I have a matrix A, a list of indices is and js, and a list of values to add to A, ws. Originally I was simply iterating through A by a nested for loop:
for idx = 1:N
i = is(idx);
j = js(idx);
w = ws(idx);
A(i,j) = A(i,j) + w;
end
However, I would like to vectorize this to increase efficiency. I thought something simple like
A(is,js) = A(is,js) + ws
would work, and it does as long as the is and js don't repeat. Said differently, if I generate idx = sub2ind(size(A),is,js);, so long as idx has no repeat values, all is well. If it does, then only the last value is added, all previous values are left out. A concrete example:
A = zeros(3,3);
indices = [1,2,3,1];
additions = [5,5,5,5];
A(indices) = A(indices) + additions;
This results in the first column having values of 5, not 5,5,10.
This is a small example, but in my actual application the lists of indices are really long and filled with redundant values. I'm hoping to vectorize this to save time, so going through and eliminating redundancies isn't really an option. So my main question is, how do I add to a matrix from a given set of redundant indices? Alternatively, is there another way of working through this without any sort of iteration?
To emphasize a nice property of accumarray (accumarray actually works with two indices)
With the example from Luis Mendo:
is = [2 3 3 1 1 2].';
js = [1 3 3 2 2 4].';
ws = [10 20 30 40 50 60].';
A3 = accumarray([is js],ws);
%% A3 =
%% 0 90 0 0
%% 10 0 0 60
%% 0 0 50 0
If I understand correctly, you only need full(sparse(is, js, ws)). This works because sparse accumulates values at matching indices.
% Example data
is = [2 3 3 1 1 2];
js = [1 3 3 2 2 4];
ws = [10 20 30 40 50 60];
% With loop
N = numel(is);
A = zeros(max(is), max(js));
for idx = 1:N
i = is(idx);
j = js(idx);
w = ws(idx);
A(i,j) = A(i,j) + w;
end
% With `sparse`
A2 = full(sparse(is, js, ws));
% Check
isequal(A, A2)
Related
I have a matrix whose columns which was shuffled according to some index. I know want to find the index that 'unshuffles' the array back into its original state.
For example:
myArray = [10 20 30 40 50 60]';
myShuffledArray = nan(6,3)
myShufflingIndex = nan(6,3)
for x = 1:3
myShufflingIndex(:,x) = randperm(length(myArray))';
myShuffledArray(:,x) = myArray(myShufflingIndex(:,x));
end
Now I want to find a matrix myUnshufflingIndex, which reverses the shuffling to get an array myUnshuffledArray = [10 20 30 40 50 60; 10 20 30 40 50 60; 10 20 30 40 50 60]'
I expect to use myUnshufflingIndex in the following way:
for x = 1:3
myUnShuffledArray(:,x) = myShuffledArray(myUnshufflingIndex(:,x), x);
end
For example, if one column in myShufflingIndex = [2 4 6 3 5 1]', then the corresponding column in myUnshufflingIndex is [6 1 4 2 5 3]'
Any ideas on how to get myUnshufflingIndex in a neat vectorised way? Also, is there a better way to unshuffle the array columnwise than in a loop?
You can get myUnshufflingIndex with a single call to sort:
[~, myUnshufflingIndex] = sort(myShufflingIndex, 1);
Alternatively, you don't even need to compute myUnshufflingIndex, since you can just use myShufflingIndex on the left hand side of the assignment to unshuffle the data:
for x = 1:3
myUnShuffledArray(myShufflingIndex(:, x), x) = myShuffledArray(:, x);
end
And if you'd like to avoid a for loop while unshuffling, you can vectorize it by adding an offset to each column of your index, turning it into a matrix of linear indices instead of just row indices:
[nRows, nCols] = size(myShufflingIndex);
myUnshufflingIndex = myShufflingIndex+repmat(0:nRows:(nRows*(nCols-1)), nRows, 1);
myUnShuffledArray = nan(nRows, nCols); % Preallocate
myUnShuffledArray(myUnshufflingIndex) = myShuffledArray;
Assume we have a 1-d Matrix, with random length:
M = [102,4,12,6,8,3,4,65,23,43,111,4]
Moreover I have a vector with values that are linked to the index of M:
V = [1,5]
What i want is a simple code of:
counter = 1;
NewM = zeros(length(V)*3,1);
for i = 1:length(V)
NewM(counter:(counter+2)) = M(V(i):(V(i)+2))
counter = counter+3;
end
So, the result will be
NewM = [102,4,12,8,3,4]
In other words, I want the V to V+2 values from M in a new array.
