weighted correlation matrix - matlab

in general i know that i can easily calculate correlation matrix in matlab,there is a lot of function for this,but what about weighted correlation?i found this matlab file
http://www.mathworks.com/matlabcentral/fileexchange/20846-weighted-correlation-matrix/content/weightedcorrs.m
but how does choosing weights depend on persons intuition or it is standard?
let say we have
x = randn(30,4)
x =
0.5377 0.8884 -1.0891 -1.1480
1.8339 -1.1471 0.0326 0.1049
-2.2588 -1.0689 0.5525 0.7223
0.8622 -0.8095 1.1006 2.5855
0.3188 -2.9443 1.5442 -0.6669
-1.3077 1.4384 0.0859 0.1873
-0.4336 0.3252 -1.4916 -0.0825
0.3426 -0.7549 -0.7423 -1.9330
3.5784 1.3703 -1.0616 -0.4390
2.7694 -1.7115 2.3505 -1.7947
-1.3499 -0.1022 -0.6156 0.8404
3.0349 -0.2414 0.7481 -0.8880
0.7254 0.3192 -0.1924 0.1001
-0.0631 0.3129 0.8886 -0.5445
0.7147 -0.8649 -0.7648 0.3035
-0.2050 -0.0301 -1.4023 -0.6003
-0.1241 -0.1649 -1.4224 0.4900
1.4897 0.6277 0.4882 0.7394
1.4090 1.0933 -0.1774 1.7119
1.4172 1.1093 -0.1961 -0.1941
0.6715 -0.8637 1.4193 -2.1384
-1.2075 0.0774 0.2916 -0.8396
0.7172 -1.2141 0.1978 1.3546
1.6302 -1.1135 1.5877 -1.0722
0.4889 -0.0068 -0.8045 0.9610
1.0347 1.5326 0.6966 0.1240
0.7269 -0.7697 0.8351 1.4367
-0.3034 0.3714 -0.2437 -1.9609
0.2939 -0.2256 0.2157 -0.1977
-0.7873 1.1174 -1.1658 -1.2078
and we have done
x(:,4) = sum(x,2); % Introduce correlation.
[r,p] = corrcoef(x) % Compute sample correlation and p-values.
and got
r =
1.0000 -0.0352 0.2673 0.6901
-0.0352 1.0000 -0.5101 0.2617
0.2673 -0.5101 1.0000 0.3504
0.6901 0.2617 0.3504 1.0000
it is unweighted correlation,but how can i do weighted correlation with help of matlab file?please help me


This function needs the weights of each observation as input. How you choose them is upto you.
If these were outputs of a simulation for example, you could let the weights be the number of performed iterations. If they were stock results, consider using the value in the portfolio. However, there is no standard way to get the 'best' weights in general. Just consider that a value that is more reliable should typically get more weight.

Related

How Can fit a curve to step function?

