Suppose I have a CuArray with random zeros and ones and I want to get a random index of CuArray corresponding to value one. For instance,
m = 100;
A = CuArray(rand([0, 1], m));
i = rand(1:m);
while A[i]!=1
i = rand(1:m);
end
Is there a function so that I can not use while looping?
Your construction of A has the following equivalent representation:
using Distributions
n_ones = rand(Binomial(m, 0.5))
one_inds = shuffle(1:m)[1:n_ones]
A = zeros(Int, m)
A[one_inds] .= 1
That is, you first choose the number of ones you are going to set (from a binomial distribution, since you have m independent choices), and then select without repetition that many indices (by just taking the init of all indices, shuffled).
Written this way, choosing a random index of a one is just
rand(one_inds)
In my models, one of the most repeated tasks to be done is counting the number of each element within an array. The counting is from a closed set, so I know there are X types of elements, and all or some of them populate the array, along with zeros that represent 'empty' cells. The array is not sorted in any way, and could by quite long (about 1M elements), and this task is done thousands of times during one simulation (which is also part of hundreds of simulations). The result should be a vector r of size X, so r(k) is the amount of k in the array.
Example:
For X = 9, if I have the following input vector:
v = [0 7 8 3 0 4 4 5 3 4 4 8 3 0 6 8 5 5 0 3]
I would like to get this result:
r = [0 0 4 4 3 1 1 3 0]
Note that I don't want the count of zeros, and that elements that don't appear in the array (like 2) have a 0 in the corresponding position of the result vector (r(2) == 0).
What would be the fastest way to achieve this goal?
tl;dr: The fastest method depend on the size of the array. For array smaller than 214 method 3 below (accumarray) is faster. For arrays larger than that method 2 below (histcounts) is better.
UPDATE: I tested this also with implicit broadcasting, that was introduced in 2016b, and the results are almost equal to the bsxfun approach, with no significant difference in this method (relative to the other methods).
Let's see what are the available methods to perform this task. For the following examples we will assume X has n elements, from 1 to n, and our array of interest is M, which is a column array that can vary in size. Our result vector will be spp1, such that spp(k) is the number of ks in M. Although I write here about X, there is no explicit implementation of it in the code below, I just define n = 500 and X is implicitly 1:500.
The naive for loop
The most simple and straightforward way to cope this task is by a for loop that iterate over the elements in X and count the number of elements in M that equal to it:
function spp = loop(M,n)
spp = zeros(n,1);
for k = 1:size(spp,1);
spp(k) = sum(M==k);
end
end
This is off course not so smart, especially if only little group of elements from X is populating M, so we better look first for those that are already in M:
function spp = uloop(M,n)
u = unique(M); % finds which elements to count
spp = zeros(n,1);
for k = u(u>0).';
spp(k) = sum(M==k);
end
end
Usually, in MATLAB, it is advisable to take advantage of the built-in functions as much as possible, since most of the times they are much faster. I thought of 5 options to do so:
1. The function tabulate
The function tabulate returns a very convenient frequency table that at first sight seem to be the perfect solution for this task:
function tab = tabi(M)
tab = tabulate(M);
if tab(1)==0
tab(1,:) = [];
end
end
The only fix to be done is to remove the first row of the table if it counts the 0 element (it could be that there are no zeros in M).
2. The function histcounts
Another option that can be tweaked quite easily to our need it histcounts:
function spp = histci(M,n)
spp = histcounts(M,1:n+1);
end
here, in order to count all different elements between 1 to n separately, we define the edges to be 1:n+1, so every element in X has it's own bin. We could write also histcounts(M(M>0),'BinMethod','integers'), but I already tested it, and it takes more time (though it makes the function independent of n).
3. The function accumarray
The next option I'll bring here is the use of the function accumarray:
function spp = accumi(M)
spp = accumarray(M(M>0),1);
end
here we give the function M(M>0) as input, to skip the zeros, and use 1 as the vals input to count all unique elements.
4. The function bsxfun
We can even use binary operation #eq (i.e. ==) to look for all elements from each type:
function spp = bsxi(M,n)
spp = bsxfun(#eq,M,1:n);
spp = sum(spp,1);
end
if we keep the first input M and the second 1:n in different dimensions, so one is a column vector the other is a row vector, then the function compares each element in M with each element in 1:n, and create a length(M)-by-n logical matrix than we can sum to get the desired result.
5. The function ndgrid
Another option, similar to the bsxfun, is to explicitly create the two matrices of all possibilities using the ndgrid function:
function spp = gridi(M,n)
[Mx,nx] = ndgrid(M,1:n);
spp = sum(Mx==nx);
end
then we compare them and sum over columns, to get the final result.
