I seem to currently be banging my head against a brick wall as try as I might, I can not see my error here.
I am attempting to write a for loop in MATLAB that uses the equation below (adiabatic compression) to calculate the new pressure after one degree of crankshaft rotation in a four stroke engine cycle.
P2 = P1 * (V2 / V1) ^2
I am using the calculated volume from the crank-slider model as an input. I have tried this is Excel and it works as expected and gives the overall max output correctly.
The for loop in question is below;
Cyl_P = ones(720,1)
for i = (2:1:length(Cyl_V))'
Cyl_P(i,:) = Cyl_P(i-1,:) .* (Cyl_V(i,:) ./ Cyl_V(i-1,:)).^1.35
end
my aim is to use the first element of the vector Cyl_P which is equal to one, as an input to the equation above, and multiply it by the second element of Cyl_V divided by the first, and multiply the volume terms by 1.35. that should calculate the second element of Cyl_P. I would then like to feed that value back in to the same equation to calculate the third element and so on.
What am I missing?
I've put the full code below
Theta = deg2rad(1:1:720)'
Stroke = 82 / 1000
R = Stroke / 2
L = 90.5 / 1000
Bore = 71.9 / 1000
d_h = (R+L) - (R.*cos(Theta)) - sqrt(L.^2 - (R.*sin(Theta)).^2)
Pist_h = d_h
figure
plot(Pist_h)
Bore_A = (pi*Bore^2)/4
Swept_V = (Pist_h .* Bore_A)
Clear_V = max(Swept_V) / 10
Total_V = max(Swept_V) + Clear_V
Cyl_V = (Swept_V + Clear_V)
figure
plot(Cyl_V)
for ii = (2:1:length(Cyl_V))'
div_V(ii,:) = (Cyl_V(ii) ./ Cyl_V(ii-1,:)).^1.35
end
Cyl_P = ones(720,1)
for i = (2:1:length(Cyl_V))'
Cyl_P(i,:) = Cyl_P(i-1,:) .* (Cyl_V(i,:) ./ Cyl_V(i-1,:)).^1.35
end
figure
plot(Cyl_P)
Your problem is transposing the arrays you feed as argument to for loops. MATLAB reads for arguments per row, thus only the first iteration will be used when you feed it a column. General comments:
' is the complex transpose, .' is the regular transpose.
i is the imaginary unit in MATLAB, it's common practise not to use it as a variable name.
2:1:4 does the same as 2:4, as 1 is the default step size.
Please use semi-colons, ;, after each row, so as to prevent MATLAB from echoing the result of each line to the command window. This makes the script easier to run, and if you have matrices with >1M entries, echoing the contents might even crash the program all together. Even in this case, you are echoing 720 entries of Cyl_P 720 times. For checking variable contents, just break the script where necessary (or run it in parts) and examine the content where warranted (e.g. Cyl_P(1:3) would suffice here to check whether the loop fills the vector as intended).