I am trying to fit curve to a step function. I tried no. of approaches like using sigmoid function,using ratio of polynomials, fitting Gauss function to derivative of step, but none of them are looking okay. Now, I came up with the idea of creating a perfect step and compute convolution of perfect step to a Gauss function and find best fit parameter using non-linear regression.
But this is also not looking good.
I am writing here code both for Sigmoid and convolution approach.
First with Sigmoid Function fit:
Function:
function d=fit_sig(param,x,y)
a=param(1);
b=param(2);
d=(a./(1+exp(-b*x)))-y;
end
main code:
a=1, b=0.09;
p0=[a,b];
sig_coff=lsqnonlin(#fit_sig,p0,[],[],[],xavg_40s1,havg_40s1);
% Plot the original and experimental data.
sig_new = sig_coff(1)./(1+exp(-sig_coff(2)*xavg_40s1));
d= havg_40s1-step_new;
figure;
plot(xavg_40s1,havg_40s1,'.-r',xavg_40s1,sig_new,'.-b');
xlabel('x-pixel'); ylabel('dz/dx (mm/pixel)'); axis square;
This is not working at all. I think my initial guesses are wrong. I tried multiple numbers but could not get it correct. I tried using curve fitting tool too but that is also not working.
Code for creating perfect step:
h=ones(1,numel(havg_40s1)); %height=1mm
h(1:81)=-0.038;
h(82:end)=1.002; %or 1.0143
figure;
plot(xavg_40s1,havg_40s1,'k.-', 'linewidth',1.5, 'markersize',16);
hold on
plot(xavg_40s1,h,'.-r','linewidth',1.5,'markersize',12);
Code using convolution approach:
Function:
function d=fit_step(param,h,x,y)
A=param(1);
mu=param(2);
sigma=param(3);
d=conv(h,A*exp(-((x-mu)/sigma).^2),'same')-y;
end
main code:
param1=[0.2247 8.1884 0.0802];
step_coff=lsqnonlin(#fit_step,param1,[],[],[],h,dx_40s1,havg_40s1);
% Plot the original and experimental data.
step_new = conv(h,step_coff(1)*exp(-((dx_40s1-step_coff(2))/step_coff(3)).^2),'same');
figure;
plot(xavg_40s1,havg_40s1,'.-r',xavg_40s1,step_new,'.-b');
This is close but edge of step has been shifted plus corners are looking sharper than measured step.
Could someone please help me out the best way to fit a step function or any suggestion to improve the code??
X data:
12.6400 12.6720 12.7040 12.7360 12.7680 12.8000 12.8320 12.8640 12.8960 12.9280 12.9600 12.9920 13.0240 13.0560 13.0880 13.1200 13.1520 13.1840 13.2160 13.2480 13.2800 13.3120 13.3440 13.3760 13.4080 13.4400 13.4720 13.5040 13.5360 13.5680 13.6000 13.6320 13.6640 13.6960 13.7280 13.7600 13.7920 13.8240 13.8560 13.8880 13.9200 13.9520 13.9840 14.0160 14.0480 14.0800 14.1120 14.1440 14.1760 14.2080 14.2400 14.272 14.3040 14.3360 14.3680 14.4000 14.4320 14.4640 14.4960 14.5280 14.5600 14.5920 14.6240 14.6560 14.6880 14.7200 14.7520 14.7840 14.8160 14.8480 14.8800 14.9120 14.9440 14.9760 15.0080 15.0400 15.0720 15.1040 15.1360 15.1680 15.2000 15.2320 15.2640 15.2960 15.3280 15.3600 15.3920 15.4240 15.4560 15.4880 15.5200 15.5520 15.5840 15.6160 15.6480 15.6800 15.7120 15.7440 15.7760 15.8080 15.8400 15.8720 15.9040 15.9360
15.9680 16.0000 16.0320 16.0640 16.0960 16.1280 16.1600 16.1920 16.2240 16.2560 16.2880 16.3200 16.3520 16.3840 16.4160 16.4480 16.4800 16.5120 16.5440 16.5760 16.6080 16.6400 16.6720 16.7040 16.7360 16.7680 16.8000 16.8320 16.8640 16.8960 16.9280 16.9600 16.9920 17.0240 17.0560 17.0880 17.1200 17.1520 17.1840 17.2160 17.2480 17.2800 17.3120 17.3440 17.3760 17.4080 17.4400 17.4720 17.5040 17.5360 17.5680 17.6000 17.6320 17.6640 17.6960 17.7280 17.7600
Y Data:
-0.0404 -0.0405 -0.0350 -0.0406 -0.0412 -0.0407 -0.0378 -0.0405 -0.0337 -0.0417 -0.0413 -0.0387 -0.0352 -0.0373 -0.0369 -0.0388 -0.0384 -0.0351 -0.0401 -0.0314 -0.0375 -0.0390 -0.0330 -0.0343 -0.0341 -0.0369 -0.0424 -0.0369 -0.0309 -0.0387 -0.0346 -0.0433 -0.0410 -0.0355 -0.0343 -0.0396 -0.0369 -0.0400 -0.0377 -0.0330 -0.0416 -0.0348 -0.0380 -0.0338 -0.0349 -0.0359 -0.0418 -0.0336 -0.0375 -0.0309 -0.0362 -0.0422 -0.0437 -0.0352 -0.0303 -0.0335 -0.0358 -0.0467 -0.0341 -0.0306 -0.0322 -0.0338 -0.0418 -0.0417 -0.0299 -0.0264 -0.0308 -0.0352 -0.0330 -0.0261 -0.0088 -0.0071 0.0013 0.0012 0.0151 0.0352 0.0475 0.0764 0.1423 0.2617 0.4057 0.6241 0.8076 0.8872 0.9248 0.9340 0.9395 0.9514 0.9650 0.9708 0.9875 0.9852 0.9955 0.9971 0.9966 0.9981 0.9983 0.9932 1.0013 1.0011 0.9961 1.0044 0.9994 1.0028 1.0028 0.9996 1.0009 1.0024 1.0027 1.0075 1.0017 1.0001 1.0033 1.0062 1.0071 1.0032 1.0026 1.0027 1.0062 1.0063 0.9981 1.0025 0.9994 1.0075 1.0026 1.0035 1.0018 0.9999 1.0045 1.0067 0.9980 1.0044 0.9976 0.9976 1.0087 1.0026 1.0010 0.9997 1.0025 0.9943 1.0098 0.9964 0.9994 0.9973 0.9997 1.0084 1.0035 0.9974 0.9967 0.9967 1.0013 1.0060 1.0026 0.9960 0.9970 0.9987 1.0054 1.0048 0.9952 0.9937 0.9972
Attached are the images of the measured step and fitted curve.

