Didn't know how to paraphrase the question well.
Function for example:
Data:https://www.dropbox.com/s/wr61qyhhf6ujvny/data.mat?dl=0
In this case how do I calculate that the rest point of this function is ~1? I have access to the vector that makes the plot.
I guess the mean is an approximation but in some cases it can be pretty bad.
Under the assumption that the "rest" point is the steady-state value in your data and the fact that the steady-state value happens the majority of the times in your data, you can simply bin all of the points and use each unique value as a separate bin. The bin with the highest count should correspond to the steady-state value.
You can do this by a combination of histc and unique. Assuming your data is stored in y, do this:
%// Find all unique values in your data
bins = unique(y);
%// Find the total number of occurrences per unique value
counts = histc(y, bins);
%// Figure out which bin has the largest count
[~,max_bin] = max(counts);
%// Figure out the corresponding y value
ss_value = bins(max_bin);
ss_value contains the steady-state value of your data, corresponding to the most occurring output point with the assumptions I laid out above.
A minor caveat with the above approach is that this is not friendly to floating point data whose unique values are generated by floating point values whose decimal values beyond the first few significant digits are different.
Here's an example of your data from point 2300 to 2320:
>> format long g;
>> y(2300:2320)
ans =
0.99995724232555
0.999957488454868
0.999957733165346
0.999957976465197
0.999958218362579
0.999958458865564
0.999958697982251
0.999958935720613
0.999959172088623
0.999959407094224
0.999959640745246
0.999959873049548
0.999960104014889
0.999960333649014
0.999960561959611
0.999960788954326
0.99996101464076
0.999961239026462
0.999961462118947
0.999961683925704
0.999961904454139
Therefore, what I'd recommend is to perhaps round so that the first 5 or so significant digits are maintained.
You can do this to your dataset before you continue:
num_digits = 5;
y_round = round(y*(10^num_digits))/(10^num_digits);
This will first multiply by 10^n where n is the number of digits you desire so that the decimal point is shifted over by n positions. We round this result, then divide by 10^n to bring it back to the scale that it was before. If you do this, for those points that were 0.9999... where there are n decimal places, these will get rounded to 1, and it may help in the above calculations.
However, more recent versions of MATLAB have this functionality already built-in to round, and you can just do this:
num_digits = 5;
y_round = round(y,num_digits);
Minor Note
More recent versions of MATLAB discourage the use of histc and recommend you use histcounts instead. Same function definition and expected inputs and outputs... so just replace histc with histcounts if your MATLAB version can handle it.
Using the above logic, you could also use the median too. If the majority of data is fluctuating around 1, then the median would have a high probability that the steady-state value is chosen... so try this too:
ss_value = median(y_round);
Related
I am relatively new to matlab. I found the consecutive mean of a set of 1E6 random numbers that has mean and standard deviation. Initially the calculated mean fluctuate and then converges to a certain value.
I will like to know the index (i.e 100th position) at which the mean converges. I have no idea how to do that.
I tried using the logical operator but i have to go through 1e6 data points. Even with that i still can't find the index.
Y_c= sigma_c * randn(n_r, 1) + mu_c; %Random number creation
Y_f=sigma_f * randn(n_r, 1) + mu_f;%Random number creation
P_u=gamma*(B*B)/2.*N_gamma+q*B.*N_q + Y_c*B.*N_c; %Calculation of Ultimate load
prog_mu=cumsum(P_u)./cumsum(ones(size(P_u))); %Progressive Cumulative Mean of system response
logical(diff(prog_mu==0)); %Find index
I suspect the issue is that the mean will never truly be constant, but will rather fluctuate around the "true mean". As such, you'll most likely never encounter a situation where the two consecutive values of the cumulative mean are identical. What you should do is determine some threshold value, below which you consider fluctuations in the mean to be approximately equal to zero, and compare the difference of the cumulative mean to that value. For instance:
epsilon = 0.01;
const_ind = find(abs(diff(prog_mu))<epsilon,1,'first');
where epsilon will be the threshold value you choose. The find command will return the index at which the variation in the cumulative mean first drops below this threshold value.
