Let us say I have the following:
M = randn(10,20);
T = randn(1,20);
I would like to threshold each column of M, by each entry of T. For example, find all indicies of all elements of M(:,1) that are greater than T(1). Find all indicies of all elements in M(:,2) that are greater than T(2), etc etc.
Of course, I would like to do this without a for-loop. Is this possible?
You can use bsxfun like this:
I = bsxfun(#gt, M, T);
Then I will be a logcial matrix of size(M) with ones where M(:,i) > T(i).
You can use bsxfun to do things like this, but it may not be faster than a for loop (more below on this).
result = bsxfun(#gt,M,T)
This will do an element wise comparison and return you a logical matrix indicating the relationship governed by the first argument. I have posted code below to show the direct comparison, indicating that it does return what you are looking for.
%var declaration
M = randn(10,20);
T = randn(1,20);
% quick method
fastres = bsxfun(#gt,M,T);
% looping method
res = false(size(M));
for i = 1:length(T)
res(:,i) = M(:,i) > T(i);
end
% check to see if the two matrices are identical
isMatch = all(all(fastres == res))
This function is very powerful and can be used to help speed up processes, but keep in mind that it will only speed things up if there is a lot of data. There is a bit of background work that bsxfun must do, which can actually cause it to be slower.
I would only recommend using it if you have several thousand data points. Otherwise, the traditional for-loop will actually be faster. Try it out for yourself by changing the size of the M and T variables.
You can replicate the threshold vector and use matrix comparison:
s=size(M);
T2=repmat(T, s(1), 1);
M(M<T2)=0;
Indexes=find(M);
Related
I'm trying to generate a 100-by-5 matrix where every line is a permutation of 1..100 (that is, every line is 5 random numbers from [1..100] without repetitions).
So far I've only been able to do it iteratively with a for-loop. Is there a way to do it more efficiently (using fewer lines of code), without loops?
N = 100;
T = zeros(N, 5);
for i = 1:N
T(i, :) = randperm(100, 5);
end
Let
N = 100; % desired number of rows
K = 5; % desired number of columns
M = 100; % size of population to sample from
Here's an approach that's probably fast; but memory-expensive, as it generates an intermediate M×N matrix and then discards N-K rows:
[~, result] = sort(rand(N, M), 2);
result = result(:, 1:K);
There is very little downside to using a loop here, at least in this minimal example. Indeed, it may well be the best-performing solution for MATLAB's execution engine. But perhaps you don't like assigning the temporary variable i or there are other advantages to vectorization in your non-minimal implementation. Consider this carefully before blindly implementing a solution.
You need to call randperm N times, but each call has no dependency on its position in the output. Without a loop index you will need something else to regulate the number of calls, but this can be just N empty cells cell(N,1). You can use this cell array to evaluate a function that calls randperm but ignores the contents (or, rather, lack of contents) of the cells, and then reassemble the function outputs into one matrix with cell2mat:
T = cell2mat(cellfun(#(~) {randperm(100,5)}, cell(N,1)));
I'm trying to insert multiple values into an array using a 'values' array and a 'counter' array. For example, if:
a=[1,3,2,5]
b=[2,2,1,3]
I want the output of some function
c=somefunction(a,b)
to be
c=[1,1,3,3,2,5,5,5]
Where a(1) recurs b(1) number of times, a(2) recurs b(2) times, etc...
Is there a built-in function in MATLAB that does this? I'd like to avoid using a for loop if possible. I've tried variations of 'repmat()' and 'kron()' to no avail.
This is basically Run-length encoding.
Problem Statement
We have an array of values, vals and runlengths, runlens:
vals = [1,3,2,5]
runlens = [2,2,1,3]
We are needed to repeat each element in vals times each corresponding element in runlens. Thus, the final output would be:
output = [1,1,3,3,2,5,5,5]
Prospective Approach
One of the fastest tools with MATLAB is cumsum and is very useful when dealing with vectorizing problems that work on irregular patterns. In the stated problem, the irregularity comes with the different elements in runlens.