Why not take a simple approach? Smooth your data, compute it’s derivative, then find the max of that derivative. You can do the first two steps by convolving with the derivative of a Gaussian, which is easy to generate.
The location of the max is the shift of the step function you’re trying to fit. The mean of the values to the left and the mean of the values to the right are the low and high values of the step function.
From first principles (toolboxes will make all of these steps simpler), the Gaussian gradient is computed like this:
x = [12.6400 12.6720 12.7040 12.7360 12.7680 12.8000 12.8320 12.8640 12.8960 12.9280 12.9600 12.9920 13.0240 13.0560 13.0880 13.1200 13.1520 13.1840 13.2160 13.2480 13.2800 13.3120 13.3440 13.3760 13.4080 13.4400 13.4720 13.5040 13.5360 13.5680 13.6000 13.6320 13.6640 13.6960 13.7280 13.7600 13.7920 13.8240 13.8560 13.8880 13.9200 13.9520 13.9840 14.0160 14.0480 14.0800 14.1120 14.1440 14.1760 14.2080 14.2400 14.272 14.3040 14.3360 14.3680 14.4000 14.4320 14.4640 14.4960 14.5280 14.5600 14.5920 14.6240 14.6560 14.6880 14.7200 14.7520 14.7840 14.8160 14.8480 14.8800 14.9120 14.9440 14.9760 15.0080 15.0400 15.0720 15.1040 15.1360 15.1680 15.2000 15.2320 15.2640 15.2960 15.3280 15.3600 15.3920 15.4240 15.4560 15.4880 15.5200 15.5520 15.5840 15.6160 15.6480 15.6800 15.7120 15.7440 15.7760 15.8080 15.8400 15.8720 15.9040 15.9360 15.9680 16.0000 16.0320 16.0640 16.0960 16.1280 16.1600 16.1920 16.2240 16.2560 16.2880 16.3200 16.3520 16.3840 16.4160 16.4480 16.4800 16.5120 16.5440 16.5760 16.6080 16.6400 16.6720 16.7040 16.7360 16.7680 16.8000 16.8320 16.8640 16.8960 16.9280 16.9600 16.9920 17.0240 17.0560 17.0880 17.1200 17.1520 17.1840 17.2160 17.2480 17.2800 17.3120 17.3440 17.3760 17.4080 17.4400 17.4720 17.5040 17.5360 17.5680 17.6000 17.6320 17.6640 17.6960 17.7280 17.7600];
y = [-0.0404 -0.0405 -0.0350 -0.0406 -0.0412 -0.0407 -0.0378 -0.0405 -0.0337 -0.0417 -0.0413 -0.0387 -0.0352 -0.0373 -0.0369 -0.0388 -0.0384 -0.0351 -0.0401 -0.0314 -0.0375 -0.0390 -0.0330 -0.0343 -0.0341 -0.0369 -0.0424 -0.0369 -0.0309 -0.0387 -0.0346 -0.0433 -0.0410 -0.0355 -0.0343 -0.0396 -0.0369 -0.0400 -0.0377 -0.0330 -0.0416 -0.0348 -0.0380 -0.0338 -0.0349 -0.0359 -0.0418 -0.0336 -0.0375 -0.0309 -0.0362 -0.0422 -0.0437 -0.0352 -0.0303 -0.0335 -0.0358 -0.0467 -0.0341 -0.0306 -0.0322 -0.0338 -0.0418 -0.0417 -0.0299 -0.0264 -0.0308 -0.0352 -0.0330 -0.0261 -0.0088 -0.0071 0.0013 0.0012 0.0151 0.0352 0.0475 0.0764 0.1423 0.2617 0.4057 0.6241 0.8076 0.8872 0.9248 0.9340 0.9395 0.9514 0.9650 0.9708 0.9875 