Related
I have a list of coordinates, coord, which looks like this when plotted:
I want to remove the long string of points that goes completely from 0 to 1 from the data set, shown on this plot starting at (0, 11) and ending at (1, 11) and the other one that begins at (0, 24) and ends at (1, 28).
So far, I have tried using kmeans to group the data by height using this code:
jet = colormap('jet');
amount = 20;
step = floor(numel(jet(:,1))/amount);
idxOIarr = cell(numel(terp));
scale = 100;
for ii = 1:numel(terp)
figure;
hold on;
expandDat = [stretched{ii}(:,1), scale.*log(terp{ii}(:,2))];
[idx, cent] = kmeans(expandDat(:,1:2), amount, 'Distance', 'cityblock');
idxOIarr{ii} = idx;
for jj = 1:amount
scatter(stretched{ii}(idx == jj,1), FREQ(terp{ii}(idx == jj,2)), 10, jet(step*jj,:), 'filled');
end
end
resulting in this image: Although it does separate the higher rows quite well, it breaks the line in the middle in two and groups the line that begins at (0,20) with some data points below it.
Is there any other way to group and remove these points?
The most efficient way to solve this involves building a graph where each point is a vertex. You join points that you consider "connected" or "closed" with an edge. Thus, the graph will connected components. Now you need to look for the connected components that span the whole range from 0 to 1.
Build the graph. Finding neighbors is most efficient using an R-tree. Here are some suggestions. You can also use a k-d tree, for example. However, this is not strictly necessary, it just can get really slow without a proper spatial indexing structure, because you'll have to compare distances between each pair of points.
Given a Nx2 matrix coord, you can find the square distances between each pair:
D = sum((reshape(coord,[],1,2) - reshape(coord,1,[],2)).^2,3);
(note again that this is expensive if N is large, and in that case using an R-tree will speed things up significantly). D(i,j) is the distance between points with indices i and j (i.e. coord(i,:) and coord(j,:).
Next, build the graph, G, nodes i and j are connected if G(i,j)==1. G is a symmetric matrix:
G = D <= max_distance;
Find connected components. A connected component is just a set of nodes that you can reach from each other by following edges. You don't really need to find all connected components, you just need to find the set of points that have x=0, and starting from each, recursively visit all elements in its connected component to see if you can reach a point that has x=1.
This next code is not tested, but helpfully it gives a starting point:
start_indices = find(coord(:,1)==0); % Is exact equality appropriate here?
end_indices = find(coord(:,1)==1);
to_remove = [];
visited = false(size(coord,1), 1);
for ii=start_indices.'
% For each point with x=0, see if we can reach any of the points at x=1
[res, visited] = can_reach(ii, end_indices, G, visited);
if res
% For this point we can, remove it!
to_remove(end+1) = ii;
end
end
% Iterative function to visit all nodes in a connected component
function [res, visited] = can_reach(start, end_indices, G, visited)
visited(start) = true;
if any(start==end_indices)
% We've reach an end point, stop iterating and return true.
res = true;
return;
end
next = find(G(start,:)); % find neighbors
next(visited(next)) = []; % remove visited neighbors
for ii=next
[res, visited] = can_reach(ii, end_indices, G, visited);
if res
% Yes, we can visit an end point, stop iterating now.
return
end
end
end
I'm using scala-graph for some graph computations on Scala, and I can't seem to understand how to do one simple thing: how do I update a certain weight?
Let's say I have:
import scalax.collection.Graph
val g = Graph(1~>2 % 1, 2~>3 % 1, 1~>3 % 3)
and now I'd like to create g2 which will be the same as g but with 1~>2 % 2. How do I do that?
There doesn't seem to be any native method to update the weight of an edge. What you can do is to remove an edge and add a new one with a different weight:
scala> g - 1~>3 % 3 + 1~>3 % 1337
res = Graph(1, 2, 3, 1~>2 %1, 1~>3 %1337, 2~>3 %1)
Edit: Note that the weight of the edge that is being removed, 1~>3 % <weight>, can have any value, since edges aren't identified by their weight.
See this thread for more details.
I have the following data that predicts a curve in the middle of the two curves which has different equation and datas.I also need to spline and smothen the curve of the middle curve
I've tried searching other codes here in stackoverflow but this is the most close to the right solution. So far the plot for the two curve is right but the interpolated point gives me wrong plot.
