I'm quite new to R, so I still struggle a bit with what may be some of the basics. I have movement data for several individuals, and some of them have been monitored for over a year. In these cases, I'm trying to partition the data into bins of max. one year (i.e. 365 consecutive days) per individual. For example, if one individual was tracked for 800 days, I would need to partition the data into the first 365 days, the second 365 days, and the remaining 70 days. Or, if one individual started on 2019/04/17, I'd like to subset the data until 2020/04/16, with the next subset beginning on 2020/04/17.
Individuals were added in different moments so not all monitoring periods begin on the same day, and per individual there can be more than one observation per day, so many rows share the same date. Naturally I want to use the timestamp column for this, but have been looking for ways and can't seem to find one. Is there a way to tell R to pick the first date and extract the next 365 days?
I could manually calculate each bin and try to partition it by hand, but I was wondering if there was a simpler way for this. I can however separate the data per individual.
Thanks!
My data looks something like this
Date.and.time Ind Lat Long
2019-04-02 08:54:03 Animal_1 Y X
2019-04-02 09:01:13 Animal_2 Y X
2019-04-02 15:45:22 Animal_1 Y X
2019-04-03 17:31:50 Animal_1 Y X
.
.
.
2021-10-14 12:34:56 Animal_1 Y X
2021-10-15 16:05:50 Animal_20 Y X
2021-10-15 22:29:37 Animal_15 Y X
I have daily time series data and I want to calculate 5-day averages of that data while also retrieving the corresponding start date for each of the 5-day averages. For example:
x = [732099 732100 732101 732102 732103 732104 732105 732106 732107 732108];
y= [1 5 3 4 6 2 3 5 6 8];
Where x and y are actually size 92x1.
Firstly, how do I compute the 5-day mean when this time series data is not divisible by 5? Ultimately, I want to compute the 'jumping mean', where the average is not computed continuously (e.g., June 1-5, June 6-10, and so on).
I've tried doing the following:
Pentad_avg = mean(reshape(y(1:90),5,[]))'; %manually adjusted to be divisible by 5
Pentad_dt = x(1:5:90); %select every 5th day for time
However, Pentad_dt gives me dates 01-Jun-2004 and 06-Jun-2004 as output. And, that brings me to my second point.
I am looking to find 5-day averages for x and y that correspond to 5-day averages of another time series. This second time series has 5-day averaged data starting from 15-Jun-2004 until 29-Aug-2004 (instead of starting at 01-Jun-2004). Ultimately, how do I align the dates and 5-day averages between these two time series?
Synchronization between two time series can be accomplished using the timeseries object. Placing your data into an object allows Matlab to intelligently process it. The most useful thing is adds for your usage is the synchronize method.
You'll want to make sure to properly set the time vector on each of the timeseries objects.
An example of what this might look like is as follows:
ts1 = timeseries(y,datestr(x));
ts2 = timeseries(OtherData,OtherTimes);
[ts1 ts2] = synchronize(ts1,ts2,'Uniform','Interval',5);
This should return to you each timeseries aligned to be with the same times. You could also specify a specific time vector to align a timeseries to using the resample method.
I am writing a for loop to average 10 years of hourly measurements made on the hour. The dates of the measurements are recorded as MATLAB datenums.
I am trying to iterate through using 0.0417 as it is the datenum for 1AM 00/00/00 but it is adding in a couple of seconds of error each time I iterate.
Can anyone recommend a better way for me to iterate by hour?
date = a(:,1);
load = a(:,7);
%loop for each hour of the year
for i=0:0.0417:366
%set condition
%condition removes year from current date
c = date(:)-datenum(year(date(:)),0,0)==i;
%evaluate condition on load vector and find mean
X(i,2)=mean(load(c==1));
end
An hour has a duration of 1/24 day, not 0.0417. Use 1/24 and the precision is sufficient high for a year.
For an even higher precision, use something like datenum(y,1,1,1:24*365,0,0) to generate all timestamps.
To avoid error drift entirely, specify the index using integers, and divide the result down inside the loop:
for hour_index=1:365*24
hour_datenum = (hour_index - 1) / 24;
end
This question already has an answer here:
Summing up till a certain interval
(1 answer)
Closed 8 years ago.
I have a dataset (which is an hdf5 file) that contains a list of 250000 values, all quite small (sub 10). I want to cut this into 5000 pieces, so 50 each, and I want to individually sum the values in each of these 5000 pieces. I then want to create a histogram of these 5000 pieces, so I need to store them as well.
I am trying to do this using MATLAB, as my very limited programming skills have been developed using this, and it seems suitable for these purposes. Now, I haven't gotten very far, but what I have done so far is:
for n = 1:50:249951
% CR check before (before pumping?)
ROdata = h5read('hdf5file', '/data', [n], [n+49]);
sum(ROdata)
end
Of course, this does not yet store the values of the sum for each n. But more importantly, it does not work. For n = 1, all is fine, and I get the correct value. But already for n = 51, (so summing 51-100), I do not get the correct sum. What's going wrong here?
How should I store these (not working) sums?
Are you looking for something like this?
I assumed you already read your data and you have a 250000x1 vector.
%example data
data = randi(42,1,250000);
% rearranges your vector in groups of 50
A = reshape(data,[],5000);
% sums every column of your reshaped vector
sums = sum(A,1);
hist(sums,5000)
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.