What matlab command, or combination of commands (using 25 characters or less), could be used to create the following matrix?
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
Hint: Look for a lower triangluar matrix that is repeated many times. First try to produce that lower triangular matrix with as few characters as possible.
You can use the following code:
A = ones(5); % create 5x5 matrix with all elements 1
B = tril(A); % return the lower triangle matrix of A
C = repmat(B, 3, 2); % repeat the matrix B, 3 times in a row and 2 times in a cloumn as you want.
In more details:
A = ones(5);
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
B = tril(A);
1 0 0 0 0
1 1 0 0 0
1 1 1 0 0
1 1 1 1 0
1 1 1 1 1
C = repmat(B, 3, 2);
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 1 0 0 0 0
1 1 0 0 0 1 1 0 0 0
1 1 1 0 0 1 1 1 0 0
1 1 1 1 0 1 1 1 1 0
1 1 1 1 1 1 1 1 1 1
A solution using rempat and implicit expansion:
repmat(1:5<(2:6)',3,2)
Related
I have created this map of Jamaica using matrix A. I want to insert text labels on this image at specific indexes for cities. For Example kingston on this map is at point 15, 38, where 15 is the row and 38 the column, this point I would label "kingston". My matrix is below and the image of it generated from imagesc is below as well. I was playing around with get(gca, 'position') but that was not successful.
% Cities
kingston = [15 38];
montegoBay = [4 15];
portRoyal = [18 31];
stThomas = [10 55];
mandeville = [13 21];
ochoRios = [2 29];
A = [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 3 3 0 0 0 0 3 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 3 0 0 0 0 3 0 3 3 3 0 0 0 0 3 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 3 3 3 0 3 0 3 3 3 3 0 3 0 3 0 3 3 3 0 3 3 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 3 3 3 0 0 0 3 3 3 3 3 3 0 3 0 3 0 0 0 0 0 0 3 0 0 0 3 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 0 0 3 3 0 0 0 3 3 3 3 3 0 3 3 3 3 0 0 0 0 0 0 0 0 0 3 0 0 0 3 3 3 0 3 3 3 3 0 0 3 0 0 1 1 1 1 1 1 1 1 1 1;
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 3 3 3 0 0 3 3 0 3 0 0 0 0 3 3 3 3 0 0 0 0 0 0 0 0 1 1 1 1 1 1;
1 0 3 0 3 3 0 0 0 3 3 3 0 3 0 3 3 3 3 3 0 3 3 3 0 3 3 3 3 3 3 0 0 0 0 0 0 0 0 0 0 0 3 3 3 0 0 0 0 0 0 0 0 0 0 1;
1 0 0 0 0 0 3 0 0 0 0 0 0 0 0 3 3 3 0 0 0 0 0 0 0 0 3 3 3 3 3 0 3 0 3 3 0 0 3 3 3 0 0 3 0 0 3 0 3 3 0 3 0 3 0 1;
1 1 1 0 0 0 0 0 0 3 3 3 3 0 0 3 3 3 3 3 3 3 3 0 3 3 3 0 0 0 3 3 3 3 0 3 3 0 3 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1;
1 1 1 1 1 1 1 1 0 0 0 0 0 0 3 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 3 1 0 0 0 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 3 3 0 3 0 0 3 3 3 3 3 3 0 3 0 0 0 0 0 3 3 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 3 3 3 3 0 0 0 0 0 3 3 3 3 3 3 0 3 0 3 0 3 3 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 3 3 3 0 0 0 0 0 0 0 3 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 0 3 0 1 1 1 1 1 3 0 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 0 3 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1];
Cmap = [1 1 1; 0 0 1; 0 1 0];
colormap(Cmap);
imagesc(A);
axis off
You should use the function text like this:
text(38,15,'kingston')
i.e. in opposite order, because your axis is in i-j direction, and not x-y (try typing axis xy to see what I mean).
