a = [1 1 1 1 2 2 2 2 3 3 3 3; 1 2 3 4 5 6 7 8 9 10 11 12]';
What is the quickest way to sum the values in column 2 that correspond to the first and last occurrence of each number in column 1?
The desired output:
1 5
2 13
3 21
EDIT: The result should be the same if the numbers in column 1 are ordered differently.
a = [2 2 2 2 1 1 1 1 3 3 3 3; 1 2 3 4 5 6 7 8 9 10 11 12]';
2 5
1 13
3 21
You can use accumarray as follows. Not sure how fast it is, especially because it uses a custom anonymous function:
[u, ~, v] = unique(a(:,1), 'stable');
s = accumarray(v, a(:,2), [], #(x) x(1)+x(end));
result = [u s];
If the values in the first column of a are always in contiguous groups, the following approach can be used as well:
ind_diff = find(diff(a(:,1))~=0);
ind_first = [1; ind_diff+1];
ind_last = [ind_diff; size(a,1)];
s = a(ind_first,2) + a(ind_last,2);
result = [unique(a(:,1), 'stable') s];
Related
As the title says, I want to find all rows in a Matlab matrix that in certain columns the values in the row are equal with the values in the previous row, or in general, equal in some row in the matrix. For example I have a matrix
1 2 3 4
1 2 8 10
4 5 7 9
2 3 6 4
1 2 4 7
and I want to find the following rows:
1 2 3 4
1 2 3 10
1 2 4 7
How do I do something like that and how do I do it generally for all the possible pairs in columns 1 and 2, and have equal values in previous rows, that exist in the matrix?
Here's a start to see if we're headed in the right direction:
>> M = [1 2 3 4;
1 2 8 10;
4 5 7 9;
2 3 6 4;
1 2 4 7];
>> N = M; %// copy M into a new matrix so we can modify it
>> idx = ismember(N(:,1:2), N(1,1:2), 'rows')
idx =
1
1
0
0
1
>> N(idx, :)
ans =
1 2 3 4
1 2 8 10
1 2 4 7
Then you can remove those rows from the original matrix and repeat.
>> N = N(~idx,:)
N =
4 5 7 9
2 3 6 4
this will give you the results
data1 =[1 2 3 4
1 2 8 10
4 5 7 9
2 3 6 4
1 2 4 7];
data2 = [1 2 3 4
1 2 3 10
1 2 4 7];
[exists,position] = ismember(data1,data2, 'rows')
where the exists vector tells you wheter the row is on the other matrix and position gives you the position...
a less elegant and simpler version would be
array_data1 = reshape (data1',[],1);
array_data2 = reshape (data2',[],1);
matchmatrix = zeros(size(data2,1),size(data1,1));
for irow1 = 1: size(data2,1)
for irow2 = 1: size(data1,1)
matchmatrix(irow1,irow2) = min(data2(irow1,:) == data1(irow2,:))~= 0;
end
end
the matchmatrix is to read as a connectivity matrix where value of 1 indicates which row of data1 matches with which row of data2
Supose there is a Matrix
A =
1 3 2 4
4 2 5 8
6 1 4 9
and I have a Vector containing the "class" of each column of this matrix for example
v = [1 , 1 , 2 , 3]
How can I sum the columns of the matrix to a new matrix as column vectors each to the column of their class? In this example columns 1 and 2 of A would added to the first column of the new matrix, column 2 to the 3 to the 2nd, column 4 the the 3rd.
Like
SUM =
4 2 4
6 5 8
7 4 9
Is this possible without loops?
One of the perfect scenarios to combine the powers of accumarray and bsxfun -
%// Since we are to accumulate columns, first step would be to transpose A
At = A.' %//'
%// Create a vector of linear IDs for use with ACCUMARRAY later on
idx = bsxfun(#plus,v(:),[0:size(A,1)-1]*max(v))
%// Use ACCUMARRAY to accumulate rows from At, i.e. columns from A based on the IDs
out = reshape(accumarray(idx(:),At(:)),[],size(A,1)).'
Sample run -
A =
1 3 2 4 6 0
4 2 5 8 9 2
6 1 4 9 8 9
v =
1 1 2 3 3 2
out =
4 2 10
6 7 17
7 13 17
An alternative with accumarray in 2D. Generate a grid with the vector v and then apply accumarray:
A = A.';
v = [1 1 2 3];
[X, Y] = ndgrid(v,1:size(A,2));
Here X and Y look like this:
X =
1 1 1
1 1 1
2 2 2
3 3 3
Y =
1 2 3
1 2 3
1 2 3
1 2 3
Then apply accumarray:
B=accumarray([X(:) Y(:)],A(:)),
SUM = B.'