0.9852 0.9955 0.9971 0.9966 0.9981 0.9983 0.9932 1.0013 1.0011 0.9961 1.0044 0.9994 1.0028 1.0028 0.9996 1.0009 1.0024 1.0027 1.0075 1.0017 1.0001 1.0033 1.0062 1.0071 1.0032 1.0026 1.0027 1.0062 1.0063 0.9981 1.0025 0.9994 1.0075 1.0026 1.0035 1.0018 0.9999 1.0045 1.0067 0.9980 1.0044 0.9976 0.9976 1.0087 1.0026 1.0010 0.9997 1.0025 0.9943 1.0098 0.9964 0.9994 0.9973 0.9997 1.0084 1.0035 0.9974 0.9967 0.9967 1.0013 1.0060 1.0026 0.9960 0.9970 0.9987 1.0054 1.0048 0.9952 0.9937 0.9972];
sigma = 3;
cutoff = ceil(4*sigma);
kernel = -cutoff:cutoff;
kernel = -kernel .* exp(-0.5 * kernel.^2 / sigma.^2);
grad = conv(y,kernel,'same');
We can find the maximum sample with max:
[~,ii] = max(grad);
This is the sample nearest the middle of the transition point. We can refine this location by fitting a parabola to the 3 samples around the peak:
px = x(ii-1:ii+1).';
py = grad(ii-1:ii+1).';
% solve the equation: py = [px.*px, px, ones(3,1)] * params;
params = [px.*px, px, ones(3,1)] \ py;
x_max = -params(2)/(2*params(1));
Finally, we might want to include the y values before and after the transition into the fitting:
left = median(y(x<x_max));
right = median(y(x>x_max));
(Though we might want to assume that left=0 and right=1.)
Plotting:
plot(x,y)
hold on
plot([x(1),x_max,x_max,x(end)],[left,left,right,right])
To fit a full error function (which is the integral of the Gaussian) we just need one more step: above we fitted a parabola to the three samples around the maximum, now instead we fit a parabola to the logarithm of the y values (see this other Q&A for an explanation), and select all y values above 0.2 times the peak value for this fit to avoid fitting to noise. This is approximately 2 sigma, and should be enough to obtain an accurate estimate of the Gaussian peak. The parameters of the Gaussian peak are also the parameters to the smoothed error function, we can correct the estimated sigma for this additional smoothing:
% using grad from the code above (as well as x and y)
[m,ii] = max(grad);
w = sum(grad > m * 0.2) / 2;
px = x(ii-w:ii+w).';
py = log(grad(ii-w:ii+w)).';
% solve the equation: py = [px.*px, px, ones(3,1)] * params;
params = [px.*px, px, ones(size(px))] \ py;
% obtain Gaussian parameters
fitted_mu = -params(2)/(2*params(1));
fitted_sigma = sqrt(-0.5/params(1));
% correct for smoothing applied
fitted_sigma = sqrt(fitted_sigma^2 - (sigma*mean(diff(x)))^2);
% evaluated fitted function
fitted_y = (erf((x-fitted_mu)/fitted_sigma) + 1) / 2 * (right-left) + left;
clf
plot(x,y)
hold on
plot(x,fitted_y)

