I am doing simulations for
time step = 1.0 e^-7 &
total number of steps ==> nsteps = 1.0 e^8
I have to find the total time of simulation at which n# steps is achieved.
Is it ok to multiply both to get the time of simulation?
time of simulation = time step * total number of steps
time of simulation = 1.0 e^-7 * 1.0 e^8
time of simulation = 10
Is this right or wrong?
Thanks in advance.
This is a yes/no question, so:
No, (or yes, depending on how accurately you want the answer to be)!
But you are really close... You need to subtract one time_step, thus the answer is really:
time_of_simulation = time_step * total_number_of_steps - time_step;
You will see the reason if you consider counting seconds. Start with a number and see how far you get if you count one second at a time.
1, 2, 3 => Three measurements, but only 2 seconds.
However, in your case, I guess you are close enough without the last subtraction, because
time of simulation = 9.999999 is pretty close to 10
Related
I am not really an expert in Tableau. I have a need to calculate a timedifference in hours, but also want to see fraction of an hour. I am using Tableau 9.
I used the function
IF DATEDIFF("hour", [CL2_Start_Time_ST], [CL2_End_Time_ST]) > 8 then NULL
ELSE DATEDIFF("hour", [CL2_Start_Time_ST], [CL2_End_Time_ST])
END
If the time difference between CL2_Start_Time_ST and CL2_End_Time_ST is less than 1 hour (for example 30 minutes) the result is 0, but I want to see 0.5 in result.
I dont want to calculate in time difference in minutes since all my other calculations are in hours and hence it is easier to create a relative plot with other calculations.
Please help.
I found the answer to the above question. The simple formula below worked. I was using DIV function and that caused the issue.
IF DATEDIFF("hour", [CL2_Start_Time_ST], [CL2_End_Time_ST]) > 8 then NULL
ELSE (DATEDIFF("minute", [CL2_Start_Time_ST], [CL2_End_Time_ST])) / 60
END
I have an EMG signal of a subject walking on a treadmill.
We used footswitches to be able to see when the subject is placing his foot, so we can see how many periods (steps) there are in time.
We would like to cut the signal in periods (steps) and get one average signal (100% step cycle).
I tried the reshape function but it does not work
when I count 38 steps:
nwaves = 38;
sig2 = reshape(sig,[numel(sig)/nwaves nwaves])';
avgSig = mean(sig2,1);
plot(avgSig);
the error displayed is this: Size arguments must be real integers.
Can anyone help me with this? Thanks!
First of all, reshaping the array is a bad approach to the problem. In real world one cannot assume that the person on the treadmill will step rhythmically with millisecond-precision (i.e. for the same amount of samples).
A more realistic approach is to use the footswitch signal: assume is really a switch on a single foot (1=foot on, 0=foot off), and its actions are filtered to avoid noise (Schmidt trigger, for example), you can get the samples index when the foot is removed from the treadmill with:
foot_off = find(diff(footswitch) < 0);
then you can transform your signal in a cell array (variable lengths) of vectors of data between consecutive steps:
step_len = diff([0, foot_off, numel(footswitch)]);
sig2 = mat2cell(sig(:), step_len, 1);
The problem now is you can't apply mean() to the signal slices in order to get an "average step": you must process each step first, then average the results.
It's probably because numel(sig)/nwaves isn't an integer. You need to round it to the nearest integer with round(numel(sig)/nwaves).
EDIT based on comments:
Your problem is you can't divide 51116 by 38 (it's 1345.2), so you can't reshape your signal in chunks of 38 long. You need a signal whose length is exactly a multiple of 38 if you want to be able to reshape it in chunks of 38. Either that, or remove the last (or first) 6 values from your signal to have an exact multiple of 38 (1345 * 38 = 51110):
nwaves = 38;
n_chunks = round(numel(sig)/nwaves);
max_sig_length = n_chunks * nwaves;
sig2 = reshape(sig(1:max_sig_length),[n_chunks nwaves])';
avgSig = mean(sig2,1);
plot(avgSig);
I am processing 1Hz timestamps (variable 'timestamp_1hz') from a logger which doesn't log exactly at the same time every second (the difference varies from 0.984 to 1.094, but sometimes 0.5 or several seconds if the logger burps). The 1Hz dataset is used to build a 10 minute averaged dataset, and each 10 minute interval must have 600 records. Because the logger doesn't log exactly at the same time every second, the timestamp slowly drifts through the 1 second mark. Issues come up when the timestamp cross the 0 mark, as well as the 0.5 mark.