Benchmarking
I have done a little test to find the fastest method from all mentioned above, I defined n = 500 for all trails. For some (especially the naive for) there is a great impact of n on the time of execution, but this is not the issue here since we want to test it for a given n.
Here are the results:
We can notice several things:
Interestingly, there is a shift in the fastest method. For arrays smaller than 214 accumarray is the fastest. For arrays larger than 214 histcounts is the fastest.
As expected the naive for loops, in both versions are the slowest, but for arrays smaller than 28 the "unique & for" option is slower. ndgrid become the slowest in arrays bigger than 211, probably because of the need to store very large matrices in memory.
There is some irregularity in the way tabulate works on arrays in size smaller than 29. This result was consistent (with some variation in the pattern) in all the trials I conducted.
(the bsxfun and ndgrid curves are truncated because it makes my computer stuck in higher values, and the trend is quite clear already)
Also, notice that the y-axis is in log10, so a decrease in unit (like for arrays in size 219, between accumarray and histcounts) means a 10-times faster operation.
I'll be glad to hear in the comments for improvements to this test, and if you have another, conceptually different method, you are most welcome to suggest it as an answer.
The code
Here are all the functions wrapped in a timing function:
function out = timing_hist(N,n)
M = randi([0 n],N,1);
func_times = {'for','unique & for','tabulate','histcounts','accumarray','bsxfun','ndgrid';
timeit(#() loop(M,n)),...
timeit(#() uloop(M,n)),...
timeit(#() tabi(M)),...
timeit(#() histci(M,n)),...
timeit(#() accumi(M)),...
timeit(#() bsxi(M,n)),...
timeit(#() gridi(M,n))};
out = cell2mat(func_times(2,:));
end
function spp = loop(M,n)
spp = zeros(n,1);
for k = 1:size(spp,1);
spp(k) = sum(M==k);
end
end
function spp = uloop(M,n)
u = unique(M);
spp = zeros(n,1);
for k = u(u>0).';
spp(k) = sum(M==k);
end
end
function tab = tabi(M)
tab = tabulate(M);
if tab(1)==0
tab(1,:) = [];
end
end
function spp = histci(M,n)
spp = histcounts(M,1:n+1);
end
function spp = accumi(M)
spp = accumarray(M(M>0),1);
end
function spp = bsxi(M,n)
spp = bsxfun(#eq,M,1:n);
spp = sum(spp,1);
end
function spp = gridi(M,n)
[Mx,nx] = ndgrid(M,1:n);
spp = sum(Mx==nx);
end
And here is the script to run this code and produce the graph:
N = 25; % it is not recommended to run this with N>19 for the `bsxfun` and `ndgrid` functions.
func_times = zeros(N,5);
for n = 1:N
func_times(n,:) = timing_hist(2^n,500);
end
% plotting:
hold on
mark = 'xo*^dsp';
for k = 1:size(func_times,2)
plot(1:size(func_times,1),log10(func_times(:,k).*1000),['-' mark(k)],...
'MarkerEdgeColor','k','LineWidth',1.5);
end
hold off
xlabel('Log_2(Array size)','FontSize',16)
ylabel('Log_{10}(Execution time) (ms)','FontSize',16)
legend({'for','unique & for','tabulate','histcounts','accumarray','bsxfun','ndgrid'},...
'Location','NorthWest','FontSize',14)
grid on
1 The reason for this weird name comes from my field, Ecology. My models are a cellular-automata, that typically simulate individual organisms in a virtual space (the M above). The individuals are of different species (hence spp) and all together form what is called "ecological community". The "state" of the community is given by the number of individuals from each species, which is the spp vector in this answer. In this models, we first define a species pool (X above) for the individuals to be drawn from, and the community state take into account all species in the species pool, not only those present in M
We know that that the input vector always contains integers, so why not use this to "squeeze" a bit more performance out of the algorithm?
I've been experimenting with some optimizations of the the two best binning methods suggested by the OP, and this is what I came up with:
The number of unique values (X in the question, or n in the example) should be explicitly converted to an (unsigned) integer type.
It's faster to compute an extra bin and then discard it, than to "only process" valid values (see the accumi_new function below).
This function takes about 30sec to run on my machine. I'm using MATLAB R2016a.
function q38941694
datestr(now)
N = 25;
func_times = zeros(N,4);
for n = 1:N
func_times(n,:) = timing_hist(2^n,500);
end
% Plotting:
figure('Position',[572 362 758 608]);
hP = plot(1:n,log10(func_times.*1000),'-o','MarkerEdgeColor','k','LineWidth',2);
xlabel('Log_2(Array size)'); ylabel('Log_{10}(Execution time) (ms)')
legend({'histcounts (double)','histcounts (uint)','accumarray (old)',...
