MATLAB: How do you add all the first n rows of a matrix within the same column, and save the result to the last row? - matlab

It was hard to phrase the question, but here's an example of what I'm looking for:
1 2 3 4
2 1 1 1
2 2 3 1
0 0 0 0
and in column one, I add all the value of all of the first three rows and save it to the third and so on, so that it becomes:
1 2 3 4
2 1 1 1
2 2 3 1
5 5 7 6


I think you can use sum:
octave:23> m = [1 2 3 4; 2 1 1 1; 2 2 3 1; 0 0 0 0]
m =
1 2 3 4
2 1 1 1
2 2 3 1
0 0 0 0
octave:24> m(length(m), :) = sum(m)
m =
1 2 3 4
2 1 1 1
2 2 3 1
5 5 7 6

Related

Matlab decision tree

I am trying to make a decision tree but the outcome is strange and I can't figure out where is wrong. There are seven variables, each of which I use 1 or 2 to represent their meaning, for example, for variable 1 the number 1 is warm and 2 is cold, for variable 2 the number 1 is yes and 2 is no.
vars = {'TEMP' 'SKIN' 'BIRTH' 'AQUATIC' 'AERIAL' 'LEGS' 'HIBER'};
x = [1 1 1 2 2 1 2
2 2 2 2 2 2 1
2 2 2 1 2 2 2
1 1 1 1 2 2 2
2 1 2 1 2 1 1
2 2 2 2 2 1 2
1 1 1 2 1 1 1
1 1 2 2 1 1 2
1 1 1 2 2 1 2
2 2 1 1 2 2 2
2 2 2 1 2 1 2
1 1 2 1 2 1 2
1 1 1 2 2 1 1
2 2 2 1 2 2 2
2 1 2 1 2 1 1];
s = {'M';'R';'F';'M';'A';'R';'M';'B';'M';'F';'R';'B';'M';'F';'A'};
y = cellstr(s);
t = classregtree(x, y, 'method','classification', 'names',vars,...
'categorical',[1 7], 'prune','off');
view(t)
The outcome is only one step tree without other information. What is wrong with this?

I'm not an expert of decision trees, anyway, playing a little bit with the parameters of classregtree (minparent, to be exact):
vars = {'TEMP' 'SKIN' 'BIRTH' 'AQUATIC' 'AERIAL' 'LEGS' 'HIBER'};
x = [1 1 1 2 2 1 2
2 2 2 2 2 2 1
2 2 2 1 2 2 2
1 1 1 1 2 2 2
2 1 2 1 2 1 1
2 2 2 2 2 1 2
1 1 1 2 1 1 1
1 1 2 2 1 1 2
1 1 1 2 2 1 2
2 2 1 1 2 2 2
2 2 2 1 2 1 2
1 1 2 1 2 1 2
1 1 1 2 2 1 1
2 2 2 1 2 2 2
2 1 2 1 2 1 1];
y = {'M';'R';'F';'M';'A';'R';'M';'B';'M';'F';'R';'B';'M';'F';'A'};
t = classregtree(x,y,'method','classification','Names',vars, ...
'categorical',[1 7],'prune','off','minparent',1);
view(t);
I've been able to reproduce something that looks fine. Anyway, since Matlab release 2011A, classregtree has become obsolete and has been superseded by fitrtree (RegressionTree) and fitctree (ClassificationTree) functions (classregtree is being kept for retrocompatibility reasons only). I recommend you to update your code and use those functions instead:
t = fitctree(x,y,'PredictorNames',vars, ...
'CategoricalPredictors',{'TEMP' 'HIBER'},'Prune','off','MinParentSize',1);

