I am trying to perform a binning average. I am using the code:
Avg = mean(reshape(a,300,144,27));
AvgF = squeeze(Avg);
The last line gets rid of singleton dimensions.
So as can be seen I am averaging over 300 points. It works fine except for times when I have a total number of points not equal to a multiple of 144*300.
Is there any way to make this binning average work even when the total number of points is not a multiple of 144*300?
EDIT: Sorry if my question sounded confusing. To clarify...
I have a file with 43200 rows and 27 columns. I am averaging by binning 300 rows at a time, which means in the end I am left with a matrix of size 144-by-27.
My code as I wrote it above works only when I have exactly 43200 rows. In some cases I have 43199, 43194, etc.. The reshape function works when I have a total number of rows that is a multiple of 300 (the bin size). Is there a way to make this binning average work when my total number of rows is not a multiple of 300?
I think I understand the problem better now...
If a is the data read from your file (of size N-by-27, where N is ideally 43,200), then I think you would want to do the following:
nRemove = rem(size(a,1),300); %# Find the number of points to remove
a = a(1:end-nRemove,:); %# Trim points to make an even multiple of 300
Avg = mean(reshape(a,300,[],27));
AvgF = squeeze(Avg);
This will remove points such that the number of rows in a will be a multiple of 300. Then your reshape and average should work. Note that I use [] in the call to RESHAPE, which lets it figure out what the number of column should be.
Related
I am using MATLAB to find the number of peaks of a signal.
I'm trying to plot the number of peaks of a signal filtered with N-point moving average filter, N goes from 2 to 30.(I also consider the number of peaks when no filter has applied at the beginning of the resulting array) My data array(imported from csv and has double values between 0 and 1) has around 50k points. When I give part of the data i.e 100, 500 or 1000 points, using array slicing, # of peaks decrease as expected. However, when I give the whole data or even 2000 points, the number of peaks stays same at 127.
I changed the number of data given to the filter to find out why this happens. I changed the commented lines like showed in the comment and tried. When less than 1000 data points given plot was fine.
Here is the signal
https://www.dropbox.com/s/e1bkcjn5ta5q610/exampleSignal.csv?dl=0
Please import it from 4th element to end, it has some strange data at the beginning, I have not taken them, VarName1 is the imported column vector's name
numberOfPeaks = zeros(30,1,'int8');
pks = findpeaks(VarName1); % VarName1(1:1000,:) (when no filter applied)
numberOfPeaks(1) = size(pks,1);
for i=2:30
h = 1/i*ones(1,i,'double');
y = filter(h,1,VarName1); % VarName1(1:1000,:)
numberOfPeaks(i) = size(findpeaks(y),1);
end
plot(1:30,numberOfPeaks);
I expect a plot like this when whole the data is given:
but I get:
I realised that the problem is int8 I use. It can only take up to 127 and this caused my big results to be as 127.
Turning it into double solves the problem.
I have pupil size data from an eye tracking experiment. In the experiment, to start each trial the participant looks in the centre of the screen for 1000 ms (prefixation period), then the trial begins. If they look away or blink, the 1000 ms period restarts. So, each trial has a different length prefixation period. When I create a matrix (with each row being a different trial, and each column being a pupil size sample over time) it creates the matrix with the number of columns based on the trial with the longest prefixation period, and then adds varying numbers of NaNs to the end of each other row. I need to extract the last 200 samples (columns) of each trial, but these are not the last 200 columns of the matrix because of the additional NaNs that are added.
At the moment I have this:
Row1 = PreFixBase(1,:); % extract the first row
Row1(isnan(Row1)) = []; %get rid of the NaNs
Row1Base = Row1(end-200+1:end); %extract the last 200 samples / columns
which I do for each row separately and then paste them back together. It works, but is really inefficient (I have 324 rows / trials) and I'm sure there must be a more concise way of doing this, but haven't been able to find the answer.
Any help appreciated.
Amy
You can use cumsum + logical indexing to extract the desired elements
Base = PreFixBase.'; % transpose the matrix
S=cumsum(~isnan(Base),1,'reverse'); % number last non NaN columns from 1 to 200 (from end to begin)
Result = reshape(Base(S> 0 & S<=200),200,[]).'; % extract data and reshape to the correct size
I have a set of ages (over 10000 of them) and I want to plot a graph with the age from 20 to 100 on the x axis and then the number of times each of those ages appears in the data on the y axis. I have tried several ways to do this and I can't figure it out. I also have some other data which requires me to plot values vs how many times they occur so any advice on how to do this would be much appreciated.
I'm quite new to Matlab so it would be great if you could explain how things in your answer work rather than just typing out some code.
Thanks.
EDIT:
So I typed histogram(Age, 80) because as I understand that will plot the values in Age on a histogram split up into 80 bars (1 for each age). Instead I get this:
The bars aren't aligned and it's clearly not 1 per age nor has it plotted the number of times each age occurs on the y axis.
You have to use histogram(), and that's correct.
Let's see with an example.
I extract 100 ages between 20 and 100:
ages=randsample([20:100],100,true);
Now I call histogram() in this manner:
h=histogram(ages,[20:100]);
where h is an histogram object and this will also show the following plot:
However, this might look easy due to the fact that my ages vector is in range 20:100, so it will not contain any other values. If your vector, as instead, contains also ages not in range 20:100, you can specify the additional option 'BinLimits' as third input in histogram() like this:
h=histogram(ages,length([20:100]),'BinLimits',[20:100]);
and this option plots a histogram using the values in ages that fall between 20 and 100 inclusive.
Note: by inspecting h you can actually see and/or edit some proprieties of your histogram. An attribute (field) of such object you might be interested to is Values. This is a vector of length 80 (in our case, since we work with 80 bins) in which the i-th element is the number of items is the i-th bin. This will help you count the occurrences (just in case you need them to go on with your analysis).
Like Luis said in comments, hist is the way to go. You should specify bin edges, rather than the number of bins:
ages = randi([20 100], [1 10000]);
hist(ages, [20:100])
Is this what you were looking for?
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.
I have a time series of measurements taken at different depths of a water column. I have divided these into individual cells (for later) and require some help on how to complete the following: e.g.
time = [733774,733774,733775,733775,733775,733776,733776];
bthD = [20,10,0,15,10,20,10];
bthA = (1000:100:1600);
%Hypsographic
Hypso = [(10:1:20)',(1000:100:2000)'];
d = [1,1.3,1,2.5,2.5,1,1.2];
data = horzcat(time',bthD',d');
uniqueTimes = unique(time);
counts = hist(time,uniqueTimes);
newData = mat2cell(data,counts,length(uniqueTimes));
So, in newData I have three cells, that correspond to different days of measurements, in each cell I have newData(:,1) being time, newData(:,2) being depth, and newData(:,3) being the measurement. I would like to find what the area is at each depth in the cells, the area at different depths is given in the variable 'Hypso'.
How could I achieve this?
Your problem formulation is excellent! Very easy to understand what you need here. All you need is the function interp1. Use the first column of Hypso, I assume, as your depth, and the second column as the area. You can use the vectorized ability of the interp1 function to find all values in one call:
areaAtDepth = interp1(Hypso(:,1),Hypso(:,2),bthD)
areaAtDepth =
Columns 1 through 6
2000 1000 NaN 1500 1000 2000
Column 7
1000
You'll notice the Nan in the third column of the output. This is because it's associated depth, 0, is outside the range of the data, or support of the data I believe. You'll need to decide what you want to do when data is outside the range, or perhaps it never should be, so an error should be logged; it's up to you! Let me know if you have any more questions!