I have a full dataset of lets say 50000 observations which are assigned to 16 classes.
I now want to draw a Sample of let's say 70% of the full data, but I want MATLAB to take the same number of samples from each class (if possible of course, because some classes have less numbers than needed)
Is there a MATLAB function that can do this, or do I have to program a new one for myself? I'm just trying to save time here.
I found cvpartition, but as far as I know this can be used only to take a sample with the same distribution over the classes as the original dataset and not a uniformly distributed sample.
Thank you for your help!
It shouldn't be too hard. Let's say that the observations are in a vector observations. Then you can do
fraction = 0.7;
classes = unique(observations);
nObs = length(observations);
nClasses = length(classes);
nSamples = round(nObs * fraction / nClasses);
for ii = 1:nClasses
idx = observations == classes(ii);
samples((ii-1)*nSamples+1:ii*nSamples) = randsample(observations(idx), nSamples);
end
Now samples is a vector of length nClasses * nsamples that contains your sampled observations, with an equal number from each class.
At the moment it will fail if one of the classes doesn't contain at least nSamples observations. The simplest fix is to add the additional arguments 'replace','true' to the call to randsample, which will tell it to replace each observation after being picked.
Related
I would like to identify the largest possible contiguous subsample of a large data set. My data set consists of roughly 15,000 financial time series of up to 360 periods in length. I have imported the data into MATLAB as a 360 by 15,000 numerical matrix.
This matrix contains a lot of NaNs due to some of the financial data not being available for the entire period. In the illustration, NaN entries are shown in dark blue, and non-NaN entries appear in light blue. It is these light blue non-NaN entries which I would like to ideally combine into an optimal subsample.
I would like to find the largest possible contiguous block of data that is contained in my matrix, while ensuring that my matrix contains a sufficient number of periods.
In a first step I would like to sort my matrix from left to right in descending order by the number of non-NaN entries in each column, that is, I would like to sort by the vector obtained by entering sum(~isnan(data),1).
In a second step I would like to find the sub-array of my data matrix that is at least 72 entries along the first dimension and is otherwise as large as possible, measured by the total number of entries.
What is the best way to implement this?
A big warning (may or may not apply depending on context)
As Oleg mentioned, when an observation is missing from a financial time series, it's often missing for reason: eg. the entity went bankrupt, the entity was delisted, or the instrument did not trade (i.e. illiquid). Constructing a sample without NaNs is likely equivalent to constructing a sample where none of these events occur!
For example, if this were hedge fund return data, selecting a sample without NaNs would exclude funds that blew up and ceased trading. Excluding imploded funds would bias estimates of expected returns upwards and estimates of variance or covariance downwards.
Picking a sample period with the fewest time series with NaNs would also exclude periods like the 2008 financial crisis, which may or may not make sense. Excluding 2008 could lead to an underestimate of how haywire things could get (though including it could lead to overestimate the probability of certain rare events).
Some things to do:
Pick a sample period as long as possible but be aware of the limitations.
Do your best to handle survivorship bias: eg. if NaNs represent delisting events, try to get some kind of delisting return.
You almost certainly will have an unbalanced panel with missing observations, and your algorithm will have to be deal with that.
Another general finance / panel data point, selecting a sample at some time point t and then following it into the future is perfectly ok. But selecting a sample based upon what happens during or after the sample period can be incredibly misleading.
Code that does what you asked:
This should do what you asked and be quite fast. Be aware of the problems though if whether an observation is missing is not random and orthogonal to what you care about.
Inputs are a T by n sized matrix X:
T = 360; % number of time periods (i.e. rows) in X
n = 15000; % number of time series (i.e. columns) in X
T_subsample = 72; % desired length of sample (i.e. rows of newX)
% number of possible starting points for series of length T_subsample
nancount_periods = T - T_subsample + 1;
nancount = zeros(n, nancount_periods, 'int32'); % will hold a count of NaNs
X_isnan = int32(isnan(X));
nancount(:,1) = sum(X_isnan(1:T_subsample, :))'; % 'initialize
% We need to obtain a count of nans in T_subsample sized window for each
% possible time period
j = 1;
for i=T_subsample + 1:T
% One pass: add new period in the window and subtract period no longer in the window
nancount(:,j+1) = nancount(:,j) + X_isnan(i,:)' - X_isnan(j,:)';
j = j + 1;
end
indicator = nancount==0; % indicator of whether starting_period, series
% has no NaNs
% number of nonan series of length T_subsample by starting period
max_subsample_size_by_starting_period = sum(indicator);
max_subsample_size = max(max_subsample_size_by_starting_period);
% find the best starting period
starting_period = find(max_subsample_size_by_starting_period==max_subsample_size, 1);
ending_period = starting_period + T_subsample - 1;
columns_mask = indicator(:,starting_period);
columns = find(columns_mask); %holds the column ids we are using
newX = X(starting_period:ending_period, columns_mask);
Here's an idea,
Assuming you can rearrange the series, calculate the distance (you decide the metric, but if looking at is nan vs not is nan, Hamming is ok).
