I have AB in base 16 and want to convert that to base 8.
A base 16 = 12 base 8
B base 16 = 13 base 8
we have ...
AB base 16 = 12 base 8 x 20^1 base 8 + 13 base 8 x 20^0 base 8 = 253 base 8
and now
Can someone explain where the hell 20 comes out from the base conversion?????
Thanks in advance!
Where the 20 comes from: 10 in base 16 = 16 in base 10 = 20 in base 8
So the calculation can be thought of as: AB in base 16 = A * 10^1 + B * 10^0 in base 16 = 12 * 20^1 + 13 * 20^0 in base 8 = 240 + 13 in base 8 = 253 in base 8
Related
Can any one help creating spiral matrix in matlab using only loops and if else conditions.
For example n=5, spiral matrix is:
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23 24 25
There is a function spiral in your MATLAB installation, doing exactly what you want.
>> spiral(5)
ans =
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
You can view the source code typing edit spiral
Try this:
nn = input('');
n = floor(1+(nn)/2);
a = zeros(nn,nn);
i=n;j=n;m=1;br=true;
if rem(nn,2)==0
j=n-1;
nn=nn+2;
end
for p=1:2:nn
k=0;
while k<p-2
k=k+1;
a(i,j)=m;
i=i-1;
m=m+1;
end
k=0;
while k<p-1
k=k+1;
a(i,j)=m;
j=j-1;
m=m+1;
end
k=0;
while k<p-1
if j<1
br = false;
break
end
k=k+1;
a(i,j)=m;
i=i+1;
m=m+1;
end
if ~br
break
end
k=0;
while k<p
k=k+1;
a(i,j)=m;
j=j+1;
m=m+1;
end
end
disp(a)
Here is a sample run:
Enter the number:
5
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23 24 25
Another one, this time using an even number:
Enter the number:
6
36 35 34 33 32 31
17 16 15 14 13 30
18 5 4 3 12 29
19 6 1 2 11 28
20 7 8 9 10 27
21 22 23 24 25 26
Explanation: It starts with the central cell in the case of an odd number as input, and the bottom-left central cell in the case of an even input. It then, starting with 1 as the value and taking one circulation at a time, moves outwards, traverses right, up, left, down, and right again, incrementing the value to be assigned with each step, until the entire matrix is full.
Here is a custom function SpiralMatrix to construct the spiral matrix as your requested
function M = SpiralMatrix(n)
M = zeros(n);
% start from element M(1,1)
i = 1;
j = 1;
s = 1; % first element assigned to M(1,1)
M(i,j) = s;
while true
% fill row from left to right
idx = find(M(i,:)==0,1,'last');
M(i,j:idx) = s + (0:(idx-j));
s = s + idx - j;
j = idx;
% fill column from top to bottom
idx = find(M(:,j)==0,1,'last');
M(i:idx,j) = s + (0:(idx-i));
s = s + idx - i;
i = idx;
% fill row from right to left
idx = find(M(i,:)==0,1,'first');
M(i,j:-1:idx) = s + (0:(j-idx));
s = s + j - idx;
j = idx;
% fill column from bottom to top
idx = find(M(:,j)==0,1,'first');
M(i:-1:idx,j) = s + (0:(i-idx));
s = s + i-idx;
i = idx;
% break if matrix if fully filled
if nnz(M) == n^2
break;
end
end
M = n^2+1-fliplr(flipud(M));
end
such that
>> SpiralMatrix(5)
ans =
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23 24 25
>> SpiralMatrix(7)
ans =
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
I would like to draw a bar chart with "unequal samples". Here is an example code
A = [16 20 15 17 22 19 17]';
B = [22 15 16 16 16 18]';
C = [23 9 15 18 13 27 17 14 16 15 21 19 17]';
group = [ ones(size(A));
2 * ones(size(B));
3 * ones(size(C))];
figure
boxplot([A; B; C],group)
set(gca,'XTickLabel',{'A','B','C'})
The output is as below:
However, I would like to have a distance between group1,2 with group 3. As same as what you see in the figure below:(this figure is just a copy paste from another source but the distance between box plot of each group is visible)
I tried to use 'factorgap' in such command
figure
boxplot([A; B; C ],group,'factorgap',[50,1])
However, because the number of samples in each group is different it did not work.
