I asked a since deleted question regarding how to determine Fourier coefficients from time series data. I am resubmitting this because I have better formulated the problem and have a solution that I'll give as I think others may find this very useful.
I have some time series data that I have binned into equally spaced time bins (a fact which will be crucial to my solution), and from that data I want to determine the Fourier series (or any function, really) that best describes the data. Here is a MWE with some test data to show the data I'm trying to fit:
import numpy as np
import matplotlib.pyplot as plt
# Create a dependent test variable to define the x-axis of the test data.
test_array = np.linspace(0, 1, 101) - 0.5
# Define some test data to try to apply a Fourier series to.
test_data = [0.9783883464566918, 0.979599093567252, 0.9821424606299206, 0.9857575507812502, 0.9899278899999995,
0.9941848228346452, 0.9978438300395263, 1.0003009205426352, 1.0012208923679058, 1.0017130521235522,
1.0021799664031628, 1.0027475606936413, 1.0034168260869563, 1.0040914266144825, 1.0047781181102355,
1.005520348837209, 1.0061899214145387, 1.006846206627681, 1.0074483048543692, 1.0078691461988312,
1.008318736328125, 1.008446947572815, 1.00862051262136, 1.0085134881422921, 1.008337095516569,
1.0079539881889774, 1.0074857334630352, 1.006747783037474, 1.005962048923679, 1.0049115434782612,
1.003812267822736, 1.0026427549407106, 1.001251963531669, 0.999898555335968, 0.9984976286266923,
0.996995982142858, 0.9955652088974847, 0.9941647321428578, 0.9927727076023389, 0.9914750532544377,
0.990212467710371, 0.9891098035363466, 0.9875998927875242, 0.9828093773946361, 0.9722532524271845,
0.9574084365384614, 0.9411012303149601, 0.9251820309477757, 0.9121488392156851, 0.9033119748549322,
0.9002445803921568, 0.9032760564202343, 0.91192435882353, 0.9249696964980555, 0.94071381372549,
0.957139088974855, 0.9721083392156871, 0.982955287937743, 0.9880613320235758, 0.9897455322896282,
0.9909590626223097, 0.9922601592233015, 0.9936513112840472, 0.9951442427184468, 0.9967071285988475,
0.9982921493123781, 0.9998775465116277, 1.001389230174081, 1.0029109110251453, 1.0044033691406251,
1.0057110841487276, 1.0069551867704276, 1.008118776264591, 1.0089884470588228, 1.0098663972602735,
1.0104514566473979, 1.0109849223300964, 1.0112043902912626, 1.0114717968750002, 1.0113343036750482,
1.0112205972495087, 1.0108811786407768, 1.010500276264591, 1.0099054552529192, 1.009353759223301,
1.008592596116505, 1.007887223091976, 1.0070715634615386, 1.0063525891472884, 1.0055587861271678,
1.0048733732809436, 1.0041832862669238, 1.0035913326848247, 1.0025318871595328, 1.000088536345776,
0.9963596140350871, 0.9918380684931506, 0.9873937281553398, 0.9833394624277463, 0.9803621496062999,
0.9786476100386117]
# Create a figure to view the data.
fig, ax = plt.subplots(1, 1, figsize=(6, 6))
# Plot the data.
ax.scatter(test_array, test_data, color="k", s=1)
This outputs the following:
The question is how to determine the Fourier series best describing this data. The usual formula for determining the Fourier coefficients requires inserting a function into an integral, but if I had a function to describe the data I wouldn't need the Fourier coefficients at all; the whole point of finding this series is to have a functional representation of the data. In the absence of such a function, then, how are the coefficients found?
My solution to this problem is to apply a discrete Fourier transform to the data using NumPy's implementation of the Fast Fourier Transform, numpy.fft.fft(); this is why it's critical that the data is evenly spaced in time, as FFT requires this. While the FFT is typically used to perform analysis of the frequency spectrum, the desired Fourier coefficients are directly related to the output of this function.