SUM =
4 2 4
6 5 8
7 4 9
As you see, using [X(:) Y(:)] create the following array:
ans =
1 1
1 1
2 1
3 1
1 2
1 2
2 2
3 2
1 3
1 3
2 3
3 3
in which the vector v containing the "class" is replicated 3 times since there are 3 unique classes that are to be summed up together.
EDIT:
As pointed out by knedlsepp you can get rid of the transpose to A and B like so:
[X2, Y2] = ndgrid(1:size(A,1),v);
B = accumarray([X2(:) Y2(:)],A(:))
which ends up doing the same. I find it a bit more easier to visualize with the transposes but that gives the same result.
How about a one-liner?
result = full(sparse(repmat(v,size(A,1),1), repmat((1:size(A,1)).',1,size(A,2)), A));
Don't optimize prematurely!
The for loop performs fine for your problem:
out = zeros(size(A,1), max(v));
for i = 1:numel(v)
out(:,v(i)) = out(:,v(i)) + A(:,i);
end
BTW: With fine, I mean: fast, fast, fast!
Suppose there are 2 vectors (i,j), A(10,1), B(10,1) which have obtain random values from [1,10] interval. eg.
A = [1 6 1 10 1 7 1 9 3 6]
B = [7 2 3 5 6 8 7 9 10 2].
I am interested in creating a new vector which will count how many values with the same i index occur. e.g.
1 and 7 ⇒ 2 occurrences
6 and 2 ⇒ 2 occurrences
1 and 3 ⇒ 1 occurrence
10 and 5 ⇒ 1 occurrence
1 and 6 ⇒ 1 occurrence
...
etc.
So that a final array/vector C occurs, with all the possible pairs and their counted occurrence with size C(number_of_pairs,2).
Use accumarray and then find:
A = [1 6 1 10 1 7 1 9 3 6];
B = [7 2 3 5 6 8 7 9 10 2]; %// data
aa = accumarray([A(:) B(:)], 1); %// how many times each pair occurs
[ii jj vv] = find(aa);
C = [ii jj vv]; %// only pairs which occurr at least once
In your example, this gives
C =
6 2 2
1 3 1
10 5 1
1 6 1
1 7 2
7 8 1
9 9 1
3 10 1
Or perhaps aa or vv are what you need; I'm not sure about what your desired output is.
Another possible approach, inspired by #Rody's answer:
mat = [A(:) B(:)];
[bb ii jj] = unique(mat, 'rows');
C = [mat(ii,:) accumarray(jj,1)];
Something like this?
A = [1 6 1 10 1 7 1 9 3 6];
B = [7 2 3 5 6 8 7 9 10 2];
%// Find the unique pairs
AB = [A;B].';
ABu = unique(AB, 'rows');
%// Count the number of occurrences of each unique pair
%// (pretty sure there's a better way to do this...)
C = arrayfun(#(ii) ...
[ABu(ii,:) sum(all(bsxfun(#eq, AB, ABu(ii,:)),2))], 1:size(ABu,1), ...
'UniformOutput', false);
C = cat(1,C{:});
Result:
C =
1 3 1
1 6 1
1 7 2
3 10 1
6 2 2
7 8 1
9 9 1
10 5 1
Something like this?
C = zeros(10);
for k = 1:10
C(A(k), B(k)) = C(A(k), B(k)) + 1
end
i want to control the creation of random numbers in this matrix :
Mp = floor(1+(10*rand(2,20)));
mp1 = sort(Mp,2);
i want to modify this code in order to have an output like this :
1 1 2 2 3 3 3 4 5 5 6 7 7 8 9 9 10 10 10 10
1 2 3 3 3 3 3 3 4 5 6 6 6 6 7 8 9 9 9 10
i have to fill each row with all the numbers going from 1 to 10 in an increasing order and the second matrix that counts the occurences of each number should be like this :
1 2 1 2 1 2 3 1 1 2 1 1 2 1 1 2 1 2 3 4
1 1 1 2 3 4 5 6 1 1 1 2 3 4 1 1 1 2 3 1
and the most tricky matrix that i'v been looking for since the last week is the third matrix that should skim through each row of the first matrix and returns the numbers of occurences of each number and the position of the last occcurence.here is an example of how the code should work. this example show the intended result after running through the first row of the first matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (positions)
1 2
2 2
3 3
4 1
5 2
6 1
7 2
8 1
9 2
10 4
(numbers)
this example show the intended result after running through the second row of the first matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (positions)
1 1 2
2 1 2
3 3 6
4 1 1
5 3
6 1 4
7 2 1
8 1 1
9 2 3
10 4
(numbers)
so the wanted matrix must be filled up with zeros from the beginning and each time after running through each row of the first matrix, we add the new result to the previous one...