Any suggestion?
The first solution I propose you is in fact a small workaround that consists in inserting another, invisible group between the second and the third one:
A = [16 20 15 17 22 19 17]';
B = [22 15 16 16 16 18]';
C = [23 9 15 18 13 27 17 14 16 15 21 19 17]';
group = [
ones(size(A));
2 * ones(size(B));
3;
4 * ones(size(C))
];
figure();
boxplot([A; B; NaN; C],group);
set(gca,'XTickLabel',{'A','B','','C'});
Here is the output:
Now, let's build up something serious:
% Define the sample data...
A = [16 20 15 17 22 19 17]';
B = [22 15 16 16 16 18]';
C = [23 9 15 18 13 27 17 14 16 15 21 19 17]';
% Find the length of the largest vector...
A_len = numel(A);
B_len = numel(B);
C_len = numel(C);
max_len = max([A_len B_len C_len]);
% Transform vectors into fixed size vectors of length max_len...
A = [A; NaN(max_len - A_len,1)];
B = [B; NaN(max_len - B_len,1)];
C = [C; NaN(max_len - C_len,1)];
% Define labels and groups...
L1 = [repmat('A',1,numel(A)),repmat('B',1,numel(B))];
L2 = repmat('C',1,numel(C));
L = [L1 L2];
G = [repmat('1',1,numel(L1)),repmat('2',1,numel(L2))];
% Plot the boxes...
boxplot([A B C],{G';L'},'FactorGap',50);
Here is the output:
I have a matrix train3.
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37
I want to select only those rows of whose 1st columns contain specific values (in my case 1 and 2).
I have tried the following,
>> train3
train3 =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37
>> ind1 = train3(:,1) == 1
ind1 =
1
0
0
0
>> ind2 = train3(:,1) == 2
ind2 =
0
1
0
0
>> mat1 = train3(ind1, :)
mat1 =
1 2 3 4 5 6 7
>> mat2 = train3(ind2, :)
mat2 =
2 12 13 14 15 16 17
>> mat3 = [mat1 ; mat2]
mat3 =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
>>
Is there any better way to do this?
Presumably you are trying to get mat3 in a single step which you can do with:
mat3 = train3(train3(:,1)==1 | train3(:,1)==2,:)
A more general way to do this would be to use ismember to get all of the rows that match the values in a list:
train3 =[
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37];
chooseList = [1 2];
colIndex = ismember(train3(:, 1), chooseList);
subset = train3(colIndex, :);
subset =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
I'm trying to implement the Baker map.
Is there a function that would allow one to divide a 8 x 8 matrix by providing, for example, a sequence of divisors 2, 4, 2 and rearranging pixels in the order as shown in the matrices below?
X = reshape(1:64,8,8);
After applying divisors 2,4,2 to the matrix X one should get a matrix like A shown below.
A=[31 23 15 7 32 24 16 8;
63 55 47 39 64 56 48 40;
11 3 12 4 13 5 14 6;
27 19 28 20 29 21 30 22;
43 35 44 36 45 37 46 38;
59 51 60 52 61 53 62 54;
25 17 9 1 26 18 10 2;
57 49 41 33 58 50 42 34]
The link to the document which I am working on is:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.5132&rep=rep1&type=pdf
This is what I want to achieve:
Edit: a little more generic solution:
%function Z = bakermap(X,divisors)
function Z = bakermap()
X = reshape(1:64,8,8)'
divisors = [ 2 4 2 ];
[x,y] = size(X);
offsets = sum(divisors)-fliplr(cumsum(fliplr(divisors)));
if any(mod(y,divisors)) && ~(sum(divisors) == y)
disp('invalid divisor vector')
return
end
blocks = #(div) cell2mat( cellfun(#mtimes, repmat({ones(x/div,div)},div,1),...
num2cell(1:div)',...
