I have a matrix A , I want to remove the rows that has similar values (1,1), (2,2), (3,3)
A =
1 1
2 1
3 1
1 2
2 2
1 3
3 3
so the matrix would be like this
2 1
3 1
1 2
1 3
Another approach without calling any function:
A = A(A(:,1) == A(:,2),:)
Efficiency of this approach vs the solution based on diff():
n = 10;
y = [round(rand(n,1)) round(rand(n,1))];
tic;
for i = 1:1e4
A = y;
A(diff(A,[],2)~=0,:);
end
toc
Elapsed time is 0.091990 seconds.
tic;
for i = 1:1e4
A = y;
A = A(A(:,1) == A(:,2),:);
end
toc
Elapsed time is 0.037842 seconds.
% Suggestion of #Dan in the comments
tic;
for i = 1:1e4
A = y;
A(A(:,1) == A(:,2),:) = [];
end
toc
Elapsed time is 0.147636 seconds.
One approach using diff -
A(diff(A,[],2)~=0,:)
For a general NXM case, where M is the number of columns of A, one can extend this as -
A(any(diff(A,[],2)~=0,2),:)
Thus, if you have
A= [1 1 1;
2 2 3;
3 1 4;
8 1 2;
2 2 2;
1 3 1;
3 3 3]
You would get -
2 2 3
3 1 4
8 1 2
1 3 1
Related
I'm trying to generate an n x n matrix like
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
where n = 5 or n 50. I'm at an impasse and can only generate a portion of the matrix. It is Problem 2.14 from Numerical Methods using MATLAB 3rd Edition by Penny and Lindfield. This is the best I have so far:
n = 5;
m = n;
A = zeros(m,n);
for i = 1:m
for j = 1:n
A(i,j) = m;
end
m = m - 1;
end
Any feedback is appreciated.
That was a nice brain-teaser, here’s my solution:
[x,y] = meshgrid(5:-1:1);
out = min(x,y)
Output:
ans =
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Here's one loop-based approach:
n = 5;
m = n;
A = zeros(m, n);
for r = 1:m
for c = 1:n
A(r, c) = n+1-max(r, c);
end
end
And here's a vectorized approach (probably not faster, just for fun):
n = 5;
A = repmat(n:-1:1, n, 1);
A = min(A, A.');
That's one of the matrices in Matlab's gallery, except that it needs a 180-degree rotation, which you can achieve with rot90:
n = 5;
A = rot90(gallery('minij', n), 2);
I am using this block of code to get all the possible combinations of the rows of the matrix having thee rows. The code is as follows:
sample = [1 1 ; 2 2; 3 3];
v = [];
for i = 1:size(sample,1)-1
v = [v;(sample(i,:))];
for j = 1:size(sample,1)
if isequal(ismember(sample(j,:),v,'rows'),0)
display([v;sample(j,:)]);
else
j = j+1;
end
end
end
This code gives me the following output:
ans =
1 1
2 2
ans =
1 1
3 3
ans =
1 1
2 2
3 3
But I need output like this:
ans =
1 1
ans =
2 2
ans =
3 3
ans =
1 1
2 2
ans =
1 1
3 3
ans =
2 2
3 3
ans =
1 1
2 2
3 3
Only a small change would be enough to get the desired result.
What about this:
% getting the number of rows
n_row = size(sample,1);
% calculating all possible permutations of the rows
v = perms([1:n_row]);
disp('---')
% now we iterate over the permutations, as you want to have matrixes of 1,
% 2 and 3 rows
for i = 1: size(v,2)
idx1 = v(:,1:i);
% remove repeated answers
idx1 = unique(idx1,'rows');
% now we iterate over the answers and display
for j = 1:size(idx1,1)
idx2 = idx1(j,:);
answer = sample(idx2,:);
disp(answer)
disp('---')
end
end
I have a list of numbers, [1:9], that I need to divide three groups. Each group must contain at least one number. I need to enumerate all of the combinations (i.e. order does not matter). Ideally, the output is a x by 3 array. Any ideas of how to do this in matlab?
Is this what you want:
x = 1:9;
n = length(x);
T=3;
out = {};
%// Loop over all possible solutions
for k=1:T^n
s = dec2base(k, T, n);
out{k}{T} = [];
for p=1:n
grpIndex = str2num(s(p))+1;
out{k}{grpIndex} = [out{k}{grpIndex} x(p)];
end
end
%// Print result. size of out is the number of ways to divide the input. out{k} contains 3 arrays with the values of x
out
Maybe this is what you want. I'm assuming that the division in groups is "monotonous", that is, first come the elements of the first group, then those of the second etc.
n = 9; %// how many numbers
k = 3; %// how many groups
b = nchoosek(1:n-1,k-1).'; %'// "breaking" points
c = diff([ zeros(1,size(b,2)); b; n*ones(1,size(b,2)) ]); %// result
Each column of c gives the sizes of the k groups:
c =
Columns 1 through 23
1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 5
1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 1
7 6 5 4 3 2 1 6 5 4 3 2 1 5 4 3 2 1 4 3 2 1 3
Columns 24 through 28
5 5 6 6 7
2 3 1 2 1
2 1 2 1 1
This produces what I was looking for. The function nchoosekr_rec() is shown below as well.