Specifically, this function outputs a series of i complex-valued coefficients c. The Fourier series coefficients are found using the relations:
Therefore the FFT allows the Fourier coefficients to be directly computed. Here is the MWE of my solution to this problem, expanding the example given above:
import numpy as np
import matplotlib.pyplot as plt
# Set the number of equal-time bins to create.
n_bins = 101
# Set the number of Fourier coefficients to use.
n_coeff = 51
# Define a function to generate a Fourier series based on the coefficients determined by the Fast Fourier Transform.
# This also includes a series of phases x to pass through the function.
def create_fourier_series(x, coefficients):
# Begin the series with the zeroeth-order Fourier coefficient.
fourier_series = coefficients[0][0] / 2
# Now generate the first through n_coeff'th terms. The period is defined to be 1 since we're operating in phase
# space.
for n in range(1, n_coeff):
fourier_series += (fourier_coeff[n][0] * np.cos(2 * np.pi * n * x) + fourier_coeff[n][1] *
np.sin(2 * np.pi * n * x))
return fourier_series
# Create a dependent test variable to define the x-axis of the test data.
test_array = np.linspace(0, 1, n_bins) - 0.5
# Define some test data to try to apply a Fourier series to.
test_data = [0.9783883464566918, 0.979599093567252, 0.9821424606299206, 0.9857575507812502, 0.9899278899999995,
0.9941848228346452, 0.9978438300395263, 1.0003009205426352, 1.0012208923679058, 1.0017130521235522,
1.0021799664031628, 1.0027475606936413, 1.0034168260869563, 1.0040914266144825, 1.0047781181102355,
1.005520348837209, 1.0061899214145387, 1.006846206627681, 1.0074483048543692, 1.0078691461988312,
1.008318736328125, 1.008446947572815, 1.00862051262136, 1.0085134881422921, 1.008337095516569,
1.0079539881889774, 1.0074857334630352, 1.006747783037474, 1.005962048923679, 1.0049115434782612,
1.003812267822736, 1.0026427549407106, 1.001251963531669, 0.999898555335968, 0.9984976286266923,
0.996995982142858, 0.9955652088974847, 0.9941647321428578, 0.9927727076023389, 0.9914750532544377,
0.990212467710371, 0.9891098035363466, 0.9875998927875242, 0.9828093773946361, 0.9722532524271845,
0.9574084365384614, 0.9411012303149601, 0.9251820309477757, 0.9121488392156851, 0.9033119748549322,
0.9002445803921568, 0.9032760564202343, 0.91192435882353, 0.9249696964980555, 0.94071381372549,
0.957139088974855, 0.9721083392156871, 0.982955287937743, 0.9880613320235758, 0.9897455322896282,
0.9909590626223097, 0.9922601592233015, 0.9936513112840472, 0.9951442427184468, 0.9967071285988475,
0.9982921493123781, 0.9998775465116277, 1.001389230174081, 1.0029109110251453, 1.0044033691406251,
1.0057110841487276, 1.0069551867704276, 1.008118776264591, 1.0089884470588228, 1.0098663972602735,
1.0104514566473979, 1.0109849223300964, 1.0112043902912626, 1.0114717968750002, 1.0113343036750482,
1.0112205972495087, 1.0108811786407768, 1.010500276264591, 1.0099054552529192, 1.009353759223301,
1.008592596116505, 1.007887223091976, 1.0070715634615386, 1.0063525891472884, 1.0055587861271678,
1.0048733732809436, 1.0041832862669238, 1.0035913326848247, 1.0025318871595328, 1.000088536345776,
0.9963596140350871, 0.9918380684931506, 0.9873937281553398, 0.9833394624277463, 0.9803621496062999,
0.9786476100386117]
# Determine the fast Fourier transform for this test data.
fast_fourier_transform = np.fft.fft(test_data[n_bins / 2:] + test_data[:n_bins / 2])
# Create an empty list to hold the values of the Fourier coefficients.
fourier_coeff = []