Now, to exploit cumsum, we need to do two things here: Initialize an array of zeros and place "appropriate" values at "key" positions over the zeros array, such that after "cumsum" is applied, we would end up with a final array of repeated vals of runlens times.
Steps: Let's number the above mentioned steps to give the prospective approach an easier perspective:
1) Initialize zeros array: What must be the length? Since we are repeating runlens times, the length of the zeros array must be the summation of all runlens.
2) Find key positions/indices: Now these key positions are places along the zeros array where each element from vals start to repeat.
Thus, for runlens = [2,2,1,3], the key positions mapped onto the zeros array would be:
[X 0 X 0 X X 0 0] % where X's are those key positions.
3) Find appropriate values: The final nail to be hammered before using cumsum would be to put "appropriate" values into those key positions. Now, since we would be doing cumsum soon after, if you think closely, you would need a differentiated version of values with diff, so that cumsum on those would bring back our values. Since these differentiated values would be placed on a zeros array at places separated by the runlens distances, after using cumsum we would have each vals element repeated runlens times as the final output.
Solution Code
Here's the implementation stitching up all the above mentioned steps -
% Calculate cumsumed values of runLengths.
% We would need this to initialize zeros array and find key positions later on.
clens = cumsum(runlens)
% Initalize zeros array
array = zeros(1,(clens(end)))
% Find key positions/indices
key_pos = [1 clens(1:end-1)+1]
% Find appropriate values
app_vals = diff([0 vals])
% Map app_values at key_pos on array
array(pos) = app_vals
% cumsum array for final output
output = cumsum(array)
Pre-allocation Hack
As could be seen that the above listed code uses pre-allocation with zeros. Now, according to this UNDOCUMENTED MATLAB blog on faster pre-allocation, one can achieve much faster pre-allocation with -
array(clens(end)) = 0; % instead of array = zeros(1,(clens(end)))
Wrapping up: Function Code
To wrap up everything, we would have a compact function code to achieve this run-length decoding like so -
function out = rle_cumsum_diff(vals,runlens)
clens = cumsum(runlens);
idx(clens(end))=0;
idx([1 clens(1:end-1)+1]) = diff([0 vals]);
out = cumsum(idx);
return;
Benchmarking
Benchmarking Code
Listed next is the benchmarking code to compare runtimes and speedups for the stated cumsum+diff approach in this post over the other cumsum-only based approach on MATLAB 2014B-
datasizes = [reshape(linspace(10,70,4).'*10.^(0:4),1,[]) 10^6 2*10^6]; %
fcns = {'rld_cumsum','rld_cumsum_diff'}; % approaches to be benchmarked
for k1 = 1:numel(datasizes)
n = datasizes(k1); % Create random inputs
vals = randi(200,1,n);
runs = [5000 randi(200,1,n-1)]; % 5000 acts as an aberration
for k2 = 1:numel(fcns) % Time approaches
tsec(k2,k1) = timeit(#() feval(fcns{k2}, vals,runs), 1);
end
end
figure, % Plot runtimes
loglog(datasizes,tsec(1,:),'-bo'), hold on
loglog(datasizes,tsec(2,:),'-k+')
set(gca,'xgrid','on'),set(gca,'ygrid','on'),
xlabel('Datasize ->'), ylabel('Runtimes (s)')
legend(upper(strrep(fcns,'_',' '))),title('Runtime Plot')
figure, % Plot speedups
semilogx(datasizes,tsec(1,:)./tsec(2,:),'-rx')
set(gca,'ygrid','on'), xlabel('Datasize ->')
legend('Speedup(x) with cumsum+diff over cumsum-only'),title('Speedup Plot')
Associated function code for rld_cumsum.m:
function out = rld_cumsum(vals,runlens)
index = zeros(1,sum(runlens));
index([1 cumsum(runlens(1:end-1))+1]) = 1;
out = vals(cumsum(index));
return;
Runtime and Speedup Plots
Conclusions
The proposed approach seems to be giving us a noticeable speedup over the cumsum-only approach, which is about 3x!
Why is this new cumsum+diff based approach better than the previous cumsum-only approach?