(I have changed the font size to 15: text(38,15,'kingston','FontSize',15))
If you want to go a step further, define your cities as a cell array:
Cities = {'kingston','montegoBay','portRoyal','stThomas','mandeville','ochoRios'};
and their location in a matrix:
location = [15 38;
4 15;
18 31;
10 55;
13 21;
2 29];
And then all you need is one text command:
text(location(:,2),location(:,1),Cities,'FontSize',12)
to get the final result:
You can use text function to do that as:
text(x,y,'MyText')
See the documentation for more info:
https://www.mathworks.com/help/matlab/ref/text.html
Hope it helps.
What matlab command, or combination of commands (using 25 characters or less), could be used to create the following matrix?
1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 1 0 0 1 0 0 1 0 0 1 0 0
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
I got as far as this;
repmat(tril(ones(3,3)),5)
But repmat creates a 5 by 5 matrix. I however, need a 4,5 matrix.
Thank you for taking the time to help!
Add one more argument to repmat and remove one from ones (as Divakar noted):
repmat(tril(ones(3)),4,5)
As you can see, you can specify how many replications you want for both the rows and the columns. A single value argument to either function will use that value for both rows and columns.
I'll throw the kron solution out there. Just because.
kron(ones(4,5),tril(ones(3)))
More than 25 characters, but hey:
bsxfun(#le,mod(0:3*5-1,3),mod(0:3*4-1,3).')
assume that i got matrix ,called mat.
mat(:,:,1) =
0 1 0 1 0
1 0 1 1 0
1 0 1 0 1
1 1 1 1 0
0 0 1 1 1
mat(:,:,2) =
1 1 0 1 1
0 1 1 0 1
0 1 1 0 1
1 0 1 1 1
0 0 0 0 0
mat(:,:,3) =
1 0 0 1 1
1 1 0 0 0
1 1 1 1 0
0 1 0 0 0
0 0 0 0 1
i want to change value in each layer where it has zero in all 3 layers.
mat_result(:,:,1) =
0 1 -1 1 0
1 0 1 1 0
1 0 1 0 1
1 1 1 1 0
-1 -1 1 1 1
mat_result(:,:,2) =
1 1 -1 1 1
0 1 1 0 1
0 1 1 0 1
1 0 1 1 1
-1 -1 0 0 0
mat_result(:,:,3) =
1 0 -1 1 1
1 1 0 0 0
1 1 1 1 0
0 1 0 0 0
-1 -1 0 0 1
First, i find the position
ind = find(mat(:,:,1)==0 & mat(:,:,2)==0 & mat(:,:,3)==0)
ind =
5
10
11
In the simple way, i do this
matt1 = mat(:,:,1);
matt2 = mat(:,:,2);
matt3 = mat(:,:,3);
matt1(ind) = -1;
matt2(ind) = -1;
matt3(ind) = -1;
mat_result = cat(3,matt1,matt2,matt3);
My question is
Is there a better ways to choose/change submatrix of submatrix?
for example(but it didn't work):
mat(:,:,1)(mat(:,:,1)==0 & mat(:,:,2)==0 & mat(:,:,3)==0) = -1;
Is there a shorter way to change value in each layer where it has zero in all 3 layers?
mat(repmat(~any(mat,3),[1,1,3]))=-1
Instead of using mat(:,:,1)==0 & mat(:,:,2)==0 & mat(:,:,3)==0 you may use the any function.
Use the repmat function to replicate the logical matrix you get by using
mat(:,:,1)==0 & mat(:,:,2)==0 & mat(:,:,3)==0.
Replicate it along the 3rd channel. Final statement is:
mat(repmat(mat(:,:,1)==0 & mat(:,:,2)==0 & mat(:,:,3)==0,[1 1 3]))=-1
I have a matrix that contains data of 0 & 1. I want to find groups of ones (not a specific size) in that matrix. Is it possible somehow?
Thanks in advance!