Now hierarchically cluster the series and rearrange them using either a dendrogram
or http://www.mathworks.com/help/bioinfo/examples/working-with-the-clustergram-function.html
You should probably prune any series that doesn't have a minimum number of non nan values before you start.
First I have only little insight in financial mathematics. I understood it that you want to find the longest continuous chain of non-NaN values for each time series. The time series should be sorted depending on the length of this chain and each time series, not containing a chain above a threshold, discarded. This can be done using
data = rand(360,15e3);
data(abs(data) <= 0.02) = NaN;
%% sort and chop data based on amount of consecutive non-NaN values
binary_data = ~isnan(data);
% find edges, denote their type and calculate the biggest chunk in each
% column
edges = [2*binary_data(1,:)-1; diff(binary_data, 1)];
chunk_size = diff(find(edges));
chunk_size(end+1) = numel(edges)-sum(chunk_size);
[row, ~, id] = find(edges);
num_row_elements = diff(find(row == 1));
num_row_elements(end+1) = numel(chunk_size) - sum(num_row_elements);
%a chunk of NaN has a -1 in id, a chunk of non-NaN a 1
chunks_per_row = mat2cell(chunk_size .* id,num_row_elements,1);
% sort by largest consecutive block of non-NaNs
max_size = cellfun(#max, chunks_per_row);
[max_size_sorted, idx] = sort(max_size, 'descend');
data_sorted = data(:,idx);
% remove all elements that only have block sizes smaller then some number
some_number = 20;
data_sort_chop = data_sorted(:,max_size_sorted >= some_number);
Note that this can be done a lot simpler, if the order of periods within a time series doesn't matter, aka data([1 2 3],id) and data([3 1 2], id) are identical.
What I do not know is, if you want to discard all periods within a time series that don't correspond to the biggest value, get all those chains as individual time series, ...
Feel free to drop a comment if it has to be more specific.
I'm struggling with one of my matlab assignments. I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est. Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them. The cycle repeats 10 times chosing different values each time and adding different random number. Here is a part of my code:
steps=10;
for s=1:steps
for i=1:1:5
rl(s,i)=m_est(randi(numel(m_est)));
rl_nr(s,i)=find(rl(s,i)==m_est);
a=-1;
b=1;
r(s)=(b-a)*rand(1,1)+a;
end
pert_layers(s,:)=rl(s,:)+r(s);
M=repmat(m_est',s,1);
end
for k=steps
for m=1:1:5
M_pert=M;
M_pert(1:k,rl_nr(k,1:m))=pert_layers(1:k,1:m);
end
end
In matrix M I am storing 10 initial models and want to replace the random numbers with indices from rl_nr matrix into those stored in pert_layers matrix. However, the last loop responsible for assigning values from pert_layers to rl_nr indices does not work properly.
Does anyone know how to solve this?
Best regards
Your code uses a lot of loops and in this particular circumstance, it's quite inefficient. It's better if you actually vectorize your code. As such, let me go through your problem description one point at a time and let's code up each part (if applicable):
I want to create 10 different models. Each of them is based on the same original array of dimensions 1x100 m_est.
I'm interpreting this as you having an array m_est of 100 elements, and with this array, you wish to create 10 different "models", where each model is 5 elements sampled from m_est. rl will store these values from m_est while rl_nr will store the indices / locations of where these values originated from. Also, for each model, you wish to add a random value to every element that is part of this model.
Then with for loop I am choosing 5 random values from the original model and want to add the same random value to each of them.
Instead of doing this with a for loop, generate all of your random indices in one go. Since you have 10 steps, and we wish to sample 5 points per step, you have 10*5 = 50 points in total. As such, why don't you use randperm instead? randperm is exactly what you're looking for, and we can use this to generate unique random indices so that we can ultimately use this to sample from m_est. randperm generates a vector from 1 to N but returns a random permutation of these elements. This way, you only get numbers enumerated from 1 to N exactly once and we will ensure no repeats. As such, simply use randperm to generate 50 elements, then reshape this array into a matrix of size 10 x 5, where the number of rows tells you the number of steps you want, while the number of columns is the total number of points per model. Therefore, do something like this:
num_steps = 10;
num_points_model = 5;
ind = randperm(numel(m_est));
ind = ind(1:num_steps*num_points_model);
rl_nr = reshape(ind, num_steps, num_points_model);
rl = m_est(rl_nr);
The first two lines are pretty straight forward. We are just declaring the total number of steps you want to take, as well as the total number of points per model. Next, what we will do is generate a random permutation of length 100, where elements are enumerated from 1 to 100, but they are in random order. You'll notice that this random vector uses only a value within the range of 1 to 100 exactly once. Because you only want to get 50 points in total, simply subset this vector so that we only get the first 50 random indices generated from randperm. These random indices get stored in ind.