for x=1:7
numgroups(x)=x;
end
c=nchoosekr_rec(numgroups,modules);
i=1;
d=zeros(1,modules);
for x=1:length(c(:,1))
c(x,modules+1)=sum(c(x,1:modules));
if c(x,modules+1)==length(opt_mods)
d(i,:)=c(x,1:modules);
i=i+1;
end
end
numgroups=[];
for x=1:length(opt_mods)
numgroups(x)=x;
end
count=0;
for x=1:length(d(:,1))
combos=combnk(numgroups,d(x,1));
for y=1:length(combos(:,1))
for z=1:nchoosek(9-d(x,1),d(x,2))
new_mods{count+z,1}=combos(y,:);
numgroups_temp{count+z,1}=setdiff(numgroups,new_mods{count+z,1});
end
count=count+nchoosek(9-d(x,1),d(x,2));
end
end
count=0;
for x=1:length(d(:,1))
for y=1:nchoosek(9,d(x,1))
combos=combnk(numgroups_temp{count+1},d(x,2));
for z=1:length(combos(:,1))
new_mods{count+z,2}=combos(z,:);
new_mods{count+z,3}=setdiff(numgroups_temp{count+z,1},new_mods{count+z,2});
end
count=count+length(combos(:,1));
end
end
function y = nchoosekr_rec(v, n)
if n == 1
y = v;
else
v = v(:);
y = [];
m = length(v);
if m == 1
y = zeros(1, n);
y(:) = v;
else
for i = 1 : m
y_recr = nchoosekr_rec(v(i:end), n-1);
s_repl = zeros(size(y_recr, 1), 1);
s_repl(:) = v(i);
y = [ y ; s_repl, y_recr ];
end
end
end
My input is the following:
X = [1 1; 1 2; 1 3; 1 4; 2 5; 1 6; 2 7; 1 8];
X =
1 1
1 2
1 3
1 4
2 5
1 6
2 7
1 8
I am looking to output a new matrix based on the value of the first column. If the value is equal to 1 -- the output will remain the same, when the value is equal to 2 then I would like to output two of the values contained in the second row. Like this:
Y =
1
2
3
4
5
5
6
7
7
8
Where 5 is output two times because the value in the first column is 2 and the same for 7
Here it is (vectorized):
C = cumsum(X(:,1))
A(C) = X(:,2)
D = hankel(A)
D(D==0) = inf
Y = min(D)
Edit:
Had a small bug, now it works.
% untested code:
Y = []; % would be better to pre-allocate
for ii = 1:size(X,1)
Y = [Y; X(ii,2)*ones(X(ii,1),1)];
end
I have a matrix which has the following form:
M =
[1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
I want to split this matrix based on the number given in the first column. So I want to split the matrix into this:
A =
[1 4 56 1;
1 3 5 1;
1 3 6 4]
B =
[2 3 5 0;
2 0 0 0]
C =
[3 1 2 3;
3 3 3 3]
I tried this by making the following loop, but this gave me the desired matrices with rows of zeros:
for i = 1:length(M)
if (M(i,1) == 1)
A(i,:) = M(i,:);
elseif (M(i,1) == 2)
B(i,:) = M(i,:);
elseif (M(i,1) == 3)
C(i,:) = M(i,:);
end
end
The result for matrix C is then for example:
C =
[0 0 0 0;
0 0 0 0;
0 0 0 0;
2 3 5 0;
2 0 0 0]
How should I solve this issue?
Additional information:
The actual data has a date in the first column in the form yyyymmdd. The data set spans several years and I want to split this dataset in matrices for each year and after that for each month.
You can use arrayfun to solve this task:
M = [
1 4 56 1;
1 3 5 1;
1 3 6 4;
2 3 5 0;
2 0 0 0;
3 1 2 3;
3 3 3 3]
A = arrayfun(#(x) M(M(:,1) == x, :), unique(M(:,1)), 'uniformoutput', false)
The result A is a cell array and its contents can be accessed as follows:
>> a{1}
ans =
1 4 56 1
1 3 5 1
1 3 6 4
>> a{2}
ans =
2 3 5 0
2 0 0 0
>> a{3}
ans =
3 1 2 3
3 3 3 3
To split the data based on an yyyymmdd format in the first column, you can use the following:
yearly = arrayfun(#(x) M(floor(M(:,1)/10000) == x, :), unique(floor(M(:,1)/10000)), 'uniformoutput', false)
monthly = arrayfun(#(x) M(floor(M(:,1)/100) == x, :), unique(floor(M(:,1)/100)), 'uniformoutput', false)
If you don't know how many outputs you'll have, it is most convenient to put the data into a cell array rather than into separate arrays. The command to do this is MAT2CELL. Note that this assumes your data is sorted. If it isn't use sortrows before running the code.
%# count the repetitions
counts = hist(M(:,1),unique(M(:,1));
%# split the array
yearly = mat2cell(M,counts,size(M,2))
%# if you'd like to split each cell further, but still keep
%# the data also grouped by year, you can do the following
%# assuming the month information is in column 2
yearByMonth = cellfun(#(x)...
mat2cell(x,hist(x(:,2),unique(x(:,2)),size(x,2)),...
yearly,'uniformOutput',false);
You'd then access the data for year 3, month 4 as yearByMonth{3}{4}
EDIT
If the first column of your data is yyyymmdd, I suggest splitting it into three columns yyyy,mm,dd, like below, to facilitate grouping afterward:
ymd = 20120918;
yymmdd = floor(ymd./[10000 100 1])
yymmdd(2:3) = yymmdd(2:3)-100*yymmdd(1:2)