EDIT: As was pointed out, this method may potentially fail if the first few random numbers are generated such that the difference between them is less than the epsilon value, but have not yet converged. I would like to suggest a different approach, then.
We calculate the cumulative means, as before, like so:
prog_mu=cumsum(P_u)./cumsum(ones(size(P_u)));
We also calculate the difference in these cumulative means, as before:
df_prog_mu = diff(prog_mu);
Now, to ensure that conversion has been achieved, we find the first index where the cumulative mean is below the threshold value epsilon and all subsequent means are also below the threshold value. To phrase this another way, we want to find the index after the last position in the array where the cumulative mean is above the threshold:
conv_index = find(~df_prog_mu,1,'last')+1;
In doing so, we guarantee that the value at the index, and all subsequent values, have converged below your predetermined threshold value.
I wouldn't imagine that the mean would suddenly become constant at a single index. Wouldn't it asymptotically approach a constant value? I would reccommend a for loop to calculate the mean (it sounds like maybe you've already done this part?) like this:
avg = [];
for k=1:length(x)
avg(k) = mean(x(1:k));
end
Then plot the consecutive mean:
plot(avg)
hold on % this will allow us to plot more data on the same figure later
If you're trying to find the point at which the consecutive mean comes within a certain range of the true mean, try this:
Tavg = 5; % or whatever your true mean is
err = 0.01; % the range you want the consecutive mean to reach before we say that it "became constant"
inRange = avg>(Tavg-err) & avg<(Tavg+err); % gives you a binary logical array telling you which values fell within the range
q = 1000; % set this as high as you can while still getting a value for consIndex
constIndex = [];
for k=1:length(inRange)
if(inRange(k) == sum(inRange(k:k+q))/(q-1);)
constIndex = k;
end
end
The below answer takes a similar approach but makes an unsafe assumption that the first value to fall within the range is the value where the function starts to converge. Any value could randomly fall within that range. We need to make sure that the following values also fall within that range. In the above code, you can edit "q" and "err" to optimize your result. I would recommend double checking it by plotting.
plot(avg(constIndex), '*')
I have vertically concatenated files from my directory into a matrix that is about 60000 x 15 in size (verified).
d=dir('*.log');
n=length(d);
data=[];
for k=1:n
data{k}=importdata(d(k).name);
end
total=[];
for k=1:n
total=[total;data{n}];
end
I am using a the following 32-iteration loop and the 'Find" function to locate row numbers where the final column is an integer corresponding to the integer iteration of the loop:
for i=1:32
v=[];
vn=[];
[v,vn]=find(abs(fix(i)-fix(total))<eps);
g=length(v)
end
I have tried to account for the floating point accuracy by using 'fix' on values of 'i' and values from matrix 'total', in addition to taking their absolute difference and checking it to be less than a tolerance of 'eps' (floating-point relative accuracy function), up to a tolerance of .99.
The 'Find' function is not working correctly. It is only working for certain integers (although it should be locating all of them (1-32)), and for the integers it does find the values are incomplete.
What is the problem here? If 'Find' is inadequate for this purpose, what is a suitable alternative?
You are getting a lot of zeros because you are looking not just at the 15th column of data but the entire data matrix so you are going to have a lot of non-integers.
Also, you're using fix on both numbers and since floating point errors can cause the number to be slightly above and below the desired integer, this will cause the ones that are below to round down an integer lower than what you'd expect. You should use round to round to the nearest integer instead.
Rather than using find to do this, I would use simple boolean logic to determine the value of the last column
for k = 1:32
% Compare column 15 to the current index
matches = abs(total(:,end) - k) < eps;
% Do stuff with these matches
g = sum(matches); % Count the matches
end
Depending on what you want to actually do with the data, you may be able to use the last column as an input to accumarray to perform an operation on each group.