Im trying to find the plot for val=30 given that (a25,vel25)=25 and (a50,vel50)=50. Please help me troubleshoot and get a table of data (x,y) for the generated interpolated curve. Thanks for your help
generated plot using this program
a50=[1.05
0.931818182
0.931818182
0.968181818
1.045454545
1.136363636
1.354545455
1.568181818
1.718181818
1.945454545
2.159090909
2.454545455
2.772727273
];
vel50=[0.85
0.705555556
0.605555556
0.533333333
0.472222222
0.45
0.427777778
0.45
0.477777778
0.533333333
0.611111111
0.711111111
0.827777778
];
a25=[0.5
0.613636364
0.686363636
0.795454545
0.918181818
0.963636364
1.090909091
1.236363636
1.304545455
1.431818182
1.545454545
1.659090909
1.818181818
];
vel25=[0.425555556
0.354444444
0.302222222
0.266666667
0.233333333
0.226666667
0.211111111
0.222222222
0.237777778
0.266666667
0.311111111
0.35
0.402222222
];
plot(a25,vel25,'b-');
hold on
plot(a50,vel50,'g-');
minX = min([a25 a50]);
maxX = max([a25,a50]);
xx = linspace(minX,maxX,100);
vel25_inter = interp1(a25,vel25,xx);
vel50_inter = interp1(a50,vel50,xx);
val = 30; % The interpolated point
interpVel = vel25_inter + ((val-25).*(vel50_inter-vel25_inter))./(50-25);
plot(xx,interpVel,'r-');
The question and answer linked in comment still apply and can be a solution.
In your case, it is not so direct because your data are not on the same grid and some are not monotonic, but once they are packaged properly, the easiest solution is still to use griddata.
By packaged properly, I mean finding the maximum common interval (on x, or what you call a), so the data can be interpolated between curve without producing NaNs.
This seems to work:
The red dashed line is the values interpolated at val=30, all the other lines are interpolations for values between 25 to 50.
The code to get there:
% back up original data, just for final plot
bkp_a50 = a50 ; bkp_vel50 = vel50 ;
% make second x vector monotonic
istart = find( diff(a50)>0 , 1 , 'first') ;
a50(1:istart-1) = [] ;
vel50(1:istart-1) = [] ;
% prepare a 3rd dimension vector (from 25 to 50)
T = [repmat(25,size(a25)) ; repmat(50,size(a50)) ] ;
% merge all observations together
A = [ a25 ; a50] ;
V = [vel25 ; vel50] ;
% find the minimum domain on which data can be interpolated
% (anything outside of that will return NaN)
Astart = max( [min(a25) min(a50)] ) ;
Astop = min( [max(a25) max(a50)] ) ;
% use the function 'griddata'
[TI,AI] = meshgrid( 25:50 , linspace(Astart,Astop,10) ) ;
VI = griddata(T,A,V,TI,AI) ;
% plot all the intermediate curves
plot(AI,VI)
hold on
% the original curves
plot(a25,vel25,'--k','linewidth',2)
plot(bkp_a50,bkp_vel50,'--k','linewidth',2)
% Highlight the curve at T = 30 ;
c30 = find( TI(1,:) == 30 ) ;
plot(AI(:,c30),VI(:,c30),'--r','linewidth',2)
There were quite a few issues with your code that's why it was not executing properly. I have just made very little changes to your code and made it running,
clc
%13
a50=[1.05
0.931818182
0.932
0.968181818
1.045454545
1.136363636
1.354545455
1.568181818
1.718181818
1.945454545
2.159090909
2.454545455
2.772727273
];
%13
vel50=[0.85
0.705555556
0.605555556
0.533333333
0.472222222
0.45
0.427777778
0.45
0.477777778
0.533333333
0.611111111
0.711111111
0.827777778
];
%13
a25=[0.5
0.613636364
0.686363636
0.795454545
0.918181818
0.963636364
1.090909091
1.236363636
1.304545455
1.431818182
1.545454545
1.659090909
1.818181818
];
%13
vel25=[0.425555556
0.354444444
0.302222222
0.266666667
0.233333333
0.226666667
0.211111111
0.222222222
0.237777778
0.266666667
0.311111111
0.35
0.402222222
];
plot(a25,vel25,'b-');
hold on
plot(a50,vel50,'g-');
minX = min([a25 a50])
maxX = max([a25 a50])
%xx = linspace(minX,maxX);
xx = linspace(0.5,2.7727,100);
vel25_inter = interp1(a25,vel25,xx);
vel50_inter = interp1(a50,vel50,xx);
val = 30; % The interpolated point
interpVel = vel25_inter + ((val-25).*(vel50_inter-vel25_inter))./(50-25);
plot(xx,interpVel,'r-');
There issues were
The interval on which you wanted to interpolate is xx = linspace(minX,maxX); but it gives an error of the type,
Matrix dimensions must agree.
