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:
Related
Take a 4-D matrix A with s1=size(A,1), s2=size(A,2), s3=size(A,3), s4=size(A,4).
Consider
B=reshape(A, s1*s2*s3*s4,1)
For example,
s1=2;
s2=3;
s3=2;
s4=4;
A(:,:,1,1)=[1 2 3; 4 5 6];
A(:,:,1,2)=[7 8 9; 10 11 12];
A(:,:,1,3)=[13 14 15; 16 17 18];
A(:,:,1,4)=[19 20 21; 22 23 24];
A(:,:,2,1)=[25 26 27; 28 29 30];
A(:,:,2,2)=[31 32 33; 34 35 36];
A(:,:,2,3)=[37 38 39; 40 41 42];
A(:,:,2,4)=[43 44 45; 46 47 48];
B=reshape(A, s1*s2*s3*s4,1);
%B=[1;4;2;5;3;6;25;28;26;29;27;30;7;10;8;11;9;12;...];
Given coordinates (i,j,h,k), I would like your help to write a function that gives me the position of A(i,j,h,k) in B.
I know how to do that when A is a 3-D matrix. In that case,
position_in_B= i + ( j-1 + (h-1)*s2 ) * s1;
How can I extend this result to 4-D matrices?
Here you can use sub2ind , for example:
B(sub2ind(size(A),1,1,1,4))
or you can continue with the extension you wrote:
position_in_B= #(i,j,h,k) i + ( j-1 + ( h-1 + (k-1) *s3 ) *s2 ) * s1;
B(position_in_B(1,1,1,4))
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 have the example code below:
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'});
what can I add to the code to show the mean of each vector in the box plot instead of the median (which is a Matlab default), I know how to do it for one vector but if we have multiple boxplots, I need some help how to do it.
any suggestion, please?
Try removing the median line and then plotting the relevant means
:
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]';
% Calculate means
meanOfA = mean(A);
meanOfB = mean(B);
meanOfC = mean(C);
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'});
% Find handle for median line and set visibility off
h = findobj(gca,'Tag','Median');
set(h,'Visible','off');
%plot means as black asterisks.
hold on
plot(1,meanOfA, 'k*')
plot(2,meanOfB, 'k*')
plot(4,meanOfC, 'k*')
I have two matrices in Matlab A and B respectively of dimension MxN and GxN.
M can be greater or smaller than G.
A and B do not contain identical rows.
I want to construct a matrix C of dimension Hx(N+2) in the following way
C=[];
for i=1:size(A,1)
%if A(i,1:end-1) coincides with a row in B(:,1:end-1) (it can coincide with only one row at most)
%then C=[C;A(i,1:end) B(j,end)]; %where j is the index of the identical row in B
%otherwise impose C=[C;A(i,1:end) 0]
end
for i=1:size(B,1)
%if B(i,1:end-1) does not coincide with any row in A(:,1:end-1)
%then impose C=[C; B(i,1:end-1) 0 B(i,end)];
end
For example:
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104]
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104]
As M and G can be very high, I am looking for the fastest way to perform this.
You can use ismember + indexing to do your task:
[idx1,idx2] = ismember(A(:,1:end-1), B(:,1:end-1), 'rows');
idx3 = ~ismember(B(:,1:end-1), A(:,1:end-1), 'rows');
C(idx1,:) = [A(idx1,:) B(idx2(idx1),end)];
C(~idx1,:) = [A(~idx1,:) zeros(sum(~idx1),1)];
C=[C;B(idx3,1:end-1) zeros(sum(idx3),1) B(idx3,end)];
You could also use intersect with a bit of preallocation to speed up the assignment (if M or G gets really large).
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104];
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104];
[M,N] = size(A);
G = size(B,1);
[tmp, idxA, idxB] = intersect(A(:,1:end-1),B(:,1:end-1),'rows')
idxBnotA = setdiff([1:G],idxB);
H = M + G - length(idxA);
C1 = zeros(H,N+1);
C1(1:M,1:N) = A;
C1(idxA,end) = B(idxB,end);
C1(M+1:end,1:end-2) = B(idxBnotA,1:end-1);
C1(M+1:end,end) = B(idxBnotA,end)
I would like to transform the matrix A into the matrix B without using cells (e.g. mat2cell) in Matlab, where
A=[1 2 3;
4 5 6;
7 8 9;
10 11 12;
13 14 15;
16 17 18;
19 20 21;
22 23 24;
25 26 27];
B=[1 2 3 10 11 12 19 20 21;
4 5 6 13 14 15 22 23 24;
7 8 9 16 17 18 25 26 27];
All you need is some reshape + permute magic -
N = 3; %// Cut after every N rows and this looks like the no. of columns in A
B = reshape(permute(reshape(A,N,size(A,1)/N,[]),[1 3 2]),N,[])
This builds a linear index to rearrange the entries of A and then reshapes into the desired matrix B:
m = 3; %// cut size in rows of A. Assumed to divide size(A,1)
n = size(A,2);
p = size(A,1);
ind = bsxfun(#plus, ...
bsxfun(#plus, (1:m).', (0:n-1)*p), permute((0:p/m-1)*m, [1 3 2]));
B = reshape(A(ind(:)), m, [])