# Loop through the FFT and pick out the a and b coefficients, which are the real and imaginary parts of the
# coefficients calculated by the FFT.
for n in range(0, n_coeff):
a = 2 * fast_fourier_transform[n].real / n_bins
b = -2 * fast_fourier_transform[n].imag / n_bins
fourier_coeff.append([a, b])
# Create the Fourier series approximating this data.
fourier_series = create_fourier_series(test_array, fourier_coeff)
# Create a figure to view the data.
fig, ax = plt.subplots(1, 1, figsize=(6, 6))
# Plot the data.
ax.scatter(test_array, test_data, color="k", s=1)
# Plot the Fourier series approximation.
ax.plot(test_array, fourier_series, color="b", lw=0.5)
This outputs the following:
Note that how I defined the FFT (importing the second half of the data followed by the first half) is a consequence of how this data was generated. Specifically, the data runs from -0.5 to 0.5, but the FFT assumes it runs from 0.0 to 1.0, necessitating this shift.
I've found that this works quite well for data that doesn't include very sharp and narrow discontinuities. I would be interested to hear if anyone has another suggested solution to this problem, and I hope people find this explanation clear and helpful.
Not sure if it helps you in anyway; I wrote a programme to interpoplate your data. This is done using buildingblocks==0.0.15
Please see below,
import matplotlib.pyplot as plt
from buildingblocks import bb
import numpy as np
Ydata = [0.9783883464566918, 0.979599093567252, 0.9821424606299206, 0.9857575507812502, 0.9899278899999995,
0.9941848228346452, 0.9978438300395263, 1.0003009205426352, 1.0012208923679058, 1.0017130521235522,
1.0021799664031628, 1.0027475606936413, 1.0034168260869563, 1.0040914266144825, 1.0047781181102355,
1.005520348837209, 1.0061899214145387, 1.006846206627681, 1.0074483048543692, 1.0078691461988312,
1.008318736328125, 1.008446947572815, 1.00862051262136, 1.0085134881422921, 1.008337095516569,
1.0079539881889774, 1.0074857334630352, 1.006747783037474, 1.005962048923679, 1.0049115434782612,
1.003812267822736, 1.0026427549407106, 1.001251963531669, 0.999898555335968, 0.9984976286266923,
0.996995982142858, 0.9955652088974847, 0.9941647321428578, 0.9927727076023389, 0.9914750532544377,
0.990212467710371, 0.9891098035363466, 0.9875998927875242, 0.9828093773946361, 0.9722532524271845,
0.9574084365384614, 0.9411012303149601, 0.9251820309477757, 0.9121488392156851, 0.9033119748549322,
0.9002445803921568, 0.9032760564202343, 0.91192435882353, 0.9249696964980555, 0.94071381372549,
0.957139088974855, 0.9721083392156871, 0.982955287937743, 0.9880613320235758, 0.9897455322896282,
0.9909590626223097, 0.9922601592233015, 0.9936513112840472, 0.9951442427184468, 0.9967071285988475,
0.9982921493123781, 0.9998775465116277, 1.001389230174081, 1.0029109110251453, 1.0044033691406251,
1.0057110841487276, 1.0069551867704276, 1.008118776264591, 1.0089884470588228, 1.0098663972602735,
1.0104514566473979, 1.0109849223300964, 1.0112043902912626, 1.0114717968750002, 1.0113343036750482,
1.0112205972495087, 1.0108811786407768, 1.010500276264591, 1.0099054552529192, 1.009353759223301,
1.008592596116505, 1.007887223091976, 1.0070715634615386, 1.0063525891472884, 1.0055587861271678,
1.0048733732809436, 1.0041832862669238, 1.0035913326848247, 1.0025318871595328, 1.000088536345776,
0.9963596140350871, 0.9918380684931506, 0.9873937281553398, 0.9833394624277463, 0.9803621496062999,
0.9786476100386117]
Xdata=list(range(0,len(Ydata)))
Xnew=list(np.linspace(0,len(Ydata),200))
Ynew=bb.interpolate(Xdata,Ydata,Xnew,40)
plt.figure()
plt.plot(Xdata,Ydata)
plt.plot(Xnew,Ynew,'*')
plt.legend(['Given Data', 'Interpolated Data'])
plt.show()
Should you want to further write code, I have also give code so that you can see the source code and learn:
import module
import inspect
src = inspect.getsource(module)
print(src)
I am trying to fit the parameters of a transit light curve.
