of a 2D fractal set
% % of size 512^2 (obtained by RANDCANTOR), with fractal dimension
% % DF = 2 + log(P) / log(2) = 1.68 (with P=0.8).
% c = randcantor(0.8, 512, 2);
% boxcount(c);
% figure, boxcount(c, 'slope');
%
% F. Moisy
% Revision: 2.10, Date: 2008/07/09
% History:
% 2006/11/22: v2.00, joined into a single file boxcountn (n=1,2,3).
% 2008/07/09: v2.10, minor improvements
% control input argument
narginchk(1,2);
% check for true color image (m-by-n-by-3 array)
if ndims(c)==3
if size(c,3)==3 && size(c,1)>=8 && size(c,2)>=8
c = sum(c,3);
end
end
warning off
c = logical(squeeze(c));
warning on
dim = ndims(c); % dim is 2 for a vector or a matrix, 3 for a cube
if dim>3
error('Maximum dimension is 3.');
end
% transpose the vector to a 1-by-n vector
if length(c)==numel(c)
dim=1;
if size(c,1)~=1
c = c';
end
end
width = max(size(c)); % largest size of the box
p = log(width)/log(2); % nbre of generations
% remap the array if the sizes are not all equal,
% or if they are not power of two
% (this slows down the computation!)
if p~=round(p) || any(size(c)~=width)
p = ceil(p);
width = 2^p;
switch dim
case 1
mz = zeros(1,width);
mz(1:length(c)) = c;
c = mz;
case 2
mz = zeros(width, width);
mz(1:size(c,1), 1:size(c,2)) = c;
c = mz;
case 3
mz = zeros(width, width, width);
mz(1:size(c,1), 1:size(c,2), 1:size(c,3)) = c;
c = mz;
end
end
n=zeros(1,p+1); % pre-allocate the number of box of size r
switch dim
case 1 %------------------- 1D boxcount ---------------------%
n(p+1) = sum(c);
for g=(p-1):-1:0
siz = 2^(p-g);
siz2 = round(siz/2);
for i=1:siz:(width-siz+1)
c(i) = ( c(i) || c(i+siz2));
end
n(g+1) = sum(c(1:siz:(width-siz+1)));
end
case 2 %------------------- 2D boxcount ---------------------%
n(p+1) = sum(c(:));
for g=(p-1):-1:0
siz = 2^(p-g);
siz2 = round(siz/2);
for i=1:siz:(width-siz+1)
for j=1:siz:(width-siz+1)
c(i,j) = ( c(i,j) || c(i+siz2,j) || c(i,j+siz2) || c(i+siz2,j+siz2) );
end
end
n(g+1) = sum(sum(c(1:siz:(width-siz+1),1:siz:(width-siz+1))));
end
case 3 %------------------- 3D boxcount ---------------------%
n(p+1) = sum(c(:));
for g=(p-1):-1:0
siz = 2^(p-g);
siz2 = round(siz/2);
for i=1:siz:(width-siz+1)
for j=1:siz:(width-siz+1)
for k=1:siz:(width-siz+1)
c(i,j,k)=( c(i,j,k) || c(i+siz2,j,k) || c(i,j+siz2,k) ...
|| c(i+siz2,j+siz2,k) || c(i,j,k+siz2) || c(i+siz2,j,k+siz2) ...
|| c(i,j+siz2,k+siz2) || c(i+siz2,j+siz2,k+siz2));
end
end
end
n(g+1) = sum(sum(sum(c(1:siz:(width-siz+1),1:siz:(width-siz+1),1:siz:(width-siz+1)))));
end
end
n = n(end:-1:1);
r = 2.^(0:p); % box size (1, 2, 4, 8...)
if any(strncmpi(varargin,'slope',1))
s=-gradient(log(n))./gradient(log(r));
semilogx(r, s, 's-');
ylim([0 dim]);
xlabel('r, box size'); ylabel('- d ln n / d ln r, local dimension');
title([num2str(dim) 'D box-count']);
elseif nargout==0 || any(strncmpi(varargin,'plot',1))
loglog(r,n,'s-');
xlabel('r, box size'); ylabel('n(r), number of boxes');
title([num2str(dim) 'D box-count']);
end
if nargout==0
clear r n
end

답변 (0개)

카테고리

Help CenterFile Exchange에서 Fractals에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by