Related
Apologies for the long post but this takes a bit to explain. I'm trying to make a script that finds the longest linear portion of a plot. Sample data is in a csv file here, it is stress and strain data for calculating the shear modulus of 3D printed samples. The code I have so far is the following:
x_data = [];
y_data = [];
x_data = Data(:,1);
y_data = Data(:,2);
plot(x_data,y_data);
grid on;
answer1 = questdlg('Would you like to load last attempt''s numbers?');
switch answer1
case 'Yes'
[sim_slopes,reg_data] = regr_and_longest_part(new_x_data,new_y_data,str2num(answer2{3}),str2num(answer2{2}),K);
case 'No'
disp('Take a look at the plot, find a range estimate, and press any button to continue');
pause;
prompt = {'Eliminate values ABOVE this x-value:','Eliminate values BELOW this x-value:','Size of divisions on x-axis:','Factor for similarity of slopes:'};
dlg_title = 'Point elimination';
num_lines = 1;
defaultans = {'0','0','0','0.1'};
if isempty(answer2) < 1
defaultans = {answer2{1},answer2{2},answer2{3},answer2{4}};
end
answer2 = inputdlg(prompt,dlg_title,num_lines,defaultans);
uv_of_x_range = str2num(answer2{1});
lv_of_x_range = str2num(answer2{2});
x_div_size = str2num(answer2{3});
K = str2num(answer2{4});
close all;
iB = find(x_data > str2num(answer2{1}),1,'first');
iS = find(x_data > str2num(answer2{2}),1,'first');
new_x_data = x_data(iS:iB);
new_y_data = y_data(iS:iB);
[sim_slopes, reg_data] = regr_and_longest_part(new_x_data,new_y_data,str2num(answer2{3}),str2num(answer2{2}),K);
end
[longest_section0, Midx]= max(sim_slopes(:,4)-sim_slopes(:,3));
longest_section=1+longest_section0;
long_sec_x_data_start = x_div_size*(sim_slopes(Midx,3)-1)+lv_of_x_range;
long_sec_x_data_end = x_div_size*(sim_slopes(Midx,4)-1)+lv_of_x_range;
long_sec_x_data_start_idx=find(new_x_data >= long_sec_x_data_start,1,'first');
long_sec_x_data_end_idx=find(new_x_data >= long_sec_x_data_end,1,'first');
long_sec_x_data = new_x_data(long_sec_x_data_start_idx:long_sec_x_data_end_idx);
long_sec_y_data = new_y_data(long_sec_x_data_start_idx:long_sec_x_data_end_idx);
[b_long_sec, longes_section_reg_data] = robustfit(long_sec_x_data,long_sec_y_data);
plot(long_sec_x_data,b_long_sec(1)+b_long_sec(2)*long_sec_x_data,'LineWidth',3,'LineStyle',':','Color','k');
function [sim_slopes,reg_data] = regr_and_longest_part(x_points,y_points,x_div,lv,K)
reg_data = cell(1,3);
scatter(x_points,y_points,'.');
grid on;
hold on;
uv = lv+x_div;
ii=0;
while lv <= x_points(end)
if uv > x_points(end)
uv = x_points(end);
end
ii=ii+1;
indices = find(x_points>lv & x_points<uv);
temp_x_points = x_points((indices));
temp_y_points = y_points((indices));
if length(temp_x_points) <= 2
break;
end
[b,stats] = robustfit(temp_x_points,temp_y_points);
reg_data{ii,1} = b(1);
reg_data{ii,2} = b(2);
reg_data{ii,3} = length(indices);
plot(temp_x_points,b(1)+b(2)*temp_x_points,'LineWidth',2);
lv = lv+x_div;
uv = lv+x_div;
end
sim_slopes = NaN(length(reg_data),4);
sim_slopes(1,:) = [reg_data{1,1},0,1,1];
idx=1;
for ii=2:length(reg_data)
coff =sim_slopes(idx,1);
if abs(reg_data{ii,1}-coff) <= K*coff
C=zeros(ii-sim_slopes(idx,3)+1,1);
for kk=sim_slopes(idx,3):ii
C(kk)=reg_data{kk,1};
end
sim_slopes(idx,1)=mean(C);
sim_slopes(idx,2)=std(C);
sim_slopes(idx,4)=ii;
else
idx = idx + 1;
sim_slopes(idx,1)=reg_data{ii,1};
sim_slopes(idx,2)=0;
sim_slopes(idx,3)=ii;
sim_slopes(idx,4)=ii;
end
end
end
Apologies for the code not being well optimized, I'm still relatively new to MATLAB. I did not use derivatives because my data is relatively noisy and derivation might have made it worse.
I've managed to get the get the code to find the longest straight part of the plot by splitting the data up into sections called x_div_size then performing a robustfit on each section, the results of which are written into reg_data. The code then runs through reg_data and finds which lines have the most similar slopes, determined by the K factor, by calculating the average of the slopes in a section of the plot and makes a note of it in sim_slopes. It then finds the longest interval with max(sim_slopes(:,4)-sim_slopes(:,3)) and performs a regression on it to give the final answer.