I'm pretty sure this can be done easier, but I'm struggeling with how to implement it in arrayfun /bsxfun...
bsxfun(#(x,y) x(y:(y+2)),M,V)
With bsxfun it's about getting the Vi on one dimension and the +0, +1 +2 on the other:
M = [102,4,12,6,8,3,4,65,23,43,111,4];
V = [1,5];
NewM = M( bsxfun(#plus, V(:), 0:2) )
NewM =
102 4 12
8 3 4
Or if you want a line:
NewM = reshape(NewM.', 1, [])
NewM =
102 4 12 8 3 4
Using arrayfun (note that M is used as an "external" matrix entity in this scope, whereas the arrayfun anonymous function parameter x correspond to the elements in V)
NewM = cell2mat(arrayfun(#(x) M(x:x+2), V, 'UniformOutput', false));
With result
NewM =
102 4 12 8 3 4
One fully vectorized solution is to expand V to create the correct indexing vector. In your case:
[1,2,3,5,6,7]
You can expand V to replicate each of it's elements n times and then add a repeating vector of 0:(n-1):
n = 3;
idx = kron(V, ones(1,n)) + mod(0:numel(V)*n-1,n)
So here kron(V, ones(1,n)) will return [1,1,1,5,5,5] and mod(0:numel(V)*n-1,n) will return [0,1,2,0,1,2] which add up to the required indexing vector [1,2,3,5,6,7]
and now it's just
M(idx)
Note a slightly faster alternative to kron is reshape(repmat(V,n,1),1,[])
I'm trying to elegantly split a vector. For example,
vec = [1 2 3 4 5 6 7 8 9 10]
According to another vector of 0's and 1's of the same length where the 1's indicate where the vector should be split - or rather cut:
cut = [0 0 0 1 0 0 0 0 1 0]
Giving us a cell output similar to the following:
[1 2 3] [5 6 7 8] [10]
Solution code
You can use cumsum & accumarray for an efficient solution -
%// Create ID/labels for use with accumarray later on
id = cumsum(cut)+1
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(mask).',vec(mask).',[],#(x) {x})
Benchmarking
Here are some performance numbers when using a large input on the three most popular approaches listed to solve this problem -
N = 100000; %// Input Datasize
vec = randi(100,1,N); %// Random inputs
cut = randi(2,1,N)-1;
disp('-------------------- With CUMSUM + ACCUMARRAY')
tic
id = cumsum(cut)+1;
mask = cut==0;
out = accumarray(id(mask).',vec(mask).',[],#(x) {x});
toc
disp('-------------------- With FIND + ARRAYFUN')
tic
N = numel(vec);
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
toc
disp('-------------------- With CUMSUM + ARRAYFUN')
tic
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
toc
Runtimes
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.068102 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.117953 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 12.560973 seconds.
Special case scenario: In cases where you might have runs of 1's, you need to modify few things as listed next -
%// Mask to get valid values from cut and vec corresponding to ones in cut
mask = cut==0
%// Setup IDs differently this time. The idea is to have successive IDs.
