How to apply a matlab code on a matrix if it already working on column vector.

1 次查看(过去 30 天)
MM = [3 3 3; 3 3 3; 0 0 0;0 0 0;3 3 3;3 3 3;0 0 0;0 0 0;0 0 0;3 3 0;0 3 3;0 0 3;3 0 0;3 3 0;3 3 3;0 0 0; 0 0 0;3 3 3;0 0 0];
i want to get output matrix in this shape:
output=[3 3 3;3 3 3;3 3 3;3 3 3 ;3 3 3;3 3 3;4 4 4;4 4 4;4 4 4;3 3 4;4 3 3;4 4 3;3 4 4;3 3 4;3 3 3;3 3 3;3 3 3;3 3 3;0 0 0];
% --------------------------------------------------------------
% This coding is working nicely on a column vector M.
% How to apply it on a matrix MM as given above.
%--------------------------------------------------------------
M =[3,3,0,0,3,3,0,0,0,3,3,0,0,3,3,0,0,3,0]';
m = M;
m(m == 0) = nan;
%lo - define the indices of the values that need to be changed:
lo = fillmissing(m,'previous') == 3 & fillmissing(m,'next') == 3 & isnan(m);
% ii - assign numbers to intervals:
ii = cumsum([0;diff(lo)>0]).*lo;
%{
p - count the amount of each value of vector ii:
1 element - the number of zeros (0's);
2 element the number of values in the first interval.
3 element the number of values in the second interval, etc.:
%}
p = accumarray(ii(:)+1,1);
% Let a - the values for to be changed (vector with length equal max(ii), in our case - 4):
a = [3,4,4,3]'; % HERE fixed..
% Condition for intervals with length equal 1
a(p == 1) = 3;
% Replacement:
M(lo) = a(ii(lo));
Thanks for all cooperation.
Very thankful for support in advance.
  2 个评论
M.S. Khan
M.S. Khan 2019-8-5
If we compare M and output matrix.
suppose its a column vector: [3 00 3 0003 0003 00030]' => [3 3 3 3 4 4 4 3 4 4 4 3 3 3 3 3 0]'
pattern of filling:
-- Between first and next 3, repalce 0s with 4
-- Beteeen next 3 and next coming 3, replace 0s with 4
-- next 3 and next coming 3, replace 0s with 3

请先登录,再进行评论。

采纳的回答

Andrei Bobrov
Andrei Bobrov 2019-8-12
编辑:Andrei Bobrov 2019-8-12
[r,c] = size(MM);
m = MM;
m(m == 0) = nan;
%lo -define the indices of the values that need to be changed:
lo = fillmissing(m,'previous') == 3...
& fillmissing(m,'next') == 3 & isnan(m);
% ii - assign numbers to intervals:
ii = cumsum([zeros(1,c);diff(lo)>0]).*lo;
%{
p - count the amount of each value of vector ii:
1 element - the number of zeros (0's);
2 element the number of values in the first interval.
3 element the number of values in the second interval, etc.:
%}
[y,x] = ndgrid(1:r,1:c);
p = accumarray([ii(:)+1,x(:)],1);
p = p(2:end,:);
% Let a -the values for to be changed (vector for each column
% with length equal max(ii), in our case - 4):
a = repmat([3,4,4,3]',1,c);
% Condition for intervals with length equal 1
a(p == 1) = 3;
% Replacement:
MM(lo) = a(ii(lo));

更多回答(1 个)

Chidvi Modala
Chidvi Modala 2019-8-8
I am assuming the length of vector MM is pxq
You can use the below code
for i=1:q
M=MM(:,i);
% insert your algorithm here
M(lo)=a(ii(lo));
MM(:,i)=M;
end
  4 个评论
M.S. Khan
M.S. Khan 2019-8-12
MM = [3 3 3; 3 3 3; 0 0 0;0 0 0;3 3 3;3 3 3;0 0 0;0 0 0;0 0 0;3 3 0;0 3 3;0 0 3;3 0 0;3 3 0;3 3 3;0 0 0; 0 0 0;3 3 3;0 0 0];
q = length(MM)
for i=1:q
M=MM(:,i);
m = M;
m(m == 0) = nan;
%lo - define the indices of the values that need to be changed:
lo = fillmissing(m,'previous') == 3 & fillmissing(m,'next') == 3 & isnan(m);
% ii - assign numbers to intervals:
ii = cumsum([0;diff(lo)>0]).*lo;
%{
p - count the amount of each value of vector ii:
1 element - the number of zeros (0's);
2 element the number of values in the first interval.
3 element the number of values in the second interval, etc.:
%}
p = accumarray(ii(:)+1,1);
% Let a - the values for to be changed (vector with length equal max(ii), in our case - 4):
a = [3,4,4,3]'; % HERE fixed..
% Condition for intervals with length equal 1
a(p == 1) = 3;
% Replacement:
M(lo) = a(ii(lo));
MM(:,i)=M;
end
% ----------------------------------- -----------------------
% the lenght of q is undefined.
first we should define the length of q. then we can use loop, right.
could you please explain. regards

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by