The problem is that it will only consider the first straight portion that it comes across. When the data is plotted, it has a few parts where it seems straightest:
As an example, when I run the script with answer2 = {'0.2','0','0.0038','0.3'} I get the following, where the black line is the straightest part found by the code:
I have the following questions:
It's clear that from about x = 0.04 to x = 0.2 there is a long straight part and I'm not sure why the script is not finding it. Playing around with different values the script always seems to pick the first longest straight part, ignoring subsequent ones.
MATLAB complains that Warning: Iteration limit reached. because there are more than 50 regressions to perform. Is there a way to bypass this limit on robustfit?
When generating sim_slopes there might be section of the plot whose slope is too different from the average of the previous slopes so it gets marked as the end of a long section. But that section sometimes is sandwiched between several other sections on either side which instead have similar slopes. How would it be possible to tell the script to ignore one wayward section and to continue as if it falls within the tolerance allowed by the K value?
Take a look at the Douglas-Peucker algorithm. If you think of your (x,y) values as the vertices of an (open) polygon, this algorithm will simplify it for you, such that the largest distance from the simplified polygon to the original is smaller than some threshold you can choose. The simplified polygon will be the set of straight lines. Find the two vertices that are furthest apart, and you're done.
MATLAB has an implementation in the Mapping Toolbox called reducem. You might also find an implementation on the File Exchange (but be careful, there is also really bad code on there). Or, you can roll your own, it's quite a simple algorithm.
You can also try using the ischange function to detect changes in the intercept and slope of the data, and then extract the longest portion from that.
Using the sample data you provided, here is what I see from a basic attempt:
>> T = readtable('Data.csv');
>> T = rmmissing(T); % Remove rows with NaN
>> T = groupsummary(T,'Var1','mean'); % Average duplicate timestamps
>> [tf,slopes,intercepts] = ischange(T.mean_Var2, 'linear', 'SamplePoints', T.Var1); % find changes
>> plot(T.Var1, T.mean_Var2, T.Var1, slopes.*T.Var1 + intercepts)
which generates the plot
You should be able to extract the longest segment based on the indices given by find(tf).
You can also tune the parameters of ischange to get fewer or more segments. Adding the name-value pair 'MaxNumChanges' with a value of 4 or 5 produces more linear segments with a tighter fit to the curve, for example, which effectively removes the kink in the plot that you see.
I've been trying to implement the following integral in MATLAB
Given a number n, I wrote the code that returns an array with n elements, containing approximations of each integral.
First, I tried this using a 'for' loop and the recurrence relationship on the first line. But from the 20th integral and above the values are completely wrong (correct to 0 significant figures and wrong sign).
The same goes if I use the explicit formula on the second line and two 'for' loops.
As n grows larger, so does the error on the approximations.
So the main issue here is that I haven't found a way to minimize the error as much as possible.
Any ideas? Thanks in advance.
Here is an example of the code and the resulting values, using the second formula:
This integral, for positive values of n, cannot have values >1 or <0
First attempt:
I tried the iterative method and found interesting thing. The approximation may not be true for all n. In fact if I keep track of (n-1)*I(n-1) in each loop I can see
I = zeros(20,3);
I(1,1) = 1-1/exp(1);
for ii = 2:20
I(ii,2) = ii-1;
I(ii,3) = (ii-1)*I(ii-1,1);
I(ii,1) = 1-I(ii,3);
end
There is some problem around n=18. In fact, I18 = 0.05719 and 18*I18 = 1.029 which is larger than 1. I don't think there is any numerical error or number overflow in this procedure.
Second attempt:
To make sure the maths is correct (I verified twice on paper) I used trapz to numerically evaluate the integral, and n=18 didn't cause any problem.
