I have these kind of rows
0 1 1
I would like to multiply it by let's say 2 or 4 to get this pattern
0 0 0 0 1 1 1 1 1 1 1 1
Now, I have some piece of old code, which basically does this in the case of multiplying by 5.
But I cannot convert this script to do it for example 2 or 4 times...
Can anyone help me to figure it out?
Here is the code:
sed -e 's/\([01]\)/\1\1\1\1/7g ; s/\([01]\{2,\}\)/\1\1\1/g ; s/\b\([01]\)\b/\1\1\1\1\1/g ; s/\([01]\)\B/\1 /g'
$ echo '0 1 1' | sed -r 's/\S/& & & & &/g'
0 0 0 0 0 1 1 1 1 1 1 1 1 1 1
using sed repeat 4 times:
kent$ echo "0 1 1
1 1 0"|sed 's/[01]/& & & &/g'
0 0 0 0 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 0 0 0 0
with awk, you can give the times you want to repeat as parameter: e.g. say repeat 5 times:
kent$ echo "0 1 1
dquote> 1 1 0"|awk -v t=5 '{f=1;while(f<=NF){ n=1;while(n<=t){printf "%s ",$f;n++;}f++;} print "";}'
0 0 0 0 0 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 0 0 0 0 0
This might work for you:
echo -e '0 1 1\n1 1 1 0 0' | sed "s/\S/$(echo {1..4}| sed 's/\S*/\&/g')/g"
0 0 0 0 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
This provides an OTT solution but it is programmable i.e. change 4 to any value you wish to multiply by.
Related
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)
Is there anyway to make this expression with another gates (especially with nand, xor, nor). I almost tried every combination but I couldn't find. Allways one output is wrong.
~A~BC + AB~C
You'll need 5 gates for this, 2 NANDs, 2 ANDs and 1 OR.
Run A and B into the first NAND, and put the output of that into the AND, along with C.
Then, into the second NAND, just put C on it's own. Then, the output of this NAND goes into the second AND, along with A and B.
The output of the 2 AND gates then goes into the OR, which will make the expression you're after.
Truth table
A B C ~A ~B ~C ~A~BC AB~C ~A~BC + AB~C
0 0 0 1 1 1 0 0 0
0 0 1 1 1 0 1 0 1
0 1 0 1 0 1 0 0 0
0 1 1 1 0 0 0 0 0
1 0 0 0 1 1 0 0 0
1 0 1 0 1 0 0 0 0
1 1 0 0 0 1 0 1 1
1 1 1 0 0 0 0 0 0
So you need 2 AND gates and 1 OR gate. Now replace AND gates with NAND gates:
A B C ~A ~B ~C NAND(~A,~B,C) NAND(A,B,~C) ?
0 0 0 1 1 1 1 1 0
0 0 1 1 1 0 0 1 1
0 1 0 1 0 1 1 1 0
0 1 1 1 0 0 1 1 0
1 0 0 0 1 1 1 1 0
1 0 1 0 1 0 1 1 0
1 1 0 0 0 1 1 0 1
1 1 1 0 0 0 1 1 0
Which gate make 1 1 -> 0 and 0 1 -> 1? XOR gate.
Simple answer: ~A~BC + AB~C = NAND(~A,~B,C) ⊕ NAND(A,B,~C)
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).')
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