I would like to sum specific columns of each row in a matrix using a for loop. Below I have included a simplified version of my problem. As of right now, I am calculating the column sums individually, but this is not effective as my actual problem has multiple matrices (data sets).
a = [1 2 3 4 5 6; 4 5 6 7 8 9];
b = [2 2 3 4 4 6; 3 3 3 4 5 5];
% Repeat the 3 lines of code below for row 2 of matrix a
% Repeat the entire process for matrix b
c = sum(a(1,1:3)); % Sum columns 1:3 of row 1
d = sum(a(1,4:6)); % Sum columns 4:6 of row 1
e = sum(a(1,:)); % Sum all columns of row 1
I would like to know how to create a for loop that automatically loops through and sums the specific columns of each row for each matrix that I have.
Thank you.
Here is a solution that you don't need to use for loop.
Assuming that you have a matrix a of size 2x12, and you want to do the row sums every 4 columns, then you can use reshape() and squeeze() to get the final result:
k = 4;
a = [1:12
13:24];
% 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
s = squeeze(sum(reshape(a,size(a,1),k,[]),2));
and you will get
s =
10 26 42
58 74 90
Related
If I have this matrix:
A:
X Y Z
1 1 2
0 3 4
0 5 6
2 7 8
7 9 10
8 11 12
3 13 14
12 14 16
15 17 18
How could I create new matrix B, C, D and E which contains:
B:
0 3 4
0 5 6
C:
X Y Z
1 1 2
2 7 8
3 13 14
D:
7 9 10
8 11 12
E:
12 14 16
15 17 18
The idea is to construct a loop asking if 0<A<1 else 1<A<5 else 6<A<10 else 11<A<15. and create new matrix from that condition. Any idea about how to store the results of the loop?
I suggest you an approach that uses the discretize function in order to group the matrix rows into different categories based on their range. Here is the full implementation:
A = [
1 1 2;
0 3 4;
0 5 6;
2 7 8;
7 9 10;
8 11 12;
3 13 14;
12 14 16;
15 17 18
];
A_range = [0 1 5 10 15];
bin_idx = discretize(A(:,1),A_range);
A_split = arrayfun(#(bin) A(bin_idx == bin,:),1:(numel(A_range) - 1),'UniformOutput',false);
celldisp(A_split);
Since you want to consider 5 different ranges based on the first column values, the arguments passed to discretize must be the first matrix column and a vector containing the group limits (first number inclusive left, second number exclusive right, second number inclusive left, third number exclusive right, and so on...). Since your ranges are a little bit messed up, feel free to adjust them to respect the correct output. The latter is returned in the form of a cell array of double matrices in which every element contains the rows belonging to a distinct group:
A_split{1} =
0 3 4
0 5 6
A_split{2} =
1 1 2
2 7 8
3 13 14
A_split{3} =
7 9 10
8 11 12
A_split{4} =
12 14 16
15 17 18
Instead of using a loop, use logical indexing to achieve what you want. Use the first column of A and check for the ranges that you want to look for, then use this to subset into the final matrix A to get what you want.
For example, to create the matrix C, find all locations in the first column of A that are between 1 and 5, then subset the matrix along the rows using these locations:
m = A(:,1) >= 1 & A(:,1) <= 5;
C = A(m,:);
You can repeat this in a similar way for the rest of the matrices you want to create.
I have the following vector
a = 3 3 5 5 20 20 20 4 4 4 2 2 2 10 10 10 6 6 1 1 1
does anyone know how to shuffle this vector with the same elementsnever be seperate?
something like bellow
a = 10 10 10 5 5 4 4 4 20 20 20 1 1 1 3 3 2 2 2 6 6
thank you, best regard...
You can use unique combined with accumarray to create a cell array where each group of values is placed into a separate cell element. You can then shuffle these elements and recombine them into an array.