>> x = linspace(0,1,1+1e4);
>> f = #(n) exp(-1)*exp(x).*x.^(n-1);
>> f = #(n) exp(-1)*exp(x).*x.^(n-1)*1e-4;
>> trapz(f(5))
ans =
1.708934160520510e-01
>> trapz(f(17))
ans =
5.571936009790170e-02
>> trapz(f(18))
ans =
5.277113416899408e-02
>>
A closer look is as follows. I18 is slightly different (to the 4th significant digit) between the (stable) numerical method and (unstable) iterative method. 18*I18 is therefore possible to exceed 1.
I = zeros(20,3);
I(1,1) = 1-1/exp(1);
for ii = 2:20
I(ii,2) = ii-1;
I(ii,3) = (ii-1)*I(ii-1,1);
I(ii,1) = 1-I(ii,3);
end
J = zeros(20,3);
x = linspace(0,1,1+1e4);
f = #(n) exp(-1)*exp(x).*x.^(n-1)*1e-4;
J(1,1) = trapz(f(1));
for jj = 2:20
J(jj,1) = trapz(f(jj));
J(jj,2) = jj-1;
J(jj,3) = (jj-1)*J(jj-1,1);
end
I suspect there is an error in each iterative step due to the nature of numerical computations. If the iteration is long, the error propagates and, unfortunately in this case, amplifies rapidly. In order to verify this, I combined the above two methods into a hybrid algo. For most of the time the iterative way is used, and once in a while a numerical integral is evaluated from scratch without relying on previous iterations.
K = zeros(40,4);
K(1,1) = 1-1/exp(1);
for kk = 2:40
K(kk,2) = trapz(f(kk));
K(kk,3) = (kk-1)*K(kk-1,1);
K(kk,4) = 1-K(kk,3);
if mod(kk,5) == 0
K(kk,1) = K(kk,2);
else
K(kk,1) = K(kk,4);
end
end
If the iteration lasts more than 4 steps, error amplification will be large enough to invert the sign, and starts nonrecoverable oscillation.
The code should be able to explain all the data structures. Anyway, let me put some focus here. The second column is the result of trapz, which is the numerical integral done on the non-iterative integration definition of I(n). The third column is (n-1)*I(n-1) and should be always positive and less than 1. The forth column is 1-(n-1)*I(n-1) and should always be positive. The first column is the choice I have made between the trapz result and iterative result, to be the "true" value of I(n).
As can be seen here, in each iteration there is a small error compared to the independent numerical way. The error grows in the 3rd and 4th iteration and finally breaks the thing in its 5th. This is observed around n=25, under the case that I pick the numerical result in every 5 loops since the beginning.
Conclusion: There is nothing wrong with any definition of this integral. However the numerical error when evaluating the expressions is unfortunately aggregating, hence limiting the way you can perform the computation.
Following on from: Detecting if any values are within a certain value of each other - MATLAB
I am currently using randi to generate a random number from which I then subtract and add a second number - generated using poissrnd:
for k=1:10
a = poissrnd(200,1);
b(k,1) = randi([1,20000]);
c(k,1:2) = [b(k,1)-a,b(k,1)+a];
end
c = sort(c);
c provides an output in this format:
823 1281
5260 5676
5372 5760
5379 5779
6808 7244
6869 7293
9203 9653
12197 12563
14411 14765
15302 15670
Which are essentially the boundaries +/- a around the point chosen in b.
I then want to set an additional variable (i.e. d = 2000) which is used as the threshold by which values are matched and then merged. The boundaries are taken into consideration for this - the output of the above value when d = 2000 would be:
1052
7456
13933
The boundaries 823-1281 are not within 2000 of any other value so the midpoint is taken - reflecting the original value. The next midpoint taken is between 5260 and 9653 because as you go along, each successive values is within 2000 of the one before it until 9653. The same logic is then applied to take the midpoint between 12197 and 15670.
Is there a quick and easy way to adapt the answer give in the linked question to deal with a 2 column format?
EDIT (in order to make it clearer):
The values held in c can be thought of as demarcating the boundaries of 'blocks' that sit on a line. Every single boundary is checked to see if anything lies within 2000 of it (the black lines).
