The following code should generate two random letters:
lett_array = ['abcdefghijklmnobqrstuvwxyz'];
SaveStr.random = ['_',lett_array(randi(26)),lett_array(randi(26))];
It works exactly as expected on my local machine (adds random letters to a save file name to maintain uniqueness).
However on a cluster using slurm scheduling, I get back well over 200 results that ALL CALCULATE THE SAME PAIR OF LETTERS (runs initiate within 15 mins of each other; end over the span of 45 minutes).
This seems... weird. I could probably force reasonable behaviour using
rng('shuffle');
lett_array = ['abcdefghijklmnobqrstuvwxyz'];
SaveStr.random = ['_',lett_array(randi(26)),lett_array(randi(26))];
but I don't understand how I'm having the problem in the first place.
Thoughts? Seems a big problem that matlab doesn't get reliably random numbers on linux clusters.
As stated in the rng (random number generator) documentation:
rng('default') puts the settings of the random number generator used by rand, randi, and randn to their default values. This way, the same random numbers are produced as if you restarted MATLAB.
The default settings are the Mersenne Twister with seed 0.
So you're getting the same values because the cluster nodes each act on a new instance on MATLAB, which uses rng('default') as the initialisation for the random number generator.
Using rng shuffle or rng('shuffle') is the correct thing to do if you want independent randomness on the cluster nodes.
You might also be interested to know about the char function. Since your let_array array is equivalent to ASCII 97:122, you can simply forget the letters array and do a single call to randi like so:
SaveStr.random = ['_', char(randi([97,122], 1, 2))];
Related
Why do I get different results when using the same code running in different version of MATLAB (2016 vs 2021) for sum(b.*x1) where b is single and x1 is double. How to avoid such error between MATLAB version?
MATLAB v.2021:
sum(b.*x1)
ans =
single
-0.0013286
MATLAB 2016
sum(b.*x1)
ans =
single
-0.0013283
In R2017b, they changed the behavior of sum for single-precision floats, and in R2020b they made the same changes for other data types too.
The change speeds up the computation, and improves accuracy by reducing the rounding errors. Simply put, previously the algorithm would just run through the array in sequence, adding up the values. The new behavior computes the sum over smaller portions of the array, and then adds up those results. This is more precise because the running total can become a very large number, and adding smaller numbers to it causes more rounding in those smaller numbers. The speed improvement comes from loop unrolling: the loop now steps over, say, 8 values at the time, and in the loop body, 8 running totals are computed (they don’t specify the number they use, the 8 here is an example).
Thus, your newer result is a better approximation to the sum of your array than the old one.
For more details (a better explanation of the new algorithm and the reason for the change), see this blog post.
Regarding how to avoid the difference: you could implement your own sum function, and use that instead of the builtin one. I would suggest writing it as a MEX-file for efficiency. However, do make sure you match the newer behavior of the builtin sum, as that is the better approximation.
Here is an example of the problem. Let's create an array with N+1 elements, where the first one has a value of N and the rest have a value of 1.
N = 1e8;
a = ones(N+1,1,'single');
a(1) = N;
The sum over this array is expected to be 2*N. If we set N large enough w.r.t. the data type, I see this in R2017a (before the change):
>> sum(a)
ans =
single
150331648
And I see this in R2018b (after the change for single-precision sum):
>> sum(a)
ans =
single
199998976
Both implementations make rounding errors here, but one is obviously much, much closer to the expected result (2e8, or 200000000).
I am currently trying to learn MATLAB independently and had a question about a command that used randn().
nddata = fix(8*randn(10,5,3))
I understand what the fix() function does, and the multi dimension array that is created by randn. However, I am not sure what 8 is doing here, it is not multiplying the outcome of the random numbers and it is not part of the limit. So I just want to know the purpose of the 8 here.
Thanks
randn generated a standard normally distributed matrix of random numbers (standard in this context is defined as mean = 0 and standard deviation = 1). The 8 factor simply stretches this distribution along the x-axis; a scalar multiplication for each value in the 3D matrix. The fix function then rounds each value to the nearest integer towards 0, i.e. -3.9 becomes -3.0. This effectively reduces the standard deviation of the data.
To see this for yourself, split the expression up and create temporary variables for each operation, and step through it with the debugger.