id = cumsum(cut)+1
[~,~,id] = unique(id(mask))
%// Finally get the output with accumarray using masked IDs and vec values
out = accumarray(id(:),vec(mask).',[],#(x) {x})
Sample run with such a case -
>> vec
vec =
1 2 3 4 5 6 7 8 9 10
>> cut
cut =
1 0 0 1 1 0 0 0 1 0
>> celldisp(out)
out{1} =
2
3
out{2} =
6
7
8
out{3} =
10
For this problem, a handy function is cumsum, which can create a cumulative sum of the cut array. The code that produces an output cell array is as follows:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [0 0 0 1 0 0 0 0 1 0];
cutsum = cumsum(cut);
cutsum(cut == 1) = NaN; %Don't include the cut indices themselves
sumvals = unique(cutsum); % Find the values to use in indexing vec for the output
sumvals(isnan(sumvals)) = []; %Remove NaN values from sumvals
output = {};
for i=1:numel(sumvals)
output{i} = vec(cutsum == sumvals(i)); %#ok<SAGROW>
end
As another answer shows, you can use arrayfun to create a cell array with the results. To apply that here, you'd replace the for loop (and the initialization of output) with the following line:
output = arrayfun(#(val) vec(cutsum == val), sumvals, 'UniformOutput', 0);
That's nice because it doesn't end up growing the output cell array.
The key feature of this routine is the variable cutsum, which ends up looking like this:
cutsum =
0 0 0 NaN 1 1 1 1 NaN 2
Then all we need to do is use it to create indices to pull the data out of the original vec array. We loop from zero to max and pull matching values. Notice that this routine handles some situations that may arise. For instance, it handles 1 values at the very beginning and very end of the cut array, and it gracefully handles repeated ones in the cut array without creating empty arrays in the output. This is because of the use of unique to create the set of values to search for in cutsum, and the fact that we throw out the NaN values in the sumvals array.
You could use -1 instead of NaN as the signal flag for the cut locations to not use, but I like NaN for readability. The -1 value would probably be more efficient, as all you'd have to do is truncate the first element from the sumvals array. It's just my preference to use NaN as a signal flag.
The output of this is a cell array with the results:
output{1} =
1 2 3
output{2} =
5 6 7 8
output{3} =
10
There are some odd conditions we need to handle. Consider the situation:
vec = [1 2 3 4 5 6 7 8 9 10 11 12 13 14];
cut = [1 0 0 1 1 0 0 0 0 1 0 0 0 1];
There are repeated 1's in there, as well as a 1 at the beginning and end. This routine properly handles all this without any empty sets:
output{1} =
2 3
output{2} =
6 7 8 9
output{3} =
11 12 13
You can do this with a combination of find and arrayfun:
vec = [1 2 3 4 5 6 7 8 9 10];
N = numel(vec);
cut = [0 0 0 1 0 0 0 0 1 0];
ind = find(cut);
ind_before = [ind-1 N]; ind_before(ind_before < 1) = 1;
ind_after = [1 ind+1]; ind_after(ind_after > N) = N;
out = arrayfun(#(x,y) vec(x:y), ind_after, ind_before, 'uni', 0);
We thus get:
>> celldisp(out)
out{1} =
1 2 3
out{2} =
5 6 7 8
out{3} =
10
So how does this work? Well, the first line defines your input vector, the second line finds how many elements are in this vector and the third line denotes your cut vector which defines where we need to cut in our vector. Next, we use find to determine the locations that are non-zero in cut which correspond to the split points in the vector. If you notice, the split points determine where we need to stop collecting elements and begin collecting elements.
However, we need to account for the beginning of the vector as well as the end. ind_after tells us the locations of where we need to start collecting values and ind_before tells us the locations of where we need to stop collecting values. To calculate these starting and ending positions, you simply take the result of find and add and subtract 1 respectively.
Each corresponding position in ind_after and ind_before tell us where we need to start and stop collecting values together. In order to accommodate for the beginning of the vector, ind_after needs to have the index of 1 inserted at the beginning because index 1 is where we should start collecting values at the beginning. Similarly, N needs to be inserted at the end of ind_before because this is where we need to stop collecting values at the end of the array.
Now for ind_after and ind_before, there is a degenerate case where the cut point may be at the end or beginning of the vector. If this is the case, then subtracting or adding by 1 will generate a start and stopping position that's out of bounds. We check for this in the 4th and 5th line of code and simply set these to 1 or N depending on whether we're at the beginning or end of the array.
The last line of code uses arrayfun and iterates through each pair of ind_after and ind_before to slice into our vector. Each result is placed into a cell array, and our output follows.