convert For syntax to a faster way in Matlab

I have below code and I want to convert it to a faster way but I don't how I can convert For syntax to a faster way in Matlab.
If user count is 5 and item count is 2 and time count is 4, I want to create this matrix:
1 1 1
1 1 2
1 1 3
1 1 4
1 2 1
1 2 2
1 2 3
1 2 4
2 1 1
2 1 2
2 1 3
2 1 4
...
result=zeros(userCount*itemCount*timeCount,4);
j=0;
for i=1:userCount
result(j*itemCount*timeCount+1:j*itemCount*timeCount+itemCount*timeCount,1)=ones(itemCount*timeCount,1)*i;
j=j+1;
end
j=0;
h=1;
for i=1:userCount*itemCount
result(j*timeCount+1:j*timeCount+timeCount,2)=ones(timeCount,1)*(h);
j=j+1;
h=h+1;
if h>itemCount
h=1;
end
end
j=0;
for i=1:userCount*itemCount
result(j*timeCount+1:j*timeCount+timeCount,3)=1:timeCount;
j=j+1;
end
for i=1:size(subs,1)
f=(result(:,1)==subs(i,1)& result(:,2)==subs(i,2));
result(f,:)=[];
end

What you are describing is to enumerate permutations for three independent linear sets. One way to achieve this would be to use ndgrid and unroll each output into a single vector:
userCount = 5; itemCount = 2; timeCount = 4;
[X,Y,Z] = ndgrid(1:timeCount,1:itemCount,1:userCount);
result = [Z(:) Y(:) X(:)];
We get:
result =
1 1 1
1 1 2
1 1 3
1 1 4
1 2 1
1 2 2
1 2 3
1 2 4
2 1 1
2 1 2
2 1 3
2 1 4
2 2 1
2 2 2
2 2 3
2 2 4
3 1 1
3 1 2
3 1 3
3 1 4
3 2 1
3 2 2
3 2 3
3 2 4
4 1 1
4 1 2
4 1 3
4 1 4
4 2 1
4 2 2
4 2 3
4 2 4
5 1 1
5 1 2
5 1 3
5 1 4
5 2 1
5 2 2
5 2 3
5 2 4

Every possible sum combination in a vector

Assuming I'm having a vectors of numbers A, for example: A=[1 3 5 3 9 6](A's length >= 2) and an Integer X=6. Need to find how many pairs (A[i],A[j]) where i<j exist in the vector which answer this condition: A[i]+A[j]=X. The number of pairs is printed.
Not allowed to use for/while. Allowed only ceil,floor,mod,repmat,reshape,size,length,transpose,sort,isempty,all,any,find ,sum,max,min.

With repmat, length and sum -
integer1 = 6; %// One of the paramters
A_ind = 1:length(A) %// Get the indices array
%// Expand A_ind into rows and A_ind' into columns, to form a meshgrid structure
A_ind_mat1 = repmat(A_ind,[length(A) 1])
A_ind_mat2 = repmat(A_ind',[1 length(A)]) %//'
%// Expand A into rows and A' into columns, to form a meshgrid structure
A_mat1 = repmat(A,[length(A) 1])
A_mat2 = repmat(A',[1 length(A)]) %//'
%// Form the binary matrix of -> (A[i],A[j]) where i<j
cond1 = A_ind_mat1 < A_ind_mat2
%// Use the binary matrix as a logical mask to select elements from the two
%// matrices and see which element pairs satisfy -> A[i]+A[j]=X and get a
%// count of those pairs with SUM
pairs_count = sum((A_mat1(cond1) + A_mat2(cond1))==integer1)
Outputs from code run to make it clearer -
A =
1 3 5 3 9 6
A_ind =
1 2 3 4 5 6
A_ind_mat1 =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
A_ind_mat2 =
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
A_mat1 =
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
A_mat2 =
1 1 1 1 1 1
3 3 3 3 3 3
5 5 5 5 5 5
3 3 3 3 3 3
9 9 9 9 9 9
6 6 6 6 6 6
cond1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 1 1 1 1 0
pairs_count =
2
A bit more explanation -
Taking few more steps to clarify why pairs_count must be 2 here -
Set all values in A_mat1 and A_mat2 to be zeros that do not satisfy the less than criteria
>> A_mat1(~cond1)=0
A_mat1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 3 0 0 0 0
1 3 5 0 0 0
1 3 5 3 0 0
1 3 5 3 9 0
>> A_mat2(~cond1)=0
A_mat2 =
0 0 0 0 0 0
3 0 0 0 0 0
5 5 0 0 0 0
3 3 3 0 0 0
9 9 9 9 0 0
6 6 6 6 6 0
Now, add A_mat1 and A_mat2 and see how many 6's you got -
>> A_mat1 + A_mat2
ans =
0 0 0 0 0 0
4 0 0 0 0 0
6 8 0 0 0 0
4 6 8 0 0 0
10 12 14 12 0 0
7 9 11 9 15 0
As you can see there are two 6's and thus our result is verified.