Here is an example graphical fitter using your data and a sigmoid from my equation search. This example uses the standard scipy differential_evolution genetic algorithm module to determine initial parameter estimates for curve fitting, and that module uses the Latin Hypercube algorithm to ensure a thorough search of parameter space requiring bounds within which to search. In this example the search bounds are derived from the data. Note that it is much easier to estimate ranges for the initial parameter estimates rather than specific values.
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
import warnings
xData = numpy.array([12.6400, 12.6720, 12.7040, 12.7360, 12.7680, 12.8000, 12.8320, 12.8640, 12.8960, 12.9280, 12.9600, 12.9920, 13.0240, 13.0560, 13.0880, 13.1200, 13.1520, 13.1840, 13.2160, 13.2480, 13.2800, 13.3120, 13.3440, 13.3760, 13.4080, 13.4400, 13.4720, 13.5040, 13.5360, 13.5680, 13.6000, 13.6320, 13.6640, 13.6960, 13.7280, 13.7600, 13.7920, 13.8240, 13.8560, 13.8880, 13.9200, 13.9520, 13.9840, 14.0160, 14.0480, 14.0800, 14.1120, 14.1440, 14.1760, 14.2080, 14.2400, 14.272, 14.3040, 14.3360, 14.3680, 14.4000, 14.4320, 14.4640, 14.4960, 14.5280, 14.5600, 14.5920, 14.6240, 14.6560, 14.6880, 14.7200, 14.7520, 14.7840, 14.8160, 14.8480, 14.8800, 14.9120, 14.9440, 14.9760, 15.0080, 15.0400, 15.0720, 15.1040, 15.1360, 15.1680, 15.2000, 15.2320, 15.2640, 15.2960, 15.3280, 15.3600, 15.3920, 15.4240, 15.4560, 15.4880, 15.5200, 15.5520, 15.5840, 15.6160, 15.6480, 15.6800, 15.7120, 15.7440, 15.7760, 15.8080, 15.8400, 15.8720, 15.9040, 15.9360, 15.9680, 16.0000, 16.0320, 16.0640, 16.0960, 16.1280, 16.1600, 16.1920, 16.2240, 16.2560, 16.2880, 16.3200, 16.3520, 16.3840, 16.4160, 16.4480, 16.4800, 16.5120, 16.5440, 16.5760, 16.6080, 16.6400, 16.6720, 16.7040, 16.7360, 16.7680, 16.8000, 16.8320, 16.8640, 16.8960, 16.9280, 16.9600, 16.9920, 17.0240, 17.0560, 17.0880, 17.1200, 17.1520, 17.1840, 17.2160, 17.2480, 17.2800, 17.3120, 17.3440, 17.3760, 17.4080, 17.4400, 17.4720, 17.5040, 17.5360, 17.5680, 17.6000, 17.6320, 17.6640, 17.6960, 17.7280, 17.7600])
yData = numpy.array([-0.0404, -0.0405, -0.0350, -0.0406, -0.0412, -0.0407, -0.0378, -0.0405, -0.0337, -0.0417, -0.0413, -0.0387, -0.0352, -0.0373, -0.0369, -0.0388, -0.0384, -0.0351, -0.0401, -0.0314, -0.0375, -0.0390, -0.0330, -0.0343, -0.0341, -0.0369, -0.0424, -0.0369, -0.0309, -0.0387, -0.0346, -0.0433, -0.0410, -0.0355, -0.0343, -0.0396, -0.0369, -0.0400, -0.0377, -0.0330, -0.0416, -0.0348, -0.0380, -0.0338, -0.0349, -0.0359, -0.0418, -0.0336, -0.0375, -0.0309, -0.0362, -0.0422, -0.0437, -0.0352, -0.0303, -0.0335, -0.0358, -0.0467, -0.0341, -0.0306, -0.0322, -0.0338, -0.0418, -0.0417, -0.0299, -0.0264, -0.0308, -0.0352, -0.0330, -0.0261, -0.0088, -0.0071, 0.0013, 0.0012, 0.0151, 0.0352, 0.0475, 0.0764, 0.1423, 0.2617, 0.4057, 0.6241, 0.8076, 0.8872, 0.9248, 0.9340, 0.9395, 0.9514, 0.9650, 0.9708, 0.9875, 0.9852, 0.9955, 0.9971, 0.9966, 0.9981, 0.9983, 0.9932, 1.0013, 1.0011, 0.9961, 1.0044, 0.9994, 1.0028, 1.0028, 0.9996, 1.0009, 1.0024, 1.0027, 1.0075, 1.0017, 1.0001, 1.0033, 1.0062, 1.0071, 1.0032, 1.0026, 1.0027, 1.0062, 1.0063, 0.9981, 1.0025, 0.9994, 1.0075, 1.0026, 1.0035, 1.0018, 0.9999, 1.0045, 1.0067, 0.9980, 1.0044, 0.9976, 0.9976, 1.0087, 1.0026, 1.0010, 0.9997, 1.0025, 0.9943, 1.0098, 0.9964, 0.9994, 0.9973, 0.9997, 1.0084, 1.0035, 0.9974, 0.9967, 0.9967, 1.0013, 1.0060, 1.0026, 0.9960, 0.9970, 0.9987, 1.0054, 1.0048, 0.9952, 0.9937, 0.9972])
def func(x, a, b, Offset): # Sigmoid A with Offset from zunzun.com
return 1.0 / (1.0 + numpy.exp(-1.0 * a * (x - b))) + Offset
# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
val = func(xData, *parameterTuple)
return numpy.sum((yData - val) ** 2.0)
def generate_Initial_Parameters():
# min and max used for bounds
maxX = max(xData)
minX = min(xData)
maxY = max(yData)
minY = min(yData)
parameterBounds = []
parameterBounds.append([minX, maxX]) # search bounds for a
parameterBounds.append([minX, maxX]) # search bounds for b
parameterBounds.append([minY, maxY]) # search bounds for Offset
# "seed" the numpy random number generator for repeatable results
result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3)
return result.x
# by default, differential_evolution completes by calling curve_fit() using parameter bounds
geneticParameters = generate_Initial_Parameters()
# now call curve_fit without passing bounds from the genetic algorithm,
# just in case the best fit parameters are aoutside those bounds
fittedParameters, pcov = curve_fit(func, xData, yData, geneticParameters)
print('Fitted parameters:', fittedParameters)
print()
modelPredictions = func(xData, *fittedParameters)
absError = modelPredictions - yData
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(yData))
print()
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(xData, yData, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(xData), max(xData))
yModel = func(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('X Data') # X axis data label
axes.set_ylabel('Y Data') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)