% Put each group into a separate cell of a cell array
[~, ~, ind] = unique(a);
C = accumarray(ind(:), a(:), [], #(x){x});
% Shuffle it
shuffled = C(randperm(numel(C)));
% Now make it back into a vector
out = cat(1, shuffled{:}).';
% 20 20 20 1 1 1 3 3 10 10 10 5 5 4 4 4 6 6 2 2 2
Another option is to get the values using unique and then compute the number that each occurs. You can then shuffle the values and use repelem to expand out the result
u = unique(a);
counts = histc(a, u);
% Shuffle the values
inds = randperm(numel(u));
% Now expand out the array
out = repelem(u(inds), counts(inds));
A very similar answer to #Suever, using a loop and logical matrix rather than cells
a = [3 3 5 5 20 20 20 4 4 4 2 2 2 10 10 10 6 6 1 1 1];
vals = unique(a); %find unique values
vals = vals(randperm(length(vals))); %shuffle vals matrix
aout = []; %initialize output matrix
for ii = 1:length(vals)
aout = [aout a(a==(vals(ii)))]; %add correct number of each value
end
Here's another approach:
a = [3 3 5 5 20 20 20 4 4 4 2 2 2 10 10 10 6 6 1 1 1];
[~, ~, lab] = unique(a);
r = randperm(max(lab));
[~, ind] = sort(r(lab));
result = a(ind);
Example result:
result =
2 2 2 3 3 5 5 20 20 20 4 4 4 10 10 10 1 1 1 6 6
It works as follows:
Assign unique labels to each element of a depending on their values (this is vector lab);
Apply a random bijection from the values of lab to themselves (the random bijection is represented by r; the result of applying it is r(lab));
Sort r(lab) and get the indices of the sorting (this is ind);
Apply those indices to a.
I have a vector of 13 entities in Matlab.
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]
I want to append values [1 2 3 4 5] at regular intervals at position 1 5 9 13 & 17.
The final value of a looks like this.
a=[1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2].
The values with italics show the appended values.
How can I do it?
Since you are looking for regular intervals, you can take advantage of the reshape and cat function:
a = [3 4 6 8 1 5 8 9 3 7 3 6 2];
v = [1 2 3 4 5];
l = [1 5 9 13 17];
interval = l(2)-l(1)-1; %computes the interval between inserts
amax = ceil(size(a,2)/interval) * interval; %calculating maximum size for zero padding
a(amax) = 0; %zero padding to allow `reshape`
b = reshape (a,[interval,size(v,2)]); %reshape into matrix
result = reshape(vertcat (v,b), [1,(size(b,1)+1)*size(b,2)]); %insert the values into the right position and convert back into vector
%remove padded zeros
final = result(result ~= 0) %remove the zero padding.
>>final =
Columns 1 through 16
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6
Columns 17 through 18
5 2
Here's an approach using boolean-indexing -
% Inputs
a = [3 4 6 8 1 5 8 9 3 7 3 6 2]
append_vals = [1 2 3 4 5]
append_interval = 4 % Starting at 1st index
% Find out indices of regular intervals where new elements are to be inserted.
% This should create that array [1,5,9,13,17]
N_total = numel(a) + numel(append_vals)
append_idx = find(rem(0:N_total-1,append_interval)==0)
% Get boolean array with 1s at inserting indices, 0s elsewhere
append_mask = ismember(1:N_total,append_idx)
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(~append_mask) = a
out(append_mask) = append_vals
Alternatively, we can also use linear-indexing and avoid creating append_mask, like so -
% Setup output array and insert new and old elements
out = zeros(1,N_total)
out(append_idx) = append_vals
out(setdiff(1:numel(out),append_idx)) = a
a=[3 4 6 8 1 5 8 9 3 7 3 6 2]; % // Your original values
pos = [1 5 9 13 17]; % // The position of the values you want to insert
b=[1 2 3 4 5]; % // The values you want to insert
% // Pre-allocate a vector with the total size to hold the resulting values
r = zeros(size(a,2)+size(pos,2),1);
r(pos) = b % // Insert the appended values into the resulting vector first
r3 = r.' <1 % // Find the indices of the original values. These will be zero in the variable r but 1 in r3
ans =
0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1
ind= find(r3==1) % // Find the indices of the original values
ind =
2 3 4 6 7 8 10 11 12 14 15 16 18
r(ind) = a; % // Insert those into the resulting vector.
r.'