I have observed transit light curve data and I am using a .py in python that through 4 parameters (period, a(semi-major axis), inclination, planet radius) returns a model transit light curve. I would like to minimize the residual between these two light curves. This is what I am trying to do: First - Estimate a max likelihood using method = "L-BFGS-B" and then apply the mcmc using emcee to estimate the uncertainties.
The code:
p = lmfit.Parameters()
p.add_many(('per', 2.), ('inc', 90.), ('a', 5.), ('rp', 0.1))
per_b = [1., 3.]
a_b = [4., 6.]
inc_b = [88., 90.]
rp_b = [0.1, 0.3]
bounds = [(per_b[0], per_b[1]), (inc_b[0], inc_b[1]), (a_b[0], a_b[1]), (rp_b[0], rp_b[1])]
def residual(p):
v = p.valuesdict()
eclipse.criarEclipse(v['per'], v['a'], v['inc'], v['rp'])
lc0 = numpy.array(eclipse.getCurvaLuz()) (observed flux data)
ts0 = numpy.array(eclipse.getTempoHoras()) (observed time data)
c = numpy.linspace(min(time_phased[bb]),max(time_phased[bb]),len(time_phased[bb]),endpoint=True)
nn = interpolate.interp1d(ts0,lc0)
return nn(c) - smoothed_LC[bb] (residual between the model and the data)
Inside def residual(p) I make sure that both the observed data (time_phased[bb] and smoothed_LC[bb]) have the same size of the model transit light curve. I want it to give me the best fit values for the parameters (v['per'], v['a'], v['inc'], v['rp']).
I need your help and I appreciate your time and your attention. Kindest regards, Yuri.
Your example is incomplete, with many partial concepts and some invalid Python. This makes it slightly hard to understand your intention. If the answer below is not sufficient, update your question with a complete example.
It seems pretty clear that you want to model your data smoothed_LC[bb] (not sure what bb is) with a model for some effect of an eclipse. With that assumption, I would recommend using the lmfit.Model approach. Start by writing a function that models the data, just so you check and plot your model. I'm not entirely sure I understand everything you're doing, but this model function might look like this:
import numpy
from scipy import interpolate
from lmfit import Model
# import eclipse from somewhere....
def eclipse_lc(c, per, a, inc, p):
eclipse.criarEclipse(per, a, inc, rp)
lc0 = numpy.array(eclipse.getCurvaLuz()) # observed flux data
ts0 = numpy.array(eclipse.getTempoHoras()) # observed time data
return interpolate.interp1d(ts0,lc0)(c)
With this model function, you can build a Model:
lc_model = Model(eclipse_lc)
and then build parameters for your model. This will automatically name them after the argument names of your model function. Here, you can also give them initial values:
params = lc_model.make_params(per=2, inc=90, a=5, rp=0.1)
You wanted to place upper and lower bounds on these parameters. This is done by setting min and max parameters, not making an ordered array of bounds:
params['per'].min = 1.0
params['per'].max = 3.0
and so on. But also: setting such tight bounds is usually a bad idea. Set bounds to avoid unphysical parameter values or when it becomes evident that you need to place them.
Now, you can fit your data with this model. Well, first you need to get the data you want to model. This seems less clear from your example, but perhaps:
c_data = numpy.linspace(min(time_phased[bb]), max(time_phased[bb]),
len(time_phased[bb]), endpoint=True)
lc_data = smoothed_LC[bb]
Well: why do you need to make this c_data? Why not just use time_phased as the independent variable? Anyway, now you can fit your data to your model with your parameters:
result = lc_model(lc_data, params, c=c_data)
At this point, you can print out a report of the results and/or view or get the best-fit arrays:
print(result.fit_report())
for p in result.params.items(): print(p)
import matplotlib.pyplot as plt
plt.plot(c_data, lc_data, label='data')
plt.plot(c_data. result.best_fit, label='fit')
plt.legend()
plt.show()
Hope that helps...