Strange result of spap2

I encounter strange results from spap2 on some data:
The actual data is the blue curve, red circles are the knots I am using and yellow curve is the display of the cubic spline curve.
The code is quite simple, I cannot figure out what is the problem:
spgood = spap2(knots_zY, 4, ec, Y);
plot(ec, Y);
hold on;
scatter(knots_zY, Y(ec==knots_zY));
fnplt(spgood)
ec is the vector -4.12:0.02:-0.54.
Y is the following vector:
4.1291 4.0732 4.0173 4.2624 4.3826 4.3267 4.2708 4.4367 4.3808 4.1031 4.1721 3.8152 4.1572
4.1013 4.0454 3.5916 3.8367 3.7808 3.8218 3.6690 3.9141 3.7333 3.8023 3.3204 3.5656 3.4305
3.5787 3.3978 3.3419 3.2860 3.4062 3.4753 3.5706 3.2385 3.1826 3.4947 3.5315 3.1746 3.2089
3.2276 3.1940 2.9162 3.0364 3.0263 2.8155 2.7596 2.9555 2.8996 2.9081 2.7322 2.8524 2.6397
2.7662 2.5279 2.5417 2.2005 2.3409 2.5108 2.5202 2.3359 2.3660 2.3100 2.1682 2.1123 2.2140
2.1288 2.1116 1.9856 2.0089 1.8845 1.9148 1.9308 1.7273 1.7642 1.7326 1.6606 1.7378 1.6570
1.5815 1.5701 1.4630 1.5503 1.5181 1.4385 1.3083 1.3168 1.2991 1.2523 1.1390 0.9988 1.0373
0.9913 1.0113 0.9754 0.8912 0.8790 0.7491 0.7557 0.7544 0.7119 0.7031 0.6843 0.6418 0.5938
0.5193 0.5334 0.4312 0.4839 0.4437 0.3992 0.3689 0.3287 0.3348 0.3076 0.2274 0.2174 0.1970
0.2188 0.1760 0.1384 0.1773 0.1342 0.1388 0.1097 0.0830 0.0782 0.0725 0.0863 0.0581 0.0466
0.0398 0.0431 0.0187 0.0187 0.0176 0.0167 0.0231 0.0033 -0.0117 -0.0016 0.0084 -0.0055 -0.0120
-0.0080 -0.0064 -0.0075 -0.0134 -0.0075 0.0012 -0.0077 -0.0024 0.0006 0.0010 0.0043 0.0016 0.0018
0.0042 0.0030 0.0029 0.0029 0.0021 0.0013 -0.0002 -0.0020 -0.0030 -0.0032 -0.0002 -0.0013 0.0035
0.0028 -0.0000 -0.0057 -0.0032 0.0020 0.0597 0.1835 0.5083 1.0275 1.6448 3.0549
The knots are defined with the following 12 values:
-4.1200 -3.9400 -3.5400 -3.3000 -3.1400 -2.6800 -2.3600 -2.0600 -1.5000 -1.1600 -0.7000 -0.5400
I don't expect a nice fit, but at least the spline fit sticks with the knots ... but here the result is completely erroneous. I am stuck with this, unable to see where is the problem with this data sample.
Note: the knots are computed in a separate algorithm and should be used for the interpolator, getting a good fit is not the question here. The question is why the spline fit does not pass through the knots.

I have made several errors.
First, it's a mistake to assume that the result spline will pass through the knots, as it is an approximation (see this answer). The approximation smoothes the whole original data so there is no way to stick on knots.
Second, I have forgot to extend the end knots to impose boundary conditions. The default boundary condition is to have all derivatives (including the 0th-order) to be zero, resulting in this shape. The solution is then to use augknt to get an actual cubic spline with two continuous derivatives:
spgood = spap2(augknt(knots_zY,4), 4, ec, Y);
The resulting fit is:
which is way better, given the choice of the knot sequence.