'UniformOutput',false) );
%create index matrix
I = [];
for ii = 1:numel(divisors);
I = [I, blocks(divisors(ii))+offsets(ii)];
end
%create Baker map
Y = flipud(X);
Z = [];
for jj=1:I(end)
Z = [Z; Y(I==jj)'];
end
Z = flipud(Z);
end
returns:
index matrix:
I =
1 1 3 3 3 3 7 7
1 1 3 3 3 3 7 7
1 1 4 4 4 4 7 7
1 1 4 4 4 4 7 7
2 2 5 5 5 5 8 8
2 2 5 5 5 5 8 8
2 2 6 6 6 6 8 8
2 2 6 6 6 6 8 8
Baker map:
Z =
31 23 15 7 32 24 16 8
63 55 47 39 64 56 48 40
11 3 12 4 13 5 14 6
27 19 28 20 29 21 30 22
43 35 44 36 45 37 46 38
59 51 60 52 61 53 62 54
25 17 9 1 26 18 10 2
57 49 41 33 58 50 42 34
But have a look at the if-condition, it's just possible for these cases. I don't know if that's enough. I also tried something like divisors = [ 1 4 1 2 ] - and it worked. As long as the sum of all divisors is equal the row-length and the modulus as well, there shouldn't be problems.
Explanation:
% definition of anonymous function with input parameter: div: divisor vector
blocks = #(div) cell2mat( ... % converts final result into matrix
cellfun(#mtimes, ... % multiplies the next two inputs A,B
repmat(... % A...
{ones(x/div,div)},... % cell with a matrix of ones in size
of one subblock, e.g. [1,1,1,1;1,1,1,1]
div,1),... % which is replicated div-times according
to actual by cellfun processed divisor
num2cell(1:div)',... % creates a vector [1,2,3,4...] according
to the number of divisors, so so finally
every Block A gets an increasing factor
'UniformOutput',false...% necessary additional property of cellfun
));
Have also a look at this revision to have a simpler insight in what is happening. You requested a generic solution, thats the one above, the one linked was with more manual inputs.
I got a data set in a matrix like the following (imported from Excel):
matrix =
Cat1 1 2 3 4
Cat2 9 10 11 12
Cat3 17 18 19 20
Cat1 5 6 7 8
Cat2 13 14 15 16
Cat3 21 22 23 24
I would like to reshape it into 3 vectors (one for every category) of the same size to do a stacked bar plot. Vectors should look like this after reshape operation (It would be nice if the vector had the name of the first column and the matrix could be of any size):
cat1 = [ 1 2 3 4 5 6 7 8]
cat2 = [ 9 10 11 12 13 14 15 16]
cat3 = [17 18 19 20 21 22 23 24]
I sincerely hope this is not duplicate. I couldn't produce a working solution with the help of the other reshape questions.
If your data is a matrix, you can manipulate the order of the rows when indexing, so you can do something like this:
rows = reshape(1:size(matrix, 1), n, []).';
res = reshape(matrix(rows, :).', [], n).';
The resulting matrix res is composed of the concatenated rows.
This solution holds for cell arrays as well, but you'll need an additional cell2mat to turn the result into a matrix.
Example
matrix = [1:4; 9:12; 17:20; 5:8; 13:16; 21:24];
n = 3;
rows = reshape(1:size(matrix, 1), n, []).';
res = reshape(matrix(rows, :).', [], n).';
The result is:
res =
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
EDIT:
Try the following:
%# dataset stored in a cell array
data = {
'Cat1' 1 2 3 4
'Cat2' 9 10 11 12
'Cat3' 17 18 19 20
'Cat1' 5 6 7 8
'Cat2' 13 14 15 16
'Cat3' 21 22 23 24
};
%# get all possible values of first column,
%# and map them to integer indices
[L,~,IDX] = unique(data(:,1));
%# for each possible "category"
groups = cell(max(IDX),1);
for i=1:max(IDX)
%# get the rows of numeric data matching current category
M = data(IDX==i, 2:end)';
%# flatten matrix into a vector and store in cell (row-major order)
groups{i} = [M{:}];
end
Now you can access the i-th "cat" vector as: groups{i}
>> [cat1,cat2,cat3] = deal(groups{:})
cat1 =
1 2 3 4 5 6 7 8
cat2 =
9 10 11 12 13 14 15 16
cat3 =
17 18 19 20 21 22 23 24
Note that the matching "cat" labels are stored in L{i} (the mapping keys)