As a side note, you can replace the first chunk of code with
d = dir('*.log');
data = cellfun(#importdata, {d.name}, 'UniformOutput', false);
total = cat(1, data{:});
An issue I've come across multiple times is wanting to take two similar data sets and create histograms from them where the bins are identical, so as to easily calculate things like histogram overlap.
You can define the number of bins (obviously) using
[counts, bins] = hist(data,number_of_bins)
But there's not an obvious way (as far as I can see) to make the bin size equal for several different data sets. If remember when I initially looked finding various people who seem to have the same issue, but no good solutions.
The right, easy way
As pointed out by horchler, this can easily be achieved using either histc (which lets you define your bins vector), or vectorizing your histogram input into hist.
The wrong, stupid way
I'm leaving below as a reminder to others that even stupid questions can yield worthwhile answers
I've been using the following approach for a while, so figured it might be useful for others (or, someone can very quickly point out the correct way to do this!).
The general approach relies on the fact that MATLAB's hist function defines an equally spaced number of bins between the largest and smallest value in your sample. So, if you append a start (smallest) and end (largest) value to your various samples which is the min and max for all samples of interest, this forces the histogram range to be equal for all your data sets. You can then truncate the first and last values to recreate your original data.
For example, create the following data set
A = randn(1,2000)+7
B = randn(1,2000)+9
[counts_A, bins_A] = hist(A, 500);
[counts_B, bins_B] = hist(B, 500);
Here for my specific data sets I get the following results
bins_A(1) % 3.8127 (this is also min(A) )
bins_A(500) % 10.3081 (this is also max(A) )
bins_B(1) % 5.6310 (this is also min(B) )
bins_B(500) % 13.0254 (this is also max(B) )
To create equal bins you can simply first define a min and max value which is slightly smaller than both ranges;
topval = max([max(A) max(B)])+0.05;
bottomval = min([min(A) min(B)])-0.05;
The addition/subtraction of 0.05 is based on knowledge of the range of values - you don't want your extra bin to be too far or too close to the actual range. That being said, for this example by using the joint min/max values this code will work irrespective of the A and B values generated.
Now we re-create histogram counts and bins using (note the extra 2 bins are for our new largest and smallest value)
[counts_Ae, bins_Ae] = hist([bottomval, A, topval], 502);
[counts_Be, bins_Be] = hist([bottomval, B, topval], 502);
Finally, you truncate the first and last bin and value entries to recreate your original sample exactly
bins_A = bins_Ae(2:501)
bins_B = bins_Ae(2:501)
counts_A = counts_Ae(2:501)
counts_B = counts_Be(2:501)
Now
bins_A(1) % 3.7655
bins_A(500) % 13.0735
bins_B(1) % 3.7655
bins_B(500) % 13.0735
From this you can easily plot both histograms again
bar([bins_A;bins_B]', [counts_A;counts_B]')
And also plot the histogram overlap with ease
bar(bins_A,(counts_A+counts_B)-(abs(counts_A-counts_B)))
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/
hi can anyone show how to use the modem.oqpskmod for BER. thanks!
h = modem.oqpskmod
y = modulate(h, values);
g = modem.oqpskdemod(h)
z = demodulate(g, y)
let's assume that i have array called values which contains only 1s and 0s.
my question is how would i calculate BER? of course if above my code is correct.
Based on this Wikipedia page, you simply have to compute the number of incorrect bits and divide by the total number of transferred bits to get the bit error rate (BER). If values is the unmodulated input signal and z is the output signal after modulation and demodulation, you can compute it like this:
BER = sum(logical(values(:)-z(:)))/numel(values);
EDIT: I modified the above code just in case you run into two situations:
If z has values other than 0 and 1.
If z is a different size than values (i.e. row vector versus column vector).
I don't know if you are ever likely to come across these two situations, but better safe than sorry. ;)