Potential Bug in MATLAB regress R2014a

MATLAB R2014a used to work fine w regress but now I get an error when the variables are fine and rank is satisfactory.
X = rand([10 3])
X =
0.8407 0.3517 0.0759
0.2543 0.8308 0.0540
0.8143 0.5853 0.5308
0.2435 0.5497 0.7792
0.9293 0.9172 0.9340
0.3500 0.2858 0.1299
0.1966 0.7572 0.5688
0.2511 0.7537 0.4694
0.6160 0.3804 0.0119
0.4733 0.5678 0.3371
K>> Y = rand([10 1])
Y =
0.1622
0.7943
0.3112
0.5285
0.1656
0.6020
0.2630
0.6541
0.6892
0.7482
[B,BINT] = regress(Y,X)
Subscript indices must either be real positive integers or logicals.
Error in regress (line 93)
b(perm) = R \ (Q'*y);
Obviously, X and Y are fine. There's something wrong w/ the matrix math in regress and it's that perm is, for some reason, being output as a vector (giving an ind error). A few lines above, qr is called like this with no further modification to perm:
[Q,R,perm] = qr(X,0);
Help file says qr is supposed to output a third argument that's a matrix - but how can that be if the math always expects a vector?
% [Q,R,E] = QR(A) produces unitary Q, upper triangular R and a
% permutation matrix E so that A*E = Q*R. The column permutation E is
% chosen so that ABS(DIAG(R)) is decreasing.
Very confusing considering both of these are built-in functions. I literally re-installed MATLAB R2014a and a few toolboxes and STILL get this error. It feels like qr got updated to output a different argument, but I don't understand why a fresh reinstall wouldn't take care of this, or why qr would have updated at all anyways. Everything else in my MATLAB is great.
Any ideas???

How to use Matlab for non linear least squares Michaelis–Menten parameters estimation

I have a set of measurements and I started making a linear approximation (as in this plot). A linear least squares estimation of the parameters V_{max} and K_{m} from this code in Matlab:
data=[2.0000 0.0615
2.0000 0.0527
0.6670 0.0334
0.6670 0.0334
0.4000 0.0138
0.4000 0.0258
0.2860 0.0129
0.2860 0.0183
0.2220 0.0083
0.2200 0.0169
0.2000 0.0129
0.2000 0.0087 ];
x = 1./data(:,1);
y = 1./data(:,2);
J = [x,ones(length(x),1)];
k = J\y;
vmax = 1/k(2);
km = k(1)*vmax;
lse = (vmax.*data(:,1))./(km+data(:,1));
plot(data(:,1),data(:,2),'o','color','red','linewidth',1)
line(data(:,1),lse,'linewidth',2)
This yields a fit that looks alright. Next, I wanted to do the same thing but with non-linear least squares. However, the fit always looks wrong, here is the code for that attempt:
options = optimset('MaxIter',10000,'MaxFunEvals',50000,'FunValCheck',...
'on','Algorithm',{'levenberg-marquardt',.00001});
p=lsqnonlin(#myfun,[0.1424,2.5444]);
lse = (p(1).*data(:,1))./(p(2)+data(:,1));
plot(data(:,1),data(:,2),'o','color','red','linewidth',1)
line(data(:,1),lse,'linewidth',2)
which requires this function in an M-File:
function F = myfun(x)
F = data(:,2)-(x(1).*data(:,1))./x(2)+data(:,1);
If you run the code you will see my problem. But hopefully, unlike me, you see what I'm doing wrong.

I think that you forgot some parentheses (some others are superfluous) in your nonlinear function. Using an anonymous function:
myfun = #(x)data(:,2)-x(1).*data(:,1)./(x(2)+data(:,1)); % Parentheses were missing
options = optimset('MaxIter',10000,'MaxFunEvals',50000,'FunValCheck','on',...
'Algorithm',{'levenberg-marquardt',.00001});
p = lsqnonlin(myfun,[0.1424,2.5444],[],[],options);
lse = p(1).*data(:,1)./(p(2)+data(:,1));
plot(data(:,1),data(:,2),'o','color','red','linewidth',1)
line(data(:,1),lse,'linewidth',2)
You also weren't actually applying any of your options.
You might look into using lsqcurvefit instead as it was designed for data fitting problems:
myfun = #(x,dat)x(1).*dat./(x(2)+dat);
options = optimset('MaxIter',10000,'MaxFunEvals',50000,'FunValCheck','on',...
'Algorithm',{'levenberg-marquardt',.00001});
p = lsqcurvefit(myfun,[0.1424,2.5444],data(:,1),data(:,2),[],[],options);
lse = myfun(p,data(:,1));
plot(data(:,1),data(:,2),'o','color','red','linewidth',1)
line(data(:,1),lse,'linewidth',2)