Well, the essence of the reason lies at the final step of the cumsum-only approach that needs to map the "cumsumed" values into vals. In the new cumsum+diff based approach, we are doing diff(vals) instead for which MATLAB is processing only n elements (where n is the number of runLengths) as compared to the mapping of sum(runLengths) number of elements for the cumsum-only approach and this number must be many times more than n and therefore the noticeable speedup with this new approach!
Benchmarks
Updated for R2015b: repelem now fastest for all data sizes.
Tested functions:
MATLAB's built-in repelem function that was added in R2015a
gnovice's cumsum solution (rld_cumsum)
Divakar's cumsum+diff solution (rld_cumsum_diff)
knedlsepp's accumarray solution (knedlsepp5cumsumaccumarray) from this post
Naive loop-based implementation (naive_jit_test.m) to test the just-in-time compiler
Results of test_rld.m on R2015b:
Old timing plot using R2015a here.
Findings:
repelem is always the fastest by roughly a factor of 2.
rld_cumsum_diff is consistently faster than rld_cumsum.
repelem is fastest for small data sizes (less than about 300-500 elements)
rld_cumsum_diff becomes significantly faster than repelem around 5 000 elements
repelem becomes slower than rld_cumsum somewhere between 30 000 and 300 000 elements
rld_cumsum has roughly the same performance as knedlsepp5cumsumaccumarray
naive_jit_test.m has nearly constant speed and on par with rld_cumsum and knedlsepp5cumsumaccumarray for smaller sizes, a little faster for large sizes
Old rate plot using R2015a here.
Conclusion
Use repelem below about 5 000 elements and the cumsum+diff solution above.
There's no built-in function I know of, but here's one solution:
index = zeros(1,sum(b));
index([1 cumsum(b(1:end-1))+1]) = 1;
c = a(cumsum(index));
Explanation:
A vector of zeroes is first created of the same length as the output array (i.e. the sum of all the replications in b). Ones are then placed in the first element and each subsequent element representing where the start of a new sequence of values will be in the output. The cumulative sum of the vector index can then be used to index into a, replicating each value the desired number of times.
For the sake of clarity, this is what the various vectors look like for the values of a and b given in the question:
index = [1 0 1 0 1 1 0 0]
cumsum(index) = [1 1 2 2 3 4 4 4]
c = [1 1 3 3 2 5 5 5]
EDIT: For the sake of completeness, there is another alternative using ARRAYFUN, but this seems to take anywhere from 20-100 times longer to run than the above solution with vectors up to 10,000 elements long:
c = arrayfun(#(x,y) x.*ones(1,y),a,b,'UniformOutput',false);
c = [c{:}];
There is finally (as of R2015a) a built-in and documented function to do this, repelem. The following syntax, where the second argument is a vector, is relevant here:
W = repelem(V,N), with vector V and vector N, creates a vector W where element V(i) is repeated N(i) times.
Or put another way, "Each element of N specifies the number of times to repeat the corresponding element of V."
Example:
>> a=[1,3,2,5]
a =
1 3 2 5
>> b=[2,2,1,3]
b =
2 2 1 3
>> repelem(a,b)
ans =
1 1 3 3 2 5 5 5
The performance problems in MATLAB's built-in repelem have been fixed as of R2015b. I have run the test_rld.m program from chappjc's post in R2015b, and repelem is now faster than other algorithms by about a factor 2:
i am trying to learn how to vectorise matlab loops, so im just doing a few small examples.
here is the standard loop i am trying to vectorise:
function output = moving_avg(input, N)
output = [];
for n = N:length(input) % iterate over y vector
summation = 0;
for ii = n-(N-1):n % iterate over x vector N times
summation += input(ii);
endfor
output(n) = summation/N;
endfor
endfunction
i have been able to vectorise one loop, but cant work out what to do with the second loop. here is where i have got to so far:
function output = moving_avg(input, N)
output = [];
for n = N:length(input) % iterate over y vector
output(n) = mean(input(n-(N-1):n));
endfor
endfunction
can someone help me simplify it further?
EDIT:
the input is just a one dimensional vector and probably maximum 100 data points. N is a single integer, less than the size of the input (typically probably around 5)
i don't actually intend to use it for any particular application, it was just a simple nested loop that i thought would be good to use to learn about vectorisation..