Because you were assigning two values for the starting point and two for the ending point. So I replace it with xx = linspace(0.5,2.7727,100); where the starting point is being the minimum of the two minimum minX and the same for maxX
There are values of a50 (0.931818182) that repeat which was producing the following error
The grid vectors are not strictly monotonic increasing.
I change one of the value and replaced it with 0.932
The output is not that promising but is suppose the one you want?
I am optimizing portfolio of N stocks over M levels of expected return. So after doing this I get the time series of weights (i.e. a N x M matrix where where each row is a combination of stock weights for a particular level of expected return). Weights add up to 1.
Now I want to plot something called portfolio composition map (right plot on the picture), which is a plot of these stock weights over all levels of expected return, each with a distinct color and length (at every level of return) is proportional to it's weight.
My questions is how to do this in Julia (or MATLAB)?
I came across this and the accepted solution seemed so complex. Here's how I would do it:
using Plots
#userplot PortfolioComposition
#recipe function f(pc::PortfolioComposition)
weights, returns = pc.args
weights = cumsum(weights,dims=2)
seriestype := :shape
for c=1:size(weights,2)
sx = vcat(weights[:,c], c==1 ? zeros(length(returns)) : reverse(weights[:,c-1]))
sy = vcat(returns, reverse(returns))
#series Shape(sx, sy)
end
end
# fake data
tickers = ["IBM", "Google", "Apple", "Intel"]
N = 10
D = length(tickers)
weights = rand(N,D)
weights ./= sum(weights, dims=2)
returns = sort!((1:N) + D*randn(N))
# plot it
portfoliocomposition(weights, returns, labels = tickers)
matplotlib has a pretty powerful polygon plotting capability, e.g. this link on plotting filled polygons:
ploting filled polygons in python
You can use this from Julia via the excellent PyPlot.jl package.
Note that the syntax for certain things changes; see the PyPlot.jl README and e.g. this set of examples.
You "just" need to calculate the coordinates from your matrix and build up a set of polygons to plot the portfolio composition graph. It would be nice to see the code if you get this working!
So I was able to draw it, and here's my code:
using PyPlot
using PyCall
#pyimport matplotlib.patches as patch
N = 10
D = 4
weights = Array(Float64, N,D)
for i in 1:N
w = rand(D)
w = w/sum(w)
weights[i,:] = w
end
weights = [zeros(Float64, N) weights]
weights = cumsum(weights,2)
returns = sort!([linspace(1,N, N);] + D*randn(N))
##########
# Plot #
##########
polygons = Array(PyObject, 4)
colors = ["red","blue","green","cyan"]
labels = ["IBM", "Google", "Apple", "Intel"]
fig, ax = subplots()
fig[:set_size_inches](5, 7)
title("Problem 2.5 part 2")
xlabel("Weights")
ylabel("Return (%)")
ax[:set_autoscale_on](false)
ax[:axis]([0,1,minimum(returns),maximum(returns)])
for i in 1:(size(weights,2)-1)
xy=[weights[:,i] returns;
reverse(weights[:,(i+1)]) reverse(returns)]
polygons[i] = matplotlib[:patches][:Polygon](xy, true, color=colors[i], label = labels[i])
ax[:add_artist](polygons[i])
end
legend(polygons, labels, bbox_to_anchor=(1.02, 1), loc=2, borderaxespad=0)
show()
# savefig("CompositionMap.png",bbox_inches="tight")
Can't say that this is the best way, to do this, but at least it is working.
So I am working on my Thesis and I need to calculate geometric characteristics of an airfoil.
To do this, I need to interpolate the horizontal and vertical coordinates of an airfoil. This is used for a tool which will calculate the geometric characteristics automatically which come from random airfoil geometry files.
Sometime the Y values of the airfoil are non monotonic. Hence, the interp1 command gives an error since some values in the Y vector are repeated.
Therefore, my question is: How do I recognize and subsequently interpolate non monotonic increasing data automatically in Matlab.