Iterate one vector through another in Matlab Part 2

This is a similar request to my post at Iterate one vector through another in Matlab
I am using Luis' suggestion with the following code:
E=[1 2 3 4 5 6 7 8 9 10];
A = [1 2];
s = size(E,2);
t = numel(A);
C = cell(1,s);
[C{:}] = ndgrid(A);
C = cat(s+1, C{:});
C = fliplr(reshape(C, t^s, s));
This produces a good result for C as a 1024x10 matrix with all possible permutations of 1 and 2 to a length of 10 columns. What I want to do is remove any rows that are not in increasing order. For example now I get:
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 2
1 1 1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 1 2 2
All are valid except for the third row since it goes from 2 to back to 1.
I have code to get the desired result, but it is very slow and inefficient.
E=[1 2 3 4 5 6 7 8 9 10];
A = [1 2];
s = size(E,2);
t = numel(A);
C = cell(1,s);
[C{:}] = ndgrid(A);
C = cat(s+1, C{:});
C = fliplr(reshape(C, t^s, s));
min=0;
for row=1:size(C,1)
for col=1:size(C,2)
if(C(row,col)>min)
min=C(row,col);
elseif(C(row,col)<min)
C(row,:)=0;
continue;
end
end
min=0;
end
C = C(any(C,2),:); %remove all zero rows
The desired output is now:
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 2
1 1 1 1 1 1 1 1 2 2
1 1 1 1 1 1 1 2 2 2
1 1 1 1 1 1 2 2 2 2
1 1 1 1 1 2 2 2 2 2
1 1 1 1 2 2 2 2 2 2
1 1 1 2 2 2 2 2 2 2
1 1 2 2 2 2 2 2 2 2
1 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2
Any ideas on how to optimize my code so I do not need to use nested loops?

The super-simple but not-quite-so-obvious solution via a couple of row-wise operations:
d = diff(C, [], 2);
m = min(d, [], 2);
C = C(m>=0, :);
Of course, in this particular example it would be far easier to just generate the resulting matrix directly:
C = flipud(triu(ones(s+1,s).*(max(A)-min(A))) + min(A));
but I assume you're also interested in less trivial values of A ;)

Matrix creation Octave / Matlab, loopless solution request

I want to create a matrix like
A = [0 0 0 0 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 0 1 1;
0 0 1 1 1;
0 1 1 1 1]
based on a vector indicating how many '0's should precede '1's on each row:
B = [4 3 3 3 2 1]
Is there a loopless way to do this ?

You don't mention in your question how the horizontal size of the array should be defined (the number of ones).
For predefined width you can use this code:
width = 5;
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,width-x) ], B, 'UniformOutput', false)');
If you want that A has minimal width, but still at least one 1 in every row:
A = cell2mat(arrayfun(#(x) [ zeros(1,x), ones(1,max(B)+1-x) ], B, 'UniformOutput', false)');

A shorter “old-school” way to achieve this without a loop would be as follows:
A = repmat(B',1,max(B)+1)<repmat([1:max(B)+1],size(B,2),1)
If you want to have a minimum number of ones
min_ones=1; %or whatever
A = repmat(B',1,max(B)+min_ones)<repmat([1:max(B)+min_ones],size(B,2),1)
I don’t know how this compares speedwise to #nrz’s approach (I’ve only got Octave to hand right now), but to me it's more intuitive as it’s simply comparing a max(B) + min_ones * column tiling of B:
4 4 4 4 4
3 3 3 3 3
3 3 3 3 3
3 3 3 3 3
2 2 2 2 2
1 1 1 1 1
with a row tiling of [1 : max(B) + min_ones]
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
To generate:
A =
0 0 0 0 1
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
0 0 1 1 1
0 1 1 1 1

This requires only one line, and seems to be faster than previous solutions based on repmat or arrayfun:
%// Example data
ncols = 5;
B = [4 3 3 3 2 1];
%// Generate A
A = bsxfun(#gt, 1:ncols, B(:));