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
You can use this function to append a bunch of values to an existing vector, given their positions in the new vector:
function r=append_interval(a,v,p)
% a - vector with initial values
% v - vector containing values to be inserted
% p - positions for values in v
lv=numel(v); % number of elements in v vector
la=numel(a); % number of elements in a vector
column_a=iscolumn(a); % check if a is a column- or row- wise vector
tot_elements=la+lv;
% size of r is tha max between the total number of elements in the two vectors and the higher positin in vector p (in this case missing positions in a are filled with zeros)
lr=max([max(p) tot_elements]);
% initialize r as nan vector
r=zeros(column_a*(lr-1)+1,~column_a*(lr-1)+1)/0;
% set elements in p position to the corresponding values in v
r(p)=v;
% copy values in a in the remaining positions and fill with zeros missing entries (if any)
tot_missing_values=lr-tot_elements;
if(tot_missing_values)
remaining_values=cat(2-iscolumn(a),a,zeros(column_a*(tot_missing_values-1)+1,~column_a*(tot_missing_values-1)+1));
else
remaining_values=a;
end
% insert values
r(isnan(r))=remaining_values;
You can use row-wise or column-wise vectors; the orientation of r will be the same of that of a.
Input:
a =
3 4 6 8 1 5 8 9 3 7 3 6 2
v =
1 2 3 4 5
p =
1 5 9 13 17
Output:
>> append_interval(a,v,p)
ans =
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2
Every sequence of positive positions is allowed and the function will pad for you with zeros the final vector, in case you indicate a position exceding the sum of the original vector and added items.
For example, if:
v3 =
1 2 3 4 5 6 90
p3 =
1 5 9 13 17 30 33
you get:
append_interval(a,v3,p3)
ans =
Columns 1 through 19
1 3 4 6 2 8 1 5 3 8 9 3 4 7 3 6 5 2 0
Columns 20 through 33
0 0 0 0 0 0 0 0 0 0 6 0 0 90
Hope this will help.
I would like to align and count vectors with different time stamps to count the corresponding bins.
Let's assume I have 3 matrix from [N,edges] = histcounts in the following structure. The first row represents the edges, so the bins. The second row represents the values. I would like to sum all values with the same bin.
A = [0 1 2 3 4 5;
5 5 6 7 8 5]
B = [1 2 3 4 5 6;
2 5 7 8 5 4]
C = [2 3 4 5 6 7 8;
1 2 6 7 4 3 2]
Now I want to sum all the same bins. My final result should be:
result = [0 1 2 3 4 5 6 7 8;
5 7 12 16 ...]
I could loop over all numbers, but I would like to have it fast.
You can use accumarray:
H = [A B C].'; %//' Concatenate the histograms and make them column vectors
V = [unique(H(:,1)) accumarray(H(:,1)+1, H(:,2))].'; %//' Find unique values and accumulate
V =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
Note: The H(:,1)+1 is to force the bin values to be positive, otherwise MATLAB will complain. We still use the actual bins in the output V. To avoid this, as #Daniel says in the comments, use the third output of unique (See: https://stackoverflow.com/a/27783568/2732801):
H = [A B C].'; %//' stupid syntax highlighting :/
[U, ~, IU] = unique(H(:,1));
V = [U accumarray(IU, H(:,2))].';
If you're only doing it with 3 variables as you've shown then there likely aren't going to be any performance hits with looping it.