Next, we simply reshape ind into a 10 x 5 matrix to get rl_nr. rl_nr will contain those indices that will be used to select those entries from m_est which is of size 10 x 5. Finally, rl will be a matrix of the same size as rl_nr, but it will contain the actual random values sampled from m_est. These random values correspond to those indices generated from rl_nr.
Now, the final step would be to add the same random number to each model. You can certainly use repmat to replicate a random column vector of 10 elements long, and duplicate them 5 times so that we have 5 columns then add this matrix together with rl.... so something like:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
r = repmat(r, 1, num_points_model);
M_pert = rl + r;
Now M_pert is the final result you want, where we take each model that is stored in rl and add the same random value to each corresponding model in the matrix. However, if I can suggest something more efficient, I would suggest you use bsxfun instead, which does this replication under the hood. Essentially, the above code would be replaced with:
a = -1;
b = 1;
r = (b-a)*rand(num_steps, 1) + a;
M_pert = bsxfun(#plus, rl, r);
Much easier to read, and less code. M_pert will contain your models in each row, with the same random value added to each particular model.
The cycle repeats 10 times chosing different values each time and adding different random number.
Already done in the above steps.
I hope you didn't find it an imposition to completely rewrite your code so that it's more vectorized, but I think this was a great opportunity to show you some of the more advanced functions that MATLAB has to offer, as well as more efficient ways to generate your random values, rather than looping and generating the values one at a time.
Hopefully this will get you started. Good luck!
I am attempting to create a model whereby there is a line - represented as a 1D matrix populated with 1's - and points on the line are generated at random. Every time a point is chosen (A), it creates a 'zone of exclusion' (based on an exponential function) such that choosing another point nearby has a much lower probability of occurring.
Two main questions:
(1) What is the best way to generate an exponential such that I can multiply the numbers surrounding the chosen point to create the zone of exclusion? I know of exppdf however i'm not sure if this allows me to create an exponential which terminates at 1, as I need the zone of exclusion to end and the probability to return to 1 eventually.
(2) How can I modify matrix values plus/minus a specific index (including that index)? I got as far as:
x(1:100) = 1; % Creates a 1D-matrix populated with 1's
p = randi([1 100],1,1);
x(p) =
But am not sure how to go about using the randomly generated number to alter values in the matrix.
Any help would be much appreciated,
Anna
Don't worry about exppdf, pick the width you want (how far away from the selected point does the probability return to 1?) and define some simple function that makes a small vector with zero in the middle and 1 at the edges. So here I'm just modifying a section of length 11 centred on p and doing nothing to the rest of x:
x(1:100)=1;
p = randi([1 100],1,1);
% following just scaled
somedist = (abs(-5:5).^2)/25;
% note - this will fail if p is at edges of data, but see below
x(p-5:p+5)=x(p-5:p+5).*somedist;
Then, instead of using randi to pick points you can use datasample which allows for giving weights. In this case your "data" is just the numbers 1:100. However, to make edges easier I'd suggest initialising with a "weight" vector which has zero padding - these sections of x will not be sampled from but stop you from having to make edge checks.
x = zeros([1 110]);
x(6:105)=1;
somedist = (abs(-5:5).^2)/25;
nsamples = 10;
for n = 1:nsamples
p = datasample(1:110,1,'Weights',x);
% if required store chosen p somewhere
x(p-5:p+5)=x(p-5:p+5).*somedist;
end
For an exponential exclusion zone you could do something like:
somedist = exp(abs(-5:5))/exp(5)-exp(0)/exp(5);
It doesn't quite return to 1 but fairly close. Here's the central region of x (ignoring the padding) after two separate runs:
I've organized some data into a nested structure that includes several subjects, 4-5 trials per subject, then identifying data like height, joint torque over a gait cycle, etc. So, for example:
subject(2).trial(4).torque
gives a matrix of joint torques for the 4th trial of subject 2, where the torque matrix columns represent degrees of freedom (hip, knee, etc.) and the rows represent time increments from 0 through 100% of a stride. What I want to do is take the mean of 5 trials for each degree of freedom and use that to represent the subject (for that degree of freedom). When I try to do it like this for the 1st degree of freedom:
for i = 2:24
numTrialsThisSubject = size(subject(i).trial, 2);
subject(i).torque = mean(subject(i).trial(1:numTrialsThisSubject).torque(:,1), 2);
end
I get this error:
??? Scalar index required for this type of multi-level indexing.