If you mean that you want to find all the "connected components in the matrix, say BW, simply use:
BW = logical([1 1 1 0 0 0 0 0
1 1 1 0 1 1 0 0
1 1 1 0 1 1 0 0
1 1 1 0 0 0 1 0
1 1 1 0 0 0 1 0
1 1 1 0 0 0 1 0
1 1 1 0 0 1 1 0
1 1 1 0 0 0 0 0]);
L = bwlabel(BW,4) %Result
This would yeild:
L =
1 1 1 0 0 0 0 0
1 1 1 0 2 2 0 0
1 1 1 0 2 2 0 0
1 1 1 0 0 0 3 0
1 1 1 0 0 0 3 0
1 1 1 0 0 0 3 0
1 1 1 0 0 3 3 0
1 1 1 0 0 0 0 0
Now if you want to find the size of various groups:
for ii=1:max(L(:))
length_vector(ii)=length(find(L==ii));
end
length_vector
This gives you:
length_vector =
24 4 5
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
I have a 2 d matrix like this:
0 0 0 0 0 0
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 0 1 1
0 0 0 1 0 0
0 0 0 1 0 1
0 0 0 1 1 0
0 0 0 1 1 1
0 0 1 0 0 0
0 0 1 0 0 1
0 0 1 0 1 0
0 0 1 0 1 1
0 0 1 1 0 0
0 0 1 1 0 1
0 0 1 1 1 0
0 0 1 1 1 1
0 1 0 0 0 0
0 1 0 0 0 1
0 1 0 0 1 0
0 1 0 0 1 1
0 1 0 1 0 0
0 1 0 1 0 1
0 1 0 1 1 0
0 1 0 1 1 1
0 1 1 0 0 0
0 1 1 0 0 1
0 1 1 0 1 0
0 1 1 0 1 1
0 1 1 1 0 0
0 1 1 1 0 1
0 1 1 1 1 0
0 1 1 1 1 1
1 0 0 0 0 0
1 0 0 0 0 1
1 0 0 0 1 0
1 0 0 0 1 1
1 0 0 1 0 0
1 0 0 1 0 1
1 0 0 1 1 0
1 0 0 1 1 1
1 0 1 0 0 0
1 0 1 0 0 1
1 0 1 0 1 0
1 0 1 0 1 1
1 0 1 1 0 0
1 0 1 1 0 1
1 0 1 1 1 0
1 0 1 1 1 1
1 1 0 0 0 0
1 1 0 0 0 1
1 1 0 0 1 0
1 1 0 0 1 1
1 1 0 1 0 0
1 1 0 1 0 1
1 1 0 1 1 0
1 1 0 1 1 1
1 1 1 0 0 0
1 1 1 0 0 1
1 1 1 0 1 0
1 1 1 0 1 1
1 1 1 1 0 0
1 1 1 1 0 1
1 1 1 1 1 0
1 1 1 1 1 1
I want to convert this matrix into something like the following:
100000
100001
100010
and so on...
The answer from #Shai might be what you are looking for, but I read your question to ask about how to concatenate the rows of the matrix to form a vector.
It then depends on what type of format you want for your output, but here is one way to get a vector of numbers instead of the matrix.
>> A = eye(3)
A =
1 0 0
0 1 0
0 0 1
Convert the matrix into string:
>> B = num2str(A)
B =
1 0 0
0 1 0
0 0 1
Extract the chars that correspond to digits (and leave out the spaces):
>> C = B(:,1:3:end)
C =
100
010
001
Here you've got your values as strings which might be what you want (even if it is not really a vector). You could also convert this back to numbers, but then you will get an output of the following kind:
>> D = str2num(C)
D =
100
10
1
Try mat2cell
c = mat2cell( x, ones(1, size(x,1)), size(x,2) );
Though, from your example, it seems like you are more interested in dec2bin:
n = 6; % all binary numbers with n bits
c = dec2bin( 0:(2^n - 1), n );