I have tried various ways to pre-process the timestamps. The timestamps with around 1 second between them should be considered valid. A few examples include:
% simple
% this screws up around half second and full second values
rawseconds = raw_1hz_new(:,6)+(raw_1hz_new(:,7)./1000);
rawsecondstest = rawseconds;
rawsecondstest(:,1) = floor(rawseconds(:,1))+ rawseconds(1,1);
% more complicated
% this screws up if there is missing data, then the issue compounds because k+1 timestamp is dependent on k timestamp
rawseconds = raw_1hz_new(:,6)+(raw_1hz_new(:,7)./1000);
A = diff(rawseconds);
numcheck = rawseconds(1,1);
integ = floor(numcheck);
fract = numcheck-integ;
if fract>0.5
rawseconds(1,1) = rawseconds(1,1)-0.5;
end
for k=2:length(rawseconds)
rawsecondstest(k,1) = rawsecondstest(k-1,1)+round(A(k-1,1));
end
I would like to pre-process the timestamps then compare it to a contiguous 1Hz timestamp using 'intersect' in order to find the missing, repeating, etc data such as this:
% pull out the time stamp (round to 1hz and convert to serial number)
timestamp_1hz=round((datenum(raw_1hz_new(:,[1:6])))*86400)/86400;
% calculate new start time and end time to find contig time
starttime=min(timestamp_1hz);
endtime=max(timestamp_1hz);
% determine the contig time
contigtime=round([floor(mean([starttime endtime])):1/86400:ceil(mean([starttime endtime]))-1/86400]'*86400)/86400;
% find indices where logger time stamp matches real time and puts
% the indices of a and b
clear Ia Ib Ic Id
[~,Ia,Ib]=intersect(timestamp_1hz,contigtime);
% find indices where there is a value in real time that is not in
% logger time
[~,Ic] = setdiff(contigtime,timestamp_1hz);
% finds the indices that are unique
[~,Id] = unique(timestamp_1hz);
You can download 10 days of the raw_1hz_new timestamps here. Any help or tips would be much appreciated!
The problem you have is that you can't simply match these stamps up to a list of times, because you could be expecting a set of datapoints at seconds = 1000, 1001, 1002, but if there was an earlier blip you could have entirely legitimate data at 1000.5, 1001.5, 1002.5 instead.
If all you want is a list of valid times/their location in your series, why not just something like (times in seconds):
A = diff(times); % difference between times
n = find(abs(A-1)<0.1) % change 0.1 to whatever your tolerance is
times2 = times(n+1);
times2 should then be a list of all your timestamps where the previous timestamp was approximately 1 second ago - works on a small set of fake data I constructed, didn't try it on yours. (For future reference: it would be more help to provide a small subset of your data, e.g. just a few minutes worth, that you know contains a blip).
I would then take the list of valid timestamps and split it up into 10 minute sections for averaging, counting how many valid timestamps were obtained in each section. If it's working, you should end up with no more than 600 - but not much less if the blips are occasional.
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 question to find a complexity class estimate of a algorithm. The question gives recorded times for an algorithm. So, do I just average out the times based on how it was computed?
Sorry, I missed a part.
ok, so it recorded time like N = 100, Algorithm = 300, next N = 200, Algorithm = 604, next N = 400 Algorithm = 1196, next N = 800 Algorithm 2395. So, do i calculate like 300/100, and 604/200 and find the average. Is that how I'm supposed to estimate the complexity class for the algorithm?
Try plotting running time vs. N and see if you get any insight. (e.g. if running time = f(N), is f(N) about equal to log(N), or sqrt(N), or... ?)
I don't think time will help figure out it's complexity class. Times can be very different even on exactly the same task (depends on the scheduler or other factors.)
Look at how many more steps it takes as your input get's larger. So if you had a sorting algorithm that took 100 steps to sort 10 items and 10000 steps to sort 100 items you'd say sorted in big O ( N^2 ) since
Input Steps
10 100 (which equals 10*10)
100 10000 (which equals 100*100)
It's not about averaging but looking for a function that maps the input to the number of steps and then finding what part of that function grows the fastest ( N^2 grows faster than N so if your function was N^2 + N you classify it as N^2).
At least that's how I remember it, but it's been ages!! :)
EDIT: Now that there are more details in your question, here is what I'd do, with the above in mind.
Don't think about averaging anything, just think about how f(100) = 300, f(200)=604, and f(400)=1196.
And it doesn't have to be exact, just in the ball park. So a simple linear function (such as f(x)=3*N ) where f(100)=300, f(200)=600, and f(400)=1200 that would describe the complexity of the algorithm you could say the complexity class of the algorithm was linear or big O(N).
Hope that helps!
Does it give you the inputs to the algorithm as well, which produce the recorded times? You can deduce the growth order according to the input size vs output running time.
i.e. input = 1, running time = 10
input = 100, running time = 100000
would appear to be O(N^2)
i.e.
with input = 1 and running time = 10, likely O(cn) where C = 10
with n = 1, N^2 and N are the same
with input = 10 and running time = 100000, likely O(cN^2) where C = 10
and N = 100*100 = 10000, * 10 = 100000
Hint: Calculate how much time the algorithm spent to process one single item.
How does these calculates time relate to each other?
Does the algorithm alway spent the same time to process one item, regardless how many items, is there a factor? maybe the time raises exponentially?