Here is a sample data set:
0.999974 0.002176
0.994846 0.002555
0.984945 0.003283
0.973279 0.004131
0.960914 0.005022
0.948350 0.005919
0.935739 0.006810
0.923111 0.007691
0.910478 0.008564
0.897850 0.009428
0.885229 0.010282
0.872617 0.011125
0.860009 0.011960
0.847406 0.012783
0.834807 0.013598
0.822210 0.014402
0.809614 0.015199
0.797021 0.015985
0.784426 0.016764
0.771830 0.017536
0.759236 0.018297
0.746639 0.019053
0.734038 0.019797
0.721440 0.020531
0.708839 0.021256
0.696240 0.021971
0.683641 0.022674
0.671048 0.023367
0.658455 0.024048
0.645865 0.024721
0.633280 0.025378
0.620699 0.026029
0.608123 0.026670
0.595552 0.027299
0.582988 0.027919
0.570436 0.028523
0.557889 0.029115
0.545349 0.029697
0.532818 0.030265
0.520296 0.030820
0.507781 0.031365
0.495276 0.031894
0.482780 0.032414
0.470292 0.032920
0.457812 0.033415
0.445340 0.033898
0.432874 0.034369
0.420416 0.034829
0.407964 0.035275
0.395519 0.035708
0.383083 0.036126
0.370651 0.036530
0.358228 0.036916
0.345814 0.037284
0.333403 0.037629
0.320995 0.037950
0.308592 0.038244
0.296191 0.038506
0.283793 0.038733
0.271398 0.038920
0.259004 0.039061
0.246612 0.039153
0.234221 0.039188
0.221833 0.039162
0.209446 0.039064
0.197067 0.038889
0.184693 0.038628
0.172330 0.038271
0.159986 0.037809
0.147685 0.037231
0.135454 0.036526
0.123360 0.035684
0.111394 0.034690
0.099596 0.033528
0.088011 0.032181
0.076685 0.030635
0.065663 0.028864
0.055015 0.026849
0.044865 0.024579
0.035426 0.022076
0.027030 0.019427
0.019970 0.016771
0.014377 0.014268
0.010159 0.012029
0.007009 0.010051
0.004650 0.008292
0.002879 0.006696
0.001578 0.005207
0.000698 0.003785
0.000198 0.002434
0.000000 0.001190
0.000000 0.000000
0.000258 -0.001992
0.000832 -0.003348
0.001858 -0.004711
0.003426 -0.005982
0.005568 -0.007173
0.008409 -0.008303
0.012185 -0.009379
0.017243 -0.010404
0.023929 -0.011326
0.032338 -0.012056
0.042155 -0.012532
0.052898 -0.012742
0.064198 -0.012720
0.075846 -0.012533
0.087736 -0.012223
0.099803 -0.011837
0.111997 -0.011398
0.124285 -0.010925
0.136634 -0.010429
0.149040 -0.009918
0.161493 -0.009400
0.173985 -0.008878
0.186517 -0.008359
0.199087 -0.007845
0.211686 -0.007340
0.224315 -0.006846
0.236968 -0.006364
0.249641 -0.005898
0.262329 -0.005451
0.275030 -0.005022
0.287738 -0.004615
0.300450 -0.004231
0.313158 -0.003870
0.325864 -0.003534
0.338565 -0.003224
0.351261 -0.002939
0.363955 -0.002680
0.376646 -0.002447
0.389333 -0.002239
0.402018 -0.002057
0.414702 -0.001899
0.427381 -0.001766
0.440057 -0.001656
0.452730 -0.001566
0.465409 -0.001496
0.478092 -0.001443
0.490780 -0.001407
0.503470 -0.001381
0.516157 -0.001369
0.528844 -0.001364
0.541527 -0.001368
0.554213 -0.001376
0.566894 -0.001386
0.579575 -0.001398
0.592254 -0.001410
0.604934 -0.001424
0.617614 -0.001434
0.630291 -0.001437
0.642967 -0.001443
0.655644 -0.001442
0.668323 -0.001439
0.681003 -0.001437
0.693683 -0.001440
0.706365 -0.001442
0.719048 -0.001444
0.731731 -0.001446
0.744416 -0.001443
0.757102 -0.001445
0.769790 -0.001444
0.782480 -0.001445
0.795173 -0.001446
0.807870 -0.001446
0.820569 -0.001446
0.833273 -0.001446
0.845984 -0.001448
0.858698 -0.001448
0.871422 -0.001451
0.884148 -0.001448
0.896868 -0.001446
0.909585 -0.001443
0.922302 -0.001445
0.935019 -0.001446
0.947730 -0.001446
0.960405 -0.001439
0.972917 -0.001437
0.984788 -0.001441
0.994843 -0.001441
1.000019 -0.001441
First column is X and the second column is Y. Notice how the last values of Y are repeated.