Seems like you are performing convolution operation there. So, just use conv -
output = zeros(size(input1))
output(N:end) = conv(input1,ones(1,N),'valid')./N
Please note that I have replaced the variable name input with input1, as input is already used as the name of a built-in function in MATLAB, so it's a good practice to avoid such conflicts.
Generic case: For a general case scenario, you can look into bsxfun to create such groups and then choose your operation that you intend to perform at the final stage. Here's how such a code would look like for sliding/moving average operation -
%// Create groups of indices for each sliding interval of length N
idx = bsxfun(#plus,[1:N]',[0:numel(input1)-N]) %//'
%// Index into input1 with those indices to get grouped elements from it along columns
input1_indexed = input1(idx)
%// Finally, choose the operation you intend to perform and apply along the
%// columns. In this case, you are doing average, so use mean(...,1).
output = mean(input1_indexed,1)
%// Also pre-append with zeros if intended to match up with the expected output
Matlab as a language does this type of operation poorly - you will always require an outside O(N) loop/operation involving at minimum O(K) copies which will not be worth it in performance to vectorize further because matlab is a heavy weight language. Instead, consider using the
filter function where these things are typically implemented in C which makes that type of operation nearly free.
For a sliding average, you can use cumsum to minimize the number of operations:
x = randi(10,1,10); %// example input
N = 3; %// window length
y = cumsum(x); %// compute cumulative sum of x
z = zeros(size(x)); %// initiallize result to zeros
z(N:end) = (y(N:end)-[0 y(1:end-N)])/N; %// compute order N difference of cumulative sum
Using MATLAB,
Imagine a Nx6 array of numbers which represent N segments with 3+3=6 initial and end point coordinates.
Assume I have a function Calc_Dist( Segment_1, Segment_2 ) that takes as input two 1x6 arrays, and that after some operations returns a scalar, namely the minimal euclidean distance between these two segments.
I want to calculate the pairwise minimal distance between all N segments of my list, but would like to avoid a double loop to do so.
I cannot wrap my head around the documentation of the bsxfun function of MATLAB, so I cannot make this work. For the sake of a minimal example (the distance calculation is obviously not correct):
function scalar = calc_dist( segment_1, segment_2 )
scalar = sum( segment_1 + segment_2 )
end
and the main
Segments = rand( 1500, 6 )
Pairwise_Distance_Matrix = bsxfun( #calc_dist, segments, segments' )
Is there any way to do this, or am I forced to use double loops ?
Thank you for any suggestion
I think you need pdist rather than bsxfun. pdist can be used in two different ways, the second of which is applicable to your problem:
With built-in distance functions, supplied as strings, such as 'euclidean', 'hamming' etc.
With a custom distance function, a handle to which you supply.
In the second case, the distance function
must be of the form
function D2 = distfun(XI, XJ),
taking as arguments a 1-by-N vector XI containing a single row of X, an
M2-by-N matrix XJ containing multiple rows of X, and returning an
M2-by-1 vector of distances D2, whose Jth element is the distance
between the observations XI and XJ(J,:).
Although the documentation doesn't tell, it's very likely that the second way is not as efficient as the first (a double loop might even be faster, who knows), but you can use it. You would need to define your function so that it fulfills the stated condition. With your example function it's easy: for this part you'd use bsxfun:
function scalar = calc_dist( segment_1, segment_2 )
scalar = sum(bsxfun(#plus, segment_1, segment_2), 2);
end
Note also that
pdist works with rows (not columns), which is what you need.
pdist reduces operations by exploiting the properties that any distance function must have. Namely, the distance of an element to itself is known to be zero; and the distance for each pair can be computed just once thanks to symmetry. If you want to arrange the output in the form of a matrix, use squareform.
So, after your actual distance function has been modified appropriately (which may be the hard part), use:
distances = squareform(pdist(segments, #calc_dist));
For example:
N = 4;
segments = rand(N,6);
distances = squareform(pdist(segments, #calc_dist));
produces
distances =
0 6.1492 7.0886 5.5016
6.1492 0 6.8559 5.2688
7.0886 6.8559 0 6.2082
5.5016 5.2688 6.2082 0
Unfortunately I don't see any "smarter" (i.e. read faster) solution than the double loop. For speed consideration I'd organize the points as a 6×N array, not the other way, because column access is way faster than row access in MATLAB.