I believe the following code does everything you asked for. If I didn't understand, you need to get a lot clearer in how you pose your question...
Note - I hard coded some values / sizes. In "real code" you would never do that, obviously.
% the bit of code that generates and sorts the initial matrix:
Mp = floor(1+(10*rand(2,20)));
mp1 = sort(Mp, 2);
clc
disp(mp1)
occCount = zeros(size(mp1));
for ii = 1:size(mp1,1)
for jj = 1:size(mp1,2)
if (jj == 1)
occCount(ii,jj) = 1;
else
if (mp1(ii,jj) == mp1(ii,jj-1))
occCount(ii,jj) = occCount(ii, jj-1) + 1;
else
occCount(ii,jj) = 1;
end
end
end
end
% this is the second matrix you asked for
disp(occCount)
% now the third:
big = zeros(10, 20);
for ii = 1:size(mp1,1)
for jj = 1:10
f = find(mp1(ii,:) == jj); % index of all of them
if numel(f) > 0
last = f(end);
n = numel(f);
big(jj, last) = big(jj, last) + n;
end
end
end
disp(big)
Please see if this is indeed what you had in mind.
The following code solves both the second and third matrix generation problems with a single loop. For clarity, the second matrix M2 is the 2-by-20 array in the example containing the cumulative occurrence count. The third matrix M3 is the sparse matrix of size 10-by-20 in the example that encodes the number and position of the last occurrence of each unique value. The code only loops over the rows, using accumarray to do most of the work. It is generalized to any size and content of mp1, as long as the rows are sorted first.
% data
mp1 = [1 1 2 2 3 3 3 4 5 5 6 7 7 8 9 9 10 10 10 10;
1 2 3 3 3 3 3 3 4 5 6 6 6 6 7 8 9 9 9 10]; % the example first matrix
nuniq = max(mp1(:));
% accumulate
M2 = zeros(size(mp1));
M3 = zeros(nuniq,size(mp1,2));
for ir=1:size(mp1,1),
cumSums = accumarray(mp1(ir,:)',1:size(mp1,2),[],#numel,[],true)';
segments = arrayfun(#(x)1:x,nonzeros(cumSums),'uni',false);
M2(ir,:) = [segments{:}];
countCoords = accumarray(mp1(ir,:)',1:size(mp1,2),[],#max,[],true);
[ii,jj] = find(countCoords);
nzinds = sub2ind(size(M3),ii,nonzeros(countCoords));
M3(nzinds) = M3(nzinds) + nonzeros(cumSums);
end
I won't print the outputs because they are a bit big for the answer, and the code is runnable as is.
NOTE: For new test data, I suggest using the commands Mp = randi(10,[2,20]); mp1 = sort(Mp,2);. Or based on your request to user2875617 and his response, ensure all numbers with mp1 = sort([repmat(1:10,2,1) randi(10,[2,10])],2); but that isn't really random...
EDIT: Error in code fixed.
I am editing the previous answer to check if it is fast when mp1 is large, and apparently it is:
N = 20000; M = 200; P = 100;
mp1 = sort([repmat(1:P, M, 1), ceil(P*rand(M,N-P))], 2);
tic
% Initialise output matrices
out1 = zeros(M, N); out2 = zeros(P, N);
for gg = 1:M
% Frequencies of each row
freqs(:, 1) = mp1(gg, [find(diff(mp1(gg, :))), end]);
freqs(:, 2) = histc(mp1(gg, :), freqs(:, 1));
cumfreqs = cumsum(freqs(:, 2));
k = 1;
for hh = 1:numel(freqs(:, 1))
out1(gg, k:cumfreqs(hh)) = 1:freqs(hh, 2);
out2(freqs(hh, 1), cumfreqs(hh)) = out2(freqs(hh, 1), cumfreqs(hh)) + freqs(hh, 2);
k = cumfreqs(hh) + 1;
end
end
toc
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