Maybe someone can provide me with a piece of code to do this? Or any suggestions are welcome as well.
Remember I need to automate this process.
Thanks for your time and effort I really appreciate it!
There is quick and dirty method if you do not know the exact function defining the foil profile. Split your data into 2 sets, top and bottom planes, so the 'x' data are monotonic increasing.
First I imported your data table in the variable A, then:
%// just reorganise your input in individual vectors. (this is optional but
%// if you do not do it you'll have to adjust the code below)
x = A(:,1) ;
y = A(:,2) ;
ipos = y > 0 ; %// indices of the top plane
ineg = y <= 0 ; %// indices of the bottom plane
xi = linspace(0,1,500) ; %// new Xi for interpolation
ypos = interp1( x(ipos) , y(ipos) , xi ) ; %// re-interp the top plane
yneg = interp1( x(ineg) , y(ineg) , xi ) ; %// re-interp the bottom plane
y_new = [fliplr(yneg) ypos] ; %// stiches the two half data set together
x_new = [fliplr(xi) xi] ;
%% // display
figure
plot(x,y,'o')
hold on
plot(x_new,y_new,'.r')
axis equal
As said on top, it is quick and dirty. As you can see from the detail figure, you can greatly improve the x resolution this way in the area where the profile is close to the horizontal direction, but you loose a bit of resolution at the noose of the foil where the profile is close to the vertical direction.
If it's acceptable then you're all set. If you really need the resolution at the nose, you could look at interpolating on x as above but do a very fine x grid near the noose (instead of the regular x grid I provided as example).
if your replace the xi definition above by:
xi = [linspace(0,0.01,50) linspace(0.01,1,500)] ;
You get the following near the nose:
adjust that to your needs.
To interpolate any function, there must be a function defined. When you define y=f(x), you cannot have the same x for two different values of y because then we are not talking about a function. In your example data, neither x nor y are monotonic, so anyway you slice it, you'll have two (or more) "y"s for the same "x". If you wish to interpolate, you need to divide this into two separate problems, top/bottom and define proper functions for interp1/2/n to work with, for example, slice it horizontally where x==0. In any case, you would have to provide additional info than just x or y alone, e.g.: x=0.5 and y is on top.
On the other hand, if all you want to do is to insert a few values between each x and y in your array, you can do this using finite differences:
%// transform your original xy into 3d array where x is in first slice and y in second
xy = permute(xy(85:95,:), [3,1,2]); %// 85:95 is near x=0 in your data
%// lets say you want to insert three additional points along each line between every two points on given airfoil
h = [0, 0.25, 0.5, 0.75].'; %// steps along each line - column vector
%// every interpolated h along the way between f(x(n)) and f(x(n+1)) can
%// be defined as: f(x(n) + h) = f(x(n)) + h*( f(x(n+1)) - f(x(n)) )
%// this is first order finite differences approximation in 1D. 2D is very
%// similar only with gradient (this should be common knowledge, look it up)
%// from here it's just fancy matrix play
%// 2D gradient of xy curve
gradxy = diff(xy, 1, 2); %// diff xy, first order, along the 2nd dimension, where x and y now run
h_times_gradxy = bsxfun(#times, h, gradxy); %// gradient times step size
xy_in_3d_array = bsxfun(#plus, xy(:,1:end-1,:), h_times_gradxy); %// addition of "f(x)" and there we have it, the new x and y for every step h
[x,y] = deal(xy_in_3d_array(:,:,1), xy_in_3d_array(:,:,2)); %// extract x and y from 3d matrix
xy_interp = [x(:), y(:)]; %// use Matlab's linear indexing to flatten x and y into columns
%// plot to check results
figure; ax = newplot; hold on;
plot(ax, xy(:,:,1), xy(:,:,2),'o-');
plot(ax, xy_interp(:,1), xy_interp(:,2),'+')
legend('Original','Interpolated',0);
axis tight;
grid;
%// The End
And these are the results, near x=0 for clarity of presentation:
Hope that helps.
Cheers.