Finding more than one maximum in an array

I would like to find more than one maximum value from an array using Matlab.
Here is my code that returns only one max and its position:
[peak, pos] = max(abs(coeffs));
Problem is that I want to detect more than one max in the array. In fact, I would need to detect the first two peaks and their positions in the following array:
>> abs(coeffs())
ans =
0.5442
0.5465
0.5545
0.5674
0.5862
0.6115
0.6438
0.6836
0.7333
0.7941
0.8689
0.9608
1.0751
1.2188
1.4027
1.6441
1.9701
2.4299
3.1178
4.2428
6.3792
11.8611
53.7537
24.9119
10.8982
7.3470
5.7768
4.9340
4.4489
4.1772
4.0564
4.0622
4.1949
4.4801
4.9825
5.8496
7.4614
11.1087
25.6071
53.2831
12.0029
6.4743
4.3096
3.1648
2.4631
1.9918
1.6558
1.4054
1.2129
1.0608
0.9379
0.8371
0.7532
0.6827
0.6224
0.5702
0.5255
0.4861
0.4517
0.4212
0.3941
0.3698
0.3481
0.3282
0.3105
0.2946
0.2796
0.2665
0.2541
0.2429
0.2326
0.2230
0.2141
0.2057
0.1986
0.1914
0.1848
0.1787
0.1729
0.1677
0.1627
0.1579
0.1537
0.1494
0.1456
0.1420
0.1385
0.1353
0.1323
0.1293
0.1267
0.1239
0.1216
0.1192
0.1172
0.1151
0.1132
0.1113
0.1096
0.1080
0.1064
0.1048
0.1038
0.1024
0.1011
0.1000
0.0987
0.0978
0.0967
0.0961
0.0951
0.0943
0.0936
0.0930
0.0924
0.0917
0.0913
0.0908
0.0902
0.0899
0.0894
0.0892
0.0889
0.0888
0.0885
0.0883
0.0882
0.0883
0.0882
0.0883
0.0882
0.0883
0.0885
0.0888
0.0889
0.0892
0.0894
0.0899
0.0902
0.0908
0.0913
0.0917
0.0924
0.0930
0.0936
0.0943
0.0951
0.0961
0.0967
0.0978
0.0987
0.1000
0.1011
0.1024
0.1038
0.1048
0.1064
0.1080
0.1096
0.1113
0.1132
0.1151
0.1172
0.1192
0.1216
0.1239
0.1267
0.1293
0.1323
0.1353
0.1385
0.1420
0.1456
0.1494
0.1537
0.1579
0.1627
0.1677
0.1729
0.1787
0.1848
0.1914
0.1986
0.2057
0.2141
0.2230
0.2326
0.2429
0.2541
0.2665
0.2796
0.2946
0.3105
0.3282
0.3481
0.3698
0.3941
0.4212
0.4517
0.4861
0.5255
0.5702
0.6224
0.6827
0.7532
0.8371
0.9379
1.0608
1.2129
1.4054
1.6558
1.9918
2.4631
3.1648
4.3096
6.4743
12.0029
53.2831
25.6071
11.1087
7.4614
5.8496
4.9825
4.4801
4.1949
4.0622
4.0564
4.1772
4.4489
4.9340
5.7768
7.3470
10.8982
24.9119
53.7537
11.8611
6.3792
4.2428
3.1178
2.4299
1.9701
1.6441
1.4027
1.2188
1.0751
0.9608
0.8689
0.7941
0.7333
0.6836
0.6438
0.6115
0.5862
0.5674
0.5545
0.5465
The reason I need only the two first max values is that the two last ones are reflections of the two first ones as a result of a fast fourier transform.

you can use many peak finding tools to do that. Here's some of them:
Findpeaks
The function [pks,locs] = findpeaks(data) returns local maxima or peaks, pks, in the input data at locations locs (sorted from first to last found). Data requires a row or column vector with real-valued elements with a minimum length of three. findpeaks compares each element of data to its neighboring values. If an element of data is larger than both of its neighbors or equals Inf, the element is a local peak. If there are no local maxima, pks is an empty vector.
For example:
[pks,locs] = findpeaks(abs(coeffs))
plot(abs(coeffs)); hold on
plot(locs(1:2),pks(1:2),'ro');
1D Non-derivative Peak Finder - a FEX tool that finds peaks without taking first or second derivatives, rather it uses local slope features in a given data set.
PeakFinder - another peak finder from the FEX by nate yoder.
and there are plenty more of these in the FEX...