As soon as any black line touches a red block, that entire red block is incorporated into the same merge block - in full. This is why the first midpoint value calculated is 1052 - nothing is touched by the two black lines emanating from the first two boundaries. However the next set of blocks all touch one another. This incorporates them all into the merge such that the midpoint is taken between 9653 and 5260 = 7456.
The block starting at 12197 is out of reach of it's preceding one so it remains separate. I've not shown all the blocks.
EDIT 2 #Esteban:
b =
849
1975
8336
9599
12057
12983
13193
13736
16887
18578
c =
662 1036
1764 2186
8148 8524
9386 9812
11843 12271
12809 13157
12995 13391
13543 13929
16687 17087
18361 18795
Your script then produces the result:
8980
12886
17741
When in fact it should be:
1424
8980
12886
17741
So it is just missing the first value - if no merge is occurring, the midpoint is just taken between the two values. Sometimes this seems to work - other times it doesn't.
For example here it works (when value is set to 1000 instead of 2000 as a test):
c =
2333 2789
5595 6023
6236 6664
10332 10754
11425 11865
12506 12926
12678 13114
15105 15517
15425 15797
19490 19874
result =
2561
6129
11723
15451
19682
See if this works for you -
th = 2000 %// threshold
%// Column arrays
col1 = c(:,1)
col2 = c(:,2)
%// Position of "group" shifts
grp_changes = diff([col2(1:end-1,:) col1(2:end,:)],[],2)>th
%// Start and stop positions of shifts
stops = [grp_changes ; 1]
starts = [1 ; stops(1:end-1)]
%// Finally the mean of shift positions, which is the desired output
out = floor(mean([col1(starts~=0) col2(stops~=0)],2))
Not 100% sure if it will work for all your samples... but this is the code I came up with which works with at least the data in your example:
value=2000;
indices = find(abs(c(2:end,1)-c(1:end-1,2))>value);
indices = vertcat(indices, length(c));
li = indices(1:end-1)+1;
ri = indices(2:end);
if li(1)==2
li=vertcat(1,li);
ri=vertcat(1,ri);
end
result = floor((c(ri,2)+c(li,1))/2)
it's not very clean and could surely be done in less lines, but it's easy to understand and it works, and since your c will be small, I dont see the need to further optimize this unless you will run it millions of time.
I have a matrix time-series data for 8 variables with about 2500 points (~10 years of mon-fri) and would like to calculate the mean, variance, skewness and kurtosis on a 'moving average' basis.
Lets say frames = [100 252 504 756] - I would like calculate the four functions above on over each of the (time-)frames, on a daily basis - so the return for day 300 in the case with 100 day-frame, would be [mean variance skewness kurtosis] from the period day201-day300 (100 days in total)... and so on.
I know this means I would get an array output, and the the first frame number of days would be NaNs, but I can't figure out the required indexing to get this done...
This is an interesting question because I think the optimal solution is different for the mean than it is for the other sample statistics.
I've provided a simulation example below that you can work through.
First, choose some arbitrary parameters and simulate some data:
%#Set some arbitrary parameters
T = 100; N = 5;
WindowLength = 10;
%#Simulate some data
X = randn(T, N);
For the mean, use filter to obtain a moving average:
MeanMA = filter(ones(1, WindowLength) / WindowLength, 1, X);
MeanMA(1:WindowLength-1, :) = nan;
I had originally thought to solve this problem using conv as follows:
MeanMA = nan(T, N);
for n = 1:N
MeanMA(WindowLength:T, n) = conv(X(:, n), ones(WindowLength, 1), 'valid');
end
MeanMA = (1/WindowLength) * MeanMA;
But as #PhilGoddard pointed out in the comments, the filter approach avoids the need for the loop.
Also note that I've chosen to make the dates in the output matrix correspond to the dates in X so in later work you can use the same subscripts for both. Thus, the first WindowLength-1 observations in MeanMA will be nan.