'accumarray (new)'},'FontSize',12,'Location','NorthWest')
grid on; grid minor;
set(hP([2,4]),'Marker','s'); set(gca,'Fontsize',16);
datestr(now)
end
function out = timing_hist(N,n)
% Convert n into an appropriate integer class:
if n < intmax('uint8')
classname = 'uint8';
n = uint8(n);
elseif n < intmax('uint16')
classname = 'uint16';
n = uint16(n);
elseif n < intmax('uint32')
classname = 'uint32';
n = uint32(n);
else % n < intmax('uint64')
classname = 'uint64';
n = uint64(n);
end
% Generate an input:
M = randi([0 n],N,1,classname);
% Time different options:
warning off 'MATLAB:timeit:HighOverhead'
func_times = {'histcounts (double)','histcounts (uint)','accumarray (old)',...
'accumarray (new)';
timeit(#() histci(double(M),double(n))),...
timeit(#() histci(M,n)),...
timeit(#() accumi(M)),...
timeit(#() accumi_new(M))
};
out = cell2mat(func_times(2,:));
end
function spp = histci(M,n)
spp = histcounts(M,1:n+1);
end
function spp = accumi(M)
spp = accumarray(M(M>0),1);
end
function spp = accumi_new(M)
spp = accumarray(M+1,1);
spp = spp(2:end);
end
I am working with Bank of America time series data for stock prices. I am trying to store the forecasted value for a specific step ahead (in this case 1:20 steps) in an array. I then need to subtract each value of the array from each value of the test array. Then I have to square each value of the array, sum all the squared values of the array, then divide by N (N = number of steps forecasted ahead).
I have the following so far. Also, the quantmod and fpp libraries are needed for this.
---------Bank of America----------
library(quantmod)
library(fpp)
BAC = getSymbols('BAC',from='2009-01-02',to='2014-10-15',auto.assign=FALSE)
BAC.adj = BAC$BAC.Adjusted
BAC.daily=dailyReturn(BAC.adj,type='log')
test = tail(BAC.daily, n = 20)
train = head(BAC.daily, n = 1437)
Trying to write a function to forecast, extract requisite value (point forecast for time i), then store it in an array where I can perform operations on that array (i.e. - add, multiply, exponentiate, sum the values of the array)
MSE = function(N){
for(i in 1:(N)){
x = forecast(model1, h = i)
y = x$mean
w = as.matrix(as.double(as.matrix(unclass(y))))
p = array(test[i,]-w[i,])
}
}
and we also have:
model1 = Arima(train, order = c(0,2,0))
MSE = function(N){
result = vector("list", length = (N))
for(i in 1:(N)){
x = forecast(model1, h = i)
point_forecast = as.double(as.matrix(unclass(x$mean)))
result[i] = point_forecast
}
result = as.matrix(do.call(cbind, result))
}
Neither of these functions have worked so far. When I run the MSE function, I get the following errors:
> MSE(20)
There were 19 warnings (use warnings() to see them)
> warnings()
Warning messages:
1: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
2: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
3: In result[i] = point_forecast :
number of items to replace is not a multiple of replacement length
4: In result[i] = point_forecast :
When I run MSE2 function, I get the following ouput:
MSE2(20)
[1] -0.15824
When putting a print statement inside, it printed out 'p' as a singular number, just like above (even though that had been run for i = 20). The x,y, and w variable in the MSE2 function act as vectors as far as storing the output, so I do not understand why p does not as well.
I appreciate any help in this matter, thank you.
Sincerely,
Mitchell Healy
Your question has two MSE functions: one in the first code block and one in the second code block.
Also, library(forecast) is needed to run Arima and forecast.
My understanding of what you are trying to do in the first paragraph is to compute the 20-step ahead forecast error. That is, what is the error in forecasts from model1 20 days ahead, based on your test data. This can be done in the code below:
model1 <- Arima(train, order = c(0,2,0))
y_fcst<-forecast(model1,h=20)$mean
errors<-as.vector(y_fcst)-as.vector(test)
MSE.fcst<-mean(errors^2)
However, I'm not sure what you're trying to do here: an ARIMA(0,2,0) model is simply modelling the differences in returns as a random walk. That is, this model just differences the returns twice and assumes this twice-differenced data is white noise. There's no parameters other than $\sigma^2$ being estimated.
Rob Hyndman has a blog post covering computing errors from rolling forecasts.
My solution to finding the MSE is below. I used log adjusted daily return data from Bank of America gathered through quantmod. Then I subsetted the data (which had length 1457) into training[1:1437] and testing[1438:1457].