I have smallish(?) sets (ranging in count from 0 to 100) of unsigned 32 bit integers. For a given set, I want to come up with minimal parameters to describe a minimal(istic) perfect hash of the given set. High level of the code I used to experiment with the idea ended up something like:
def murmur(key, seed=0x0):
// Implements 32bit murmur3 hash...
return theHashedKey
sampleInput = [18874481, 186646817, 201248225, 201248705, 201251025, 201251137, 201251185, 184472337, 186649073, 201248625, 201248721, 201251041, 201251153, 184473505, 186649089, 201248657, 201251009, 201251057, 201251169, 186646818, 201248226, 201248706, 201251026, 201251138, 201251186, 186649074, 201248626, 201248722, 201251042, 201251154, 186649090, 201248658, 201251010, 201251058, 201251170]
for seed in range(11111): // arbitrary upper seed limit
for modulus in range(10000):
hashSet = set((murmur(x, seed=seed) % modulus for x in sampleInput))
if len(hashSet) >= len(allValves):
print('minimal modulus', modulus, 'for seed', seed)
break
This is just basic pseudo code for a 2 axis brute force search. I add lines by keeping track of the different values, I can find seed and modulus values that give a perfect hash and then select the one with the smallest modulus.
It seems to me that there should be a more elegant/deterministic way to come up with these values? But that's where my math skills overflow.
I'm experimenting in Python right now, but ultimately want to implement something in C on a small embedded platform.
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 two arrays of data that I'm trying to amalgamate. One contains actual latencies from an experiment in the first column (e.g. 0.345, 0.455... never more than 3 decimal places), along with other data from that experiment. The other contains what is effectively a 'look up' list of latencies ranging from 0.001 to 0.500 in 0.001 increments, along with other pieces of data. Both data sets are X-by-Y doubles.
What I'm trying to do is something like...
for i = 1:length(actual_latency)
row = find(predicted_data(:,1) == actual_latency(i))
full_set(i,1:4) = [actual_latency(i) other_info(i) predicted_info(row,2) ...
predicted_info(row,3)];
end
...in order to find the relevant row in predicted_data where the look up latency corresponds to the actual latency. I then use this to created an amalgamated data set, full_set.
I figured this would be really simple, but the find function keeps failing by throwing up an empty matrix when looking for an actual latency that I know is in predicted_data(:,1) (as I've double-checked during debugging).
Moreover, if I replace find with a for loop to do the same job, I get a similar error. It doesn't appear to be systematic - using different participant data sets throws it up in different places.
Furthermore, during debugging mode, if I use find to try and find a hard-coded value of actual_latency, it doesn't always work. Sometimes yes, sometimes no.
I'm really scratching my head over this, so if anyone has any ideas about what might be going on, I'd be really grateful.
You are likely running into a problem with floating point comparisons when you do the following:
predicted_data(:,1) == actual_latency(i)
Even though your numbers appear to only have three decimal places of precision, they may still differ by very small amounts that are not being displayed, thus giving you an empty matrix since FIND can't get an exact match.
One feature of floating point numbers is that certain numbers can't be exactly represented, since they aren't an integer power of 2. This occurs with the numbers 0.1 and 0.001. If you repeatedly add or multiply one of these numbers you can see some unexpected behavior. Amro pointed out one example in his comment: 0.3 is not exactly equal to 3*0.1. This can also be illustrated by creating your look-up list of latencies in two different ways. You can use the normal colon syntax:
vec1 = 0.001:0.001:0.5;
Or you can use LINSPACE:
vec2 = linspace(0.001,0.5,500);
You'd think these two vectors would be equal to one another, but think again!:
>> isequal(vec1,vec2)
ans =
0 %# FALSE!
This is because the two methods create the vectors by performing successive additions or multiplications of 0.001 in different ways, giving ever so slightly different values for some entries in the vector. You can take a look at this technical solution for more details.
When comparing floating point numbers, you should therefore do your comparisons using some tolerance. For example, this finds the indices of entries in the look-up list that are within 0.0001 of your actual latency:
tolerance = 0.0001;
for i = 1:length(actual_latency)
row = find(abs(predicted_data(:,1) - actual_latency(i)) < tolerance);
...
The topic of floating point comparison is also covered in this related question.
You may try to do the following:
row = find(abs(predicted_data(:,1) - actual_latency(i))) < eps)
EPS is accuracy of floating-point operation.
Have you tried using a tolerance rather than == ?