For the variance, I can't see how to use either filter or conv or even a running sum to make things more efficient, so instead I perform the calculation manually at each iteration:
VarianceMA = nan(T, N);
for t = WindowLength:T
VarianceMA(t, :) = var(X(t-WindowLength+1:t, :));
end
We could speed things up slightly by exploiting the fact that we have already calculated the mean moving average. Simply replace the within loop line in the above with:
VarianceMA(t, :) = (1/(WindowLength-1)) * sum((bsxfun(#minus, X(t-WindowLength+1:t, :), MeanMA(t, :))).^2);
However, I doubt this will make much difference.
If anyone else can see a clever way to use filter or conv to get the moving window variance I'd be very interested to see it.
I leave the case of skewness and kurtosis to the OP, since they are essentially just the same as the variance example, but with the appropriate function.
A final point: if you were converting the above into a general function, you could pass in an anonymous function as one of the arguments, then you would have a moving average routine that works for arbitrary choice of transformations.
Final, final point: For a sequence of window lengths, simply loop over the entire code block for each window length.
I have managed to produce a solution, which only uses basic functions within MATLAB and can also be expanded to include other functions, (for finance: e.g. a moving Sharpe Ratio, or a moving Sortino Ratio). The code below shows this and contains hopefully sufficient commentary.
I am using a time series of Hedge Fund data, with ca. 10 years worth of daily returns (which were checked to be stationary - not shown in the code). Unfortunately I haven't got the corresponding dates in the example so the x-axis in the plots would be 'no. of days'.
% start by importing the data you need - here it is a selection out of an
% excel spreadsheet
returnsHF = xlsread('HFRXIndices_Final.xlsx','EquityHedgeMarketNeutral','D1:D2742');
% two years to be used for the moving average. (250 business days in one year)
window = 500;
% create zero-matrices to fill with the MA values at each point in time.
mean_avg = zeros(length(returnsHF)-window,1);
st_dev = zeros(length(returnsHF)-window,1);
skew = zeros(length(returnsHF)-window,1);
kurt = zeros(length(returnsHF)-window,1);
% Now work through the time-series with each of the functions (one can add
% any other functions required), assinging the values to the zero-matrices
for count = window:length(returnsHF)
% This is the most tricky part of the script, the indexing in this section
% The TwoYearReturn is what is shifted along one period at a time with the
% for-loop.
TwoYearReturn = returnsHF(count-window+1:count);
mean_avg(count-window+1) = mean(TwoYearReturn);
st_dev(count-window+1) = std(TwoYearReturn);
skew(count-window+1) = skewness(TwoYearReturn);
kurt(count-window +1) = kurtosis(TwoYearReturn);
end
% Plot the MAs
subplot(4,1,1), plot(mean_avg)
title('2yr mean')
subplot(4,1,2), plot(st_dev)
title('2yr stdv')
subplot(4,1,3), plot(skew)
title('2yr skewness')
subplot(4,1,4), plot(kurt)
title('2yr kurtosis')
I am using interp1 to inteprolate some data:
temp = 4 + (30-4).*rand(365,10);
depth = 1:10;
dz = 0.5; %define new depth interval
bthD = min(depth):dz:max(depth); %new depth vector
for i = 1:length(temp);
i_temp(i,:) = interp1(depth,temp(i,:),bthD);
end
Here, I am increasing the resolution of my measurements by interpolating the measurements from 1 m increments to 0.5 m increments. This code works fine i.e. it gives me the matrix I was looking for. However, when I apply this to my actual data, it takes a long time to run, primarily as I am running an additional loop which runs through various cells. Is there a way of achieving what is described above without using the loop, in other words, is there a faster method?
Replace your for loop with:
i_temp = interp1(depth,temp',bthD)';
You can get rid of the transposes if you change the way that temp is defined, and if you are OK with i_temp being a 19x365 array instead of 365x19.
BTW, the documentation for interp1 is very clear that you can pass in an array as the second argument.