I am trying to fit my data points. It looks like the fitting without errors are not that optimistic, therefore now I am trying to fit the data implementing the errors at each point. My fit function is below:
def fit_func(x,a,b,c):
return np.log10(a*x**b + c)
then my data points are below:
r = [ 0.00528039,0.00721161,0.00873037,0.01108928,0.01413011,0.01790143,0.02263833, 0.02886089,0.03663713,0.04659512,0.05921978,0.07540126,0.09593949, 0.12190075,0.15501736,0.19713563,0.25041524,0.3185025,0.40514023,0.51507869, 0.65489938,0.83278859,1.05865016,1.34624082]
logf = [-1.1020581079659384, -1.3966927245616112, -1.4571368537041418, -1.5032694247562564, -1.8534775558300272, -2.2715812166948304, -2.2627690390113862, -2.5275290780299331, -3.3798813619309365, -6.0, -2.6270989211307034, -2.6549656159564918, -2.9366845162570079, -3.0955026428779604, -3.2649261507250289, -3.2837123017838366, -3.0493752067042856, -3.3133647996463229, -3.0865051494299243, -3.1347499415910169, -3.1433062918466632, -3.1747394718538979, -3.1797597345585245, -3.1913094832146616]
Because my data is in log scale, logf, then the error bar for each data point is not symmetric. The upper error bar and lower error bar are below:
upper = [0.070648916083227764, 0.44346256268274886, 0.11928131794776076, 0.094260899008089094, 0.14357124858039971, 0.27236750587684311, 0.18877122991380402, 0.28707938182603066, 0.72011863806906318, 0, 0.16813325716948757, 0.13624929595316049, 0.21847915642008875, 0.25456116079315372, 0.31078368240910148, 0.23178227464741452, 0.09158189214515966, 0.14020538489677881, 0.059482730164901909, 0.051786777740678414, 0.041126467609954531, 0.034394612910981337, 0.027206248503368613, 0.021847333685597548]
lower = [0.06074797748043137, 0.21479225959441428, 0.093479845697059583, 0.077406149968278104, 0.1077175009766278, 0.16610073183912188, 0.13114254113054535, 0.17133966123838595, 0.57498950902908286, 2.9786837094190934, 0.12090437578535695, 0.10355760401838676, 0.14467588244034646, 0.15942693835964539, 0.17929440903034921, 0.15031667827534712, 0.075592499975030591, 0.10581886912443572, 0.05230849287772843, 0.04626422871423852, 0.03756658820680725, 0.03186944137872727, 0.025601929615431285, 0.02080073540367966]
I have the fitting as:
popt, pcov = optimize.curve_fit(fit_func, r, logf,sigma=[lower,upper])
logf_fit = fit_func(r,*popt)
But this is wrong, how can I implement the curve fitting from scipy to include the upper and lower errors? How could I get the fitting errors of the fitting parameters a, b, c?
You can use scipy.optimize.leastsq with custom weights:
import scipy.optimize as optimize
import numpy as np
# redefine lists as array
x=np.array(r)
y=np.array(logf)
errup=np.array(upper)
errlow=np.array(lower)
# error function
def fit_func(x,a,b,c):
return np.log10(a*x**b + c)
def my_error(V):
a,b,c=V
yfit=fit_func(x,a,b,c)
weight=np.ones_like(yfit)
weight[yfit>y]=errup[yfit>y] # if the fit point is above the measure, use upper weight
weight[yfit<=y]=errlow[yfit<=y] # else use lower weight
return (yfit-y)**2/weight**2
answer=optimize.leastsq(my_error,x0=[0.0001,-1,0.0006])
a,b,c=answer[0]
print(a,b,c)
It works, but is very sensitive to initial values, since there is a log which can go in wrