I am familiar with Matlab but am still having trouble with vectorized methods in my intuition, so I was wondering if anyone could demonstrate how they would manage this problem.
I have an array, for example A = [1 1 2 2 1 3 3 3 4 3 4 4 5].
I want to return an array B such that each element is the index of A's most 'recent' element with a different value than the previous ones.
So for our array A, B would equal [x x 2 2 4 5 5 5 8 9 10 10 12], where the x's can be any consistent value you like, because there is no previous index satisfying those characteristics.
I know how I would code it as a for-loop, and I bet the for-loop is probably faster, but can anyone vectorize this to faster than the for-loop?
Here's my for-loop:
prev=0;
B=zeros(length(A),1);
for i=2:length(A)
if A(i-1)~=A(i)
prev=i-1;
end
B(i)=prev;
end
Find the indices of the entries where the value changes:
ind = find(diff(A) ~= 0);
The values that should appear in B are therefore:
val = [0 ind];
Construct the diff of B: fill in the difference between the values that should appear at the right places:
Bd = zeros(size(B))';
Bd(ind + 1) = diff(val);
Now use cumsum to construct B:
B = cumsum(Bd)
Not sure whether this results in a speed-up though.
Related
Let say we have the vector v=[1,2,3] and we want to build the matrix of all the combinations of the numbers contained in v, i.e.
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
Since I'm not good in recursion, firstly I tried to write the code to build such a matrix by using for loops
makeLoop([1,2,3])
function A = makeLoop(v)
loops=length(v);
for i = 1:loops
dummy=v;
m=factorial(loops)/loops;
A((1+m*(i-1)):m*i,1)=v(i);
v(i)=[];
loops2=length(v);
for j = 1:loops2
dummy2=v;
m2=factorial(loops2)/loops2;
A(((1+m2*(j-1))+m*(i-1)):(m2*j+m*(i-1)),2)=v(j);
v(j)=[];
loops3=length(v);
for k = 1:loops3
m3=factorial(loops3)/loops3;
A(((1+m2*(j-1))+m*(i-1)):(m2*j+m*(i-1)),3)=v(k);
end
v=dummy2;
end
v=dummy;
end
end
it seems like it work, but obviously write it all for a bigger v would be like hell. Anyway I don't understand how to properly write the recursion, I think the recursive structure will be something like this
function A = makeLoop(v)
if length(v)==1
"do the last for loop"
else
"do a regular loop and call makeLoop(v) (v shrink at each loop)"
end
but I don't get which parts should I remove from the original code, and which to keep.
You were very close! The overall structure that you proposed is sound and your loopy-code can be inserted into it with practically no changes:
function A = makeLoop(v)
% number of (remaining) elements in the vector
loops = length(v);
if loops==1 %"do the last for loop"
A = v; %Obviously, if you input only a single number, the output has to be that number
else %"do a regular loop and call makeLoop(v) (v shrink at each loop)"
%preallocate matrix to store results
A = zeros(factorial(loops),loops);
%number of results per vector element
m = factorial(loops)/loops;
for i = 1:loops
%For each element of the vector, call the function again with that element missing.
dummy = v;
dummy(i) = [];
AOut = makeLoop(dummy);
%Then add that element back to the beginning of the output and store it.
A((1+m*(i-1)):m*i,:) = [bsxfun(#times,v(i),ones(m,1)) AOut];
end
end
Explanation bsxfun() line:
First, read the bsxfun documentation, it explains how it works way better than I could. But long story short, with bsxfun() we can replicate a scalar easily by multiplying it with a column vector of ones. E.g. bsxfun(#times,5,[1;1;1]) will result in the vector [5;5;5]. Note that since Matlab 2016b, bsxfun(#times,5,[1;1;1]) can written shorter as 5.*[1;1;1]
To the task at hand, we want to add v(i) in front (as the first column) of all permutations that may occur after it. Therefore we need to replicate the v(i) into the 1. dimension to match the number of rows of AOut, which is done with bsxfun(#times,v(i),ones(m,1)). Then we just horizontally concatenate this with AOut.
You can simply use the perms function to achieve this:
v = [1 2 3];
perms(v)
ans =
3 2 1
3 1 2
2 3 1
2 1 3
1 3 2
1 2 3
If you want them sorted using the same criterion you applied in the desired output, use the following code (refer to this page for an official documentation of the sortrows functon):
v = [1 2 3];
p = perms(v);
p = sortrows(p)
p =
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
If i have a vector (array) in matlab, is it possible to remove the values from that vector based on some restriction (e.g.. all non negative numbers).
Can you please advise me on the best approach to do that.
Yes, you can either use logical indexing to keep the values which meet a criterion or use the find function to get the indexes which hold values that meet a criterion.
logical indexing
the find function
An example of logical indexing where we want to remove all the values from a vector which are not greater than three:
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3)
x =
4 5 6
You can also ask for multiple criteria as you would expect. In the following example, we want to keep every value which is greater than three, but not five.
>> x=[1,2,3,4,5,6]
x =
1 2 3 4 5 6
>> x=x(x>3 & x~=5)
x =
4 6
Finally, the find function can come in handy when you need the indexes of values which meet a criterion.
>> x=[1,1,2,2,5,5]
x =
1 1 2 2 5 5
>> ind=find(x>3)
ind =
5 6
Logical indexing and find can also be applied to matrices with more than one row/column.
Thanks #Alan for helping me improve the answer.
You may want to look into logical indexing, as it neatly handles your problem.
To use the example you gave, if you have a vector a of numbers, and you want to remove all negative numbers you could do the following:
b = a(a >= 0);
which would create a vector b containing only the positive elements of a, or you could try:
a(a < 0) = [];
would set any elements in the vector a to []
I have two vectors with the same elements but their order is not same. For eg
A
10
9
8
B
8
9
10
I want to find the mapping between the two
B2A
3
2
1
How can I do this in matlab efficiently?