But if you are really averse to the looping idea, then you can do it using arrayfun.
rng = 0:8;
output = arrayfun(#(x)sum([A(2,A(1,:) == x), B(2,B(1,:) == x), C(2,C(1,:) == x)]), rng);
output = cat(1, rng, output);
output =
0 1 2 3 4 5 6 7 8
5 7 12 16 22 17 8 3 2
This can be beneficial for particularly large A, B, and C variables as there is no copying of data.
We have p.e. i = 1:25 iterations.
Each iteration result is a 1xlength(N) cell array, where 0<=N<=25.
iteration 1: 4 5 9 10 20
iteration 2: 3 8 9 13 14 6
...
iteration 25: 1 2 3
We evaluate the results of all iterations to one matrix sorted according to frequency each value is repeated in descending order like this example:
Matrix=
Columns 1 through 13
16 22 19 25 2 5 8 14 17 21 3 12 13
6 5 4 4 3 3 3 3 3 3 2 2 2
Columns 14 through 23
18 20 1 6 7 9 10 11 15 23
2 2 1 1 1 1 1 1 1 1
Result explanation: Column 1: N == 16 is present in 6 iterations, column 2: N == 22 is present in 5 iterations etc.
If a number N isn't displayed (in that paradigm N == 4, N == 24) in any iteration, is not listed with frequency index of zero either.
I want to associate each iteration (i) to the first N it is displayed p.e. N == 9 to be present only in first iteration i = 1 and not in i = 2 too, N == 3 only to i = 2 and not in i = 25 too etc until all i's to be unique associated to N's.
Thank you in advance.
Here's a way that uses a feature of unique (i.e. that it returns the index to the first value) that was introduced in R2012a
%# make some sample data
iteration{1} = [1 2 4 6];
iteration{2} = [1 3 6];
iteration{3} = [1 2 3 4 5 6];
nIter= length(iteration);
%# create an index vector so we can associate N's with iterations
nn = cellfun(#numel,iteration);
idx = zeros(1,sum(nn));
idx([1,cumsum(nn(1:end-1))+1]) = 1;
idx = cumsum(idx); %# has 4 ones, 3 twos, 6 threes
%# create a vector of the same length as idx with all the N's
nVec = cat(2,iteration{:});
%# run `unique` on the vector to identify the first occurrence of each N
[~,firstIdx] = unique(nVec,'first');
%# create a "cleanIteration" array, where each N only appears once
cleanIter = accumarray(idx(firstIdx)',firstIdx',[nIter,1],#(x){sort(nVec(x))},{});
cleanIter =
[1x4 double]
[ 3]
[ 5]
>> cleanIter{1}
ans =
1 2 4 6
Here is another solution using accumarray. Explanations in the comments
% example data (from your question)
iteration{1} = [4 5 9 10 20 ];
iteration{2} = [3 8 9 13 14 6];
iteration{3} = [1 2 3];
niterations = length(iteration);
% create iteration numbers
% same as Jonas did in the first part of his code, but using a short loop
for i=1:niterations
idx{i} = i*ones(size(iteration{i}));
end
% count occurences of values from all iterations
% sort them in descending order
occurences = accumarray([iteration{:}]', 1);
[occ val] = sort(occurences, 1, 'descend');
% remove zero occurences and create the Matrix
nonzero = find(occ);
Matrix = [val(nonzero) occ(nonzero)]'
Matrix =
3 9 1 2 4 5 6 8 10 13 14 20
2 2 1 1 1 1 1 1 1 1 1 1
% find minimum iteration number for all occurences
% again, using accumarray with #min function
assoc = accumarray([iteration{:}]', [idx{:}]', [], #min);
nonzero = find(assoc);
result = [nonzero assoc(nonzero)]'
result =
1 2 3 4 5 6 8 9 10 13 14 20
3 3 2 1 1 2 2 1 1 2 2 1