The solution is:
forc = function(N){
forecast = matrix(data = NA, nrow = (N) )
for(i in 1:N){
fit = Arima(BAC.adj[(1+(i-1)):(1437+(i-1))], order = c(0,0,4))
x = forecast(fit, h = 1)
forecast[i,] = as.numeric(x$mean)
}
error = test - forecast
error_squared = error^2
sum_error_squared = sum(error_squared)
MSE = sum_error_squared/N
MSE
}
I am currently looking at Binomial Option Pricing. I have written the code below, which works fine, when you enter the variables in one at a time. However, entering each set of values is very tedious, and I need to be able to analyse a large set of data. I have created arrays for each of the variables. But, I keep getting the error; A(I) = B, the number of elements in B must equal I. The function is shown below.
function C = BinC(S0,K,r,sig,T,N);
% PURPOSE:
% To return the value of a European call option using the Binomial method
%-------------------------------------------------------------------------
% INPUTS:
% S0 - The initial price of the underlying asset
% K - The strike price
% r - The risk free rate of return, expressed as a decimal
% sig - The volatility of the underlying asset, expressed as a decimal
% T - The time to maturity, expressed as a decimal
% N - The number of steps
%-------------------------------------------------------------------------
dt = T/N;
u = exp(sig*sqrt(dt));
d = 1/u;
p = (exp(r*dt) - d)/(u - d);
S = zeros(N+1,1);
% Price of underlying asset at time T
for n = 1:N+1
S(n) = S0*(d^(N+1-n))*(u^(n-1));
end
% Price of Option at time T
for n = 1:N+1
C(n) = max(S(n)- K, 0);
end
% Backtrack to get option price at time 0
for i = N:-1:1
for n = 1:i
C(n) = exp(-r*dt)*(p*C(n+1) + (1-p)*C(n));
end
end
disp(C(1))
After importing my data, I entered this in to the command window.
for i=1:20
w(i)= BinC(S0(i),K(i),r(i),sig(i),T(i),N(i));
end
When I enter w, all I get back is w = []. I have no idea how I can make A(I) = B. I apologise, if this is a very silly question, but I am new to Matlab and in need of help. Thanks
Your function computes an entire vector C, but displays only C(1). This display is deceptive: it makes you think the function is returning a scalar, but it's not: it's returning the entire vector C, which you try to store into a scalar location.
The solution is simple: Change your function definition to this (rename the output variable):
function out = BinC(S0,K,r,sig,T,N);
Then at the last line of the function, remove the disp, and replace it with
out = C(1);
To verify all of this (compare with your non-working example), try calling it by itself at the command line, and examine the output.
This question is related to matlab: find the index of common values at the same entry from two arrays.
Suppose that I have an 1000 by 10000 matrix that contains value 0,1,and 2. Each row are treated as a sample. I want to calculate the pairwise distance between those samples according to the formula d = 1-1/(2p)sum(a/c+b/d) where a,b,c,d can treated as as the row vector of length 10000 according to some definition and p=10000. c and d are probabilities such that c+d=1.
An example of how to find the values of a,b,c,d: suppose we want to find d between sample i and bj, then I look at row i and j.
If kth entry of row i and j has value 2 and 2, then a=2,b=0,c=1,d=0 (I guess I will assign 0/0=0 in this case).
If kth entry of row i and j has value 2 and 1 or vice versa, then a=1,b=0,c=3/4,d=1/4.
The similar assignment will give to the case for 2,0(a=0,b=0,c=1/2,d=1/2),1,1(a=1,b=1,c=1/2,d=1/2),1,0(a=0,b=1,c=1/4,d=3/4),0,0(a=0,b=2,c=0,d=1).
The matlab code I have so far is using for loops for i and j, then find the cases above by using find, then create two arrays for a/c and b/d. This is extremely slow, is there a way that I can improve the efficiency?
Edit: the distance d is the formula given in this paper on page 13.
Provided those coefficients are fixed, then I think I've successfully vectorised the distance function. Figuring out the formulae was fun. I flipped things around a bit to minimise division, and since I wasn't aware of pdist until #horchler's comment, you get it wrapped in loops with the constants factored out:
% m is the data
[n p] = size(m, 1);
distance = zeros(n);
for ii=1:n
for jj=ii+1:n
a = min(m(ii,:), m(jj,:));
b = 2 - max(m(ii,:), m(jj,:));
c = 4 ./ (m(ii,:) + m(jj,:));
c(c == Inf) = 0;
d = 1 - c;
distance(ii,jj) = sum(a.*c + b.*d);
% distance(jj,ii) = distance(ii,jj); % optional for the full matrix
end
end
distance = 1 - (1 / (2 * p)) * distance;