So:
N = 150000;
Segments = rand(6, N);
Pairwise_Distance_Matrix = Inf(N, N);
for i = 1:(N-1)
for j = (i+1):N
Pairwise_Distance_Matrix(i,j) = calc_dist(Segments(:,i), Segments(:,j));
end;
end;
Minimum_Pairwise_Distance = min(min(Pairwise_Distance_Matrix));
Contrary to common wisdom, explicit loops are faster now in MATLAB compared to the likes of arrayfun, cellfun or structfun; bsxfun beats everything else in terms of speed, but it doesn't apply to your case.
Context: I'm working on Project Euler Problem 23 using Matlab in order to practice my barely existing programming skills.
My Problem:
Now I have a vector with roughly 6500 numbers (ranging from 12 to 28122) as elements and want to calculate all the two element sums. That is I only need one instance of every sum, so having calculated a1 + an it's not necessary to calculate an + a1.
Edit for clarification: This includes the sums a1+a1, a2+a2,..., an+an.
The problem is that this is much too slow.
Problem specific constraints:
It's a given that sums 28123 or over aren't necessary to calculate, since those can't be used to solve the problem further.
My approach:
AbundentNumberSumsRaw=[];
for i=1:3490
AbundentNumberSumsRaw=[AbundentNumberSumRaw AbundentNumbers(i)+AbundentNumbers(i:end);
end
This works terribly :p
My Comments:
I'm pretty sure that incrementally increasing the vector AbundentNumbersRaw is bad coding, since that means memory usage will spike unnecessarily. I haven't done so, since a) I don't know what size vector to pre-allocate and b) I couldn't come up with a way to inject the sums into AbundentNumbersRaw in a orderly manner without using some ugly looking nested loops.
"for i=1:3490" is lower than the numbers of elements simply because I checked and saw that all the resulting sums for numbers whose index are above 3490 would be too large for me to use anyway.
I'm pretty sure my main issue is that the program need to do a lot of incremental increases of the vector AbundentNumbersRaw.
Any and all help and suggestions would be much appreciated :)
Cheers
Rasmus
Suppose
a = 28110*rand(6500,1)+12;
then
sums = [
a(1) + a(1:end)
a(2) + a(2:end)
...
];
is the calculation you're after.
You also state that sums whose value goes over 28123 should be discarded.
This can be generalized like so:
% Compute all 2-element sums without repetitions
C = arrayfun(#(x) a(x)+a(x:end), 1:numel(a), 'uniformoutput', false);
C = cat(1, C{:});
% discard sums exceeding threshold
C(C>28123) = [];
or using a loop
% Compute all 2-element sums without repetitions
E = cell(numel(a),1);
for ii = 1:numel(a)
E{ii} = a(ii)+a(ii:end); end
E = cat(1, E{:});
% discard sums exceeding threshold
E(E>28123) = [];
Simple testing shows that arrayfun is somewhat faster than the loop, so I'd go for the arrayfun option.
As your primary problem is to find out, which integers in a given set can be written as the sum of two integers of a different set, I'd choose a different approach:
AbundantNumbers = 1:6500; % replace with the list you generated somewhere else
maxInteger = 28122;
AbundantNumberSum(1:maxInteger) = true; % logical array
for i = 1:length(AbundantNumbers)
sumIndices = AbundantNumbers(i) + AbundantNumbers;
AbundantNumberSum(sumIndices(sumIndices <= maxInteger)) = false;
end
Unfortunantely, this is not an answer to your question but to your problem ;-) For the MatLab way to solve your original question, see the elegant answer of Rody Oldenhuis.
My approach would be the following:
v = 1:3490; % your vector here
s = length(v);
result = zeros(s); % preallocate memory
for m = 1:s
result(m,m:end) = v(m)+v(m:end);
end
You will get a matrix of 3490 x 3490 elements and more than half of them 0.