I know I can use a nested for loop to loop through the trials, store them in a temp matrix, then take the mean of the temp columns, but I'd like to avoid creating another variable for the temp matrix if I can. Is this possible?
You can use a combination of deal() and cell2mat().
Try this (use the built-in debugger to run through the code to see how it works):
for subject_k = 2:24
% create temporary cell array for holding the matrices:
temp_torques = cell(length(subject(subject_k).trial), 1);
% deal the matrices from all the trials (copy to temp_torques):
[temp_torques{:}] = deal(subject(subject_k).trial.torque);
% convert to a matrix and concatenate all matrices over rows:
temp_torques = cell2mat(temp_torques);
% calculate mean of degree of freedom number 1 for all trials:
subject(subject_k).torque = mean(temp_torques(:,1));
end
Notice that I use subject_k for the subject counter variable. Be careful with using i and j in MATLAB as names of variables, as they are already defined as 0 + 1.000i (complex number).
As mentioned above in my comment, adding another loop and temp variable turned out to be the simplest execution.
I have a Problem. I have a Matrix A with integer values between 0 and 5.
for example like:
x=randi(5,10,10)
Now I want to call a filter, size 3x3, which gives me the the most common value
I have tried 2 solutions:
fun = #(z) mode(z(:));
y1 = nlfilter(x,[3 3],fun);
which takes very long...
and
y2 = colfilt(x,[3 3],'sliding',#mode);
which also takes long.
I have some really big matrices and both solutions take a long time.
Is there any faster way?
+1 to #Floris for the excellent suggestion to use hist. It's very fast. You can do a bit better though. hist is based on histc, which can be used instead. histc is a compiled function, i.e., not written in Matlab, which is why the solution is much faster.
Here's a small function that attempts to generalize what #Floris did (also that solution returns a vector rather than the desired matrix) and achieve what you're doing with nlfilter and colfilt. It doesn't require that the input have particular dimensions and uses im2col to efficiently rearrange the data. In fact, the the first three lines and the call to im2col are virtually identical to what colfit does in your case.
function a=intmodefilt(a,nhood)
[ma,na] = size(a);
aa(ma+nhood(1)-1,na+nhood(2)-1) = 0;
aa(floor((nhood(1)-1)/2)+(1:ma),floor((nhood(2)-1)/2)+(1:na)) = a;
[~,a(:)] = max(histc(im2col(aa,nhood,'sliding'),min(a(:))-1:max(a(:))));
a = a-1;
Usage:
x = randi(5,10,10);
y3 = intmodefilt(x,[3 3]);
For large arrays, this is over 75 times faster than colfilt on my machine. Replacing hist with histc is responsible for a factor of two speedup. There is of course no input checking so the function assumes that a is all integers, etc.
Lastly, note that randi(IMAX,N,N) returns values in the range 1:IMAX, not 0:IMAX as you seem to state.
One suggestion would be to reshape your array so each 3x3 block becomes a column vector. If your initial array dimensions are divisible by 3, this is simple. If they don't, you need to work a little bit harder. And you need to repeat this nine times, starting at different offsets into the matrix - I will leave that as an exercise.
Here is some code that shows the basic idea (using only functions available in FreeMat - I don't have Matlab on my machine at home...):
N = 100;
A = randi(0,5*ones(3*N,3*N));
B = reshape(permute(reshape(A,[3 N 3 N]),[1 3 2 4]), [ 9 N*N]);
hh = hist(B, 0:5); % histogram of each 3x3 block: bin with largest value is the mode
[mm mi] = max(hh); % mi will contain bin with largest value
figure; hist(B(:),0:5); title 'histogram of B'; % flat, as expected
figure; hist(mi-1, 0:5); title 'histogram of mi' % not flat?...
Here are the plots:
The strange thing, when you run this code, is that the distribution of mi is not flat, but skewed towards smaller values. When you inspect the histograms, you will see that is because you will frequently have more than one bin with the "max" value in it. In that case, you get the first bin with the max number. This is obviously going to skew your results badly; something to think about. A much better filter might be a median filter - the one that has equal numbers of neighboring pixels above and below. That has a unique solution (while mode can have up to four values, for nine pixels - namely, four bins with two values each).
Something to think about.
Can't show you a mex example today (wrong computer); but there are ample good examples on the Mathworks website (and all over the web) that are quite easy to follow. See for example http://www.shawnlankton.com/2008/03/getting-started-with-mex-a-short-tutorial/