I think the Matlab sort is efficient. So:
[~,I]=sort(A); %sort A; we want the indices, not the values
[~,J]=sort(B); %same with B
%I(1) and J(1) both point to the smallest value, and a similar statement is true
%for other pairs, even with repeated values.
%Now, find the index vector that sorts I
[~,K]=sort(I);
%if K(1) is k, then A(k) is the kth smallest entry in A, and the kth smallest
%entry in B is J(k)
%so B2A(1)=J(k)=J(K(1)), where BSA is the desired permutation vector
% A similar statement holds for the other entries
%so finally
B2A=J(K);
if the above were in script "findB2A" the following should be a check for it
N=1e4;
M=100;
A=floor(M*rand(1,N));
[~,I]=sort(rand(1,N));
B=A(I);
findB2A;
all(A==B(B2A))
There are a couple of ways of doing this. The most efficient in terms of lines of code is probably using ismember(). The return values are [Lia,Locb] = ismember(A,B), where Locb are the indices in B which correspond to the elements of A. You can do [~, B2A] = ismember(A, B) to get the result you want. If your version of MATLAB does not allow ~, supply a throwaway argument for the first output.
You must ensure that there is a 1-to-1 mapping to get meaningful results, otherwise the index will always point to the first matching element.
Here a solution :
arrayfun(#(x)find(x == B), A)
I tried with bigger arrays :
A = [ 7 5 2 9 1];
B = [ 1 9 7 5 2];
It gives the following result :
ans =
3 4 5 2 1
Edit
Because arrayfun is usually slower than the equivalent loop, here a solution with a loop:
T = length(A);
B2A = zeros(1, length(A));
for tt = 1:T
B2A(1, tt) = find(A(tt) == B);
end
I would go for Joe Serrano's answer using three chained sort's.
Another approach is to test all combinations for equality with bsxfun:
[~, B2A] = max(bsxfun(#eq, B(:), A(:).'));
This gives B2A such that B(B2A) equals A. If you want it the other way around (not clear from your example), simply reverse A and B within bsxfun.
My question has two parts:
Split a given matrix into its columns
These columns should be stored into an array
eg,
A = [1 3 5
3 5 7
4 5 7
6 8 9]
Now, I know the solution to the first part:
the columns are obtained via
tempCol = A(:,iter), where iter = 1:end
Regarding the second part of the problem, I would like to have (something like this, maybe a different indexing into arraySplit array), but one full column of A should be stored at a single index in splitArray:
arraySplit(1) = A(:,1)
arraySplit(2) = A(:,2)
and so on...
for the example matrix A,
arraySplit(1) should give me [ 1 3 4 6 ]'
arraySplit(2) should give me [ 3 5 5 8 ]'
I am getting the following error, when i try to assign the column vector to my array.
In an assignment A(I) = B, the number of elements in B and I must be the same.
I am doing the allocation and access of arraySplit wrongly, please help me out ...
Really it sounds like A is alread what you want--I can't imagine a scenario where you gain anything by splitting them up. But if you do, then your best bet is likely a cell array, ie.
C = cell(1,3);
for i=1:3
C{i} = A(:,i);
end
Edit: See #EitanT's comment below for a more elegant way to do this. Also accessing the vector uses the same syntax as setting it, e.g. v = C{2}; will put the second column of A into v.
In a Matlab array, each element must have the same type. In most cases, that is a float type. An your example A(:, 1) is a 4 by 1 array. If you assign it to, say, B(:, 2) then B(:, 1) must also be a 4 by 1 array.
One common error that may be biting you is that a 4 by 1 array and a 1 by 4 array are not the same thing. One is a column vector and one is a row vector. Try transposing A(:, 1) to get a 1 by 4 row array.
You could try something like the following:
A = [1 3 5;
3 5 7;
4 5 7;
6 8 9]
arraySplit = zeros(4,1,3);
for i =1:3
arraySplit(:,:,i) = A(:,i);
end
and then call arraySplit(:,:,1) to get the first vector, but that seems to be an unnecessary step, since you can readily do that by accessing the exact same values as A(:,1).
I have a matrix A with size (nr,nc), a vector of column indices B (so B has size (nr,1) and every element in B is an integer between 1 and nc), and I want to do something to every element in A that is of the form A(i,B(i)) for i between 1 and nr, efficiency being the key concern.
For concreteness, say C is a vector of size (nr,1), the goal is to do
for i=1:nr
A(i,B(i))=A(i,B(i))+C(i)
end
more efficiently. The context is usually that nr>>nc (because when nr is large vectorization is efficient for many operations). I have gotten a factor 3 speedup by using an indicator function approach:
for k=1:nc
A(:,k)=A(:,k)+(k==B).*C
end
Are there other ways (more efficient hopefully) to do this?
I guess this is similar to many questions on double-indexing, but it's concretely one I run into all the time.
Use linear indexing:
idx = sub2ind(size(A), 1:nr, B');
A(idx) = A(idx) + C';
or (edited version with one less transpose)
idx = sub2ind(size(A), (1:nr)', B);
A(idx) = A(idx) + C;
One way would be to use linear indexing of the matrix. You will need a vector v holding the offsets of the first element in each line, then index using A(v + B). For example:
>A=[1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
>B = [1 2 3] % we want the 1st element of row 1, 2nd of row 2, 3rd of row 3
>ii = [0 3 6] + B
>a(ii)
1 5 9
Note: As groovingandi had shown, it is also possible (and more readable) to use sub2ind to generate the ii linear indices vector. The idea is essentially the same.