We can check for the degenerate case by placing a 1 at the beginning and end of cut and some values in between:
vec = [1 2 3 4 5 6 7 8 9 10];
cut = [1 0 0 1 0 0 0 1 0 1];
Using this example and the above code, we get:
>> celldisp(out)
out{1} =
1
out{2} =
2 3
out{3} =
5 6 7
out{4} =
9
out{5} =
10
Yet another way, but this time without any loops or accumulating at all...
lengths = diff(find([1 cut 1])) - 1; % assuming a row vector
lengths = lengths(lengths > 0);
data = vec(~cut);
result = mat2cell(data, 1, lengths); % also assuming a row vector
The diff(find(...)) construct gives us the distance from each marker to the next - we append boundary markers with [1 cut 1] to catch any runs of zeros which touch the ends. Each length is inclusive of its marker, though, so we subtract 1 to account for that, and remove any which just cover consecutive markers, so that we won't get any undesired empty cells in the output.
For the data, we mask out any elements corresponding to markers, so we just have the valid parts we want to partition up. Finally, with the data ready to split and the lengths into which to split it, that's precisely what mat2cell is for.
Also, using #Divakar's benchmark code;
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.272810 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.436276 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 17.112259 seconds.
-------------------- With mat2cell
Elapsed time is 0.084207 seconds.
...just sayin' ;)
Here's what you need:
function spl = Splitting(vec,cut)
n=1;
j=1;
for i=1:1:length(b)
if cut(i)==0
spl{n}(j)=vec(i);
j=j+1;
else
n=n+1;
j=1;
end
end
end
Despite how simple my method is, it's in 2nd place for performance:
-------------------- With CUMSUM + ACCUMARRAY
Elapsed time is 0.264428 seconds.
-------------------- With FIND + ARRAYFUN
Elapsed time is 0.407963 seconds.
-------------------- With CUMSUM + ARRAYFUN
Elapsed time is 18.337940 seconds.
-------------------- SIMPLE
Elapsed time is 0.271942 seconds.
Unfortunately there is no 'inverse concatenate' in MATLAB. If you wish to solve a question like this you can try the below code. It will give you what you looking for in the case where you have two split point to produce three vectors at the end. If you want more splits you will need to modify the code after the loop.
The results are in n vector form. To make them into cells, use num2cell on the results.
pos_of_one = 0;
% The loop finds the split points and puts their positions into a vector.
for kk = 1 : length(cut)
if cut(1,kk) == 1
pos_of_one = pos_of_one + 1;
A(1,one_pos) = kk;
end
end
F = vec(1 : A(1,1) - 1);
G = vec(A(1,1) + 1 : A(1,2) - 1);
H = vec(A(1,2) + 1 : end);
I am using MATLAB to write a code that multiplies polynomials. Most parts of my code work however there is one part where I have two row vectors a and b. I want to remove repeated elements of a and then add the corresponding elements of b. This is what I have written
c=length(a);
d=length(b);
remove=[];
for i=1:c
for j=i+1:c
if (a(i)==a(j))
remove=[remove,i];
b(j)=b(i)+b(j);
end
end
end
a(remove)=[];
b(remove)=[];
The problem with this is if there is an element in a that appears more than twice, it doesn't work properly.
For example if a=[5,6,8,9,6,7,9,10,8,9,11,12] and b=[1,7,1,-1,3,21,3,-3,-4,-28,-4,4]
then once this code is run a becomes [5,6,7,10,8,9,11,12] which is correct but b becomes [1,10,21,-3,-3,-27,-4,4] which is correct except the -27 should be a -26.
I know why this happens because the 9 in a(1,4) gets compared with the 9 in a(1,7) so b(1,7) becomes b(1,7)+b(1,4) and then a(1,4) gets compared with the 9 in a(1,10). and then later the a(1,7) compares with a(1,10) and so the new b(1,7) adds to the b(1,10) however the b(1,4) adds to the b(1,10) too. I somehow need to stop this once one repeated element has been found because here b(1,4) has been added twice when it should only be added once.
I am not supposed to use any built in functions, is there a way of resolving this easily?
I would prefer using built in functions, but assuming you have to stick to your own approach, you can try this:
a=[5,6,8,9,6,7,9,10,8,9,11,12];
b=[1,7,1,-1,3,21,3,-3,-4,-28,-4,4];
n = numel(a);
remove = zeros(1,n);
temp = a;
for ii = 1:n
for jj = ii+1:n
if temp(ii) == temp(jj)
temp(ii) = NaN;
remove(ii) = ii;
b(jj) = b(jj) + b(ii);
end
end
end
a(remove(remove>0)) = []
b(remove(remove>0)) = []
a =
5 6 7 10 8 9 11 12
b =
1 10 21 -3 -3 -26 -4 4
It's not much different from your approach, except for changing the iith value of a if it is found later. To avoid overwriting the values in a with NaN, I'm using a temporary variable for this.
Also, as you can see, I'm avoiding remove = [remove i], because this will create a growing vector, which is very slow.
It can be solved in a vectorized manner with the following indexing nightmare (perhaps someone will come up with a simpler approach):
a = [5,6,8,9,6,7,9,10,8,9,11,12];
b = [1,7,1,-1,3,21,3,-3,-4,-28,-4,4];
[sa, ind1] = sort(a);
[~, ii, jj] = unique(sa);
[ind2, ind3] = sort(ind1(ii));
a = a(ind2);
b = accumarray(jj(:),b(ind1)).';
b = b(ind3);
Anyway, to multiply polynomials you could use conv:
>> p1 = [1 3 0 2]; %// x^3 + 3x^2 + 1
>> p2 = [2 -1 4]; %// 2x^2 - x + 4
>> conv(p1,p2)
ans =
2 5 1 16 -2 8 %// 2x^5 + 5x^4 + x^3 + 16x^2 - 2x + 8
I know that in MATLAB, in the 1D case, you can select elements with indexing such as a([1 5 3]), to return the 1st, 5th, and 3rd elements of a. I have a 2D array, and would like to select out individual elements according to a set of tuples I have. So I may want to get a(1,3), a(1,4), a(2,5) and so on. Currently the best I have is diag(a(tuples(:,1), tuples(:,2)), but this requires a prohibitive amount of memory for larger a and/or tuples. Do I have to convert these tuples into linear indices, or is there a cleaner way of accomplishing what I want without taking so much memory?
Converting to linear indices seems like a legitimate way to go:
indices = tuples(:, 1) + size(a,1)*(tuples(:,2)-1);
selection = a(indices);
Note that this is also implement in the Matlab built-in solution sub2ind, as in nate'2 answer:
a(sub2ind(size(a), tuples(:,1),tuples(:,2)))
however,
a = rand(50);
tuples = [1,1; 1,4; 2,5];
start = tic;
for ii = 1:1e4
indices = tuples(:,1) + size(a,1)*(tuples(:,2)-1); end
time1 = toc(start);
start = tic;
for ii = 1:1e4
sub2ind(size(a),tuples(:,1),tuples(:,2)); end
time2 = toc(start);
round(time2/time1)
which gives
ans =
38
so although sub2ind is easier on the eyes, it's also ~40 times slower. If you have to do this operation often, choose the method above. Otherwise, use sub2ind to improve readability.
if x and y are vectors of the x y values of matrix a, then sub2und should solve your problem:
a(sub2ind(size(a),x,y))
For example
a=magic(3)
a =
8 1 6
3 5 7
4 9 2
x = [3 1];
y = [1 2];
a(sub2ind(size(a),x,y))
ans =
4 1
you can reference the 2D matlab position with a 1D number as in:
a = [3 4 5;
6 7 8;
9 10 11;];
a(1) = 3;
a(2) = 6;
a(6) = 10;
So if you can get the positions in a matrix like this:
a([(col1-1)*(rowMax)+row1, (col2-1)*(rowMax)+row2, (col3-1)*(rowMax)+row3])
note: rowmax is 3 in this case
will give you a list of the elements at col1/row1 col2/row2 and col3/row3.
so if
row1 = col1 = 1
row2 = col2 = 2
row3 = col3 = 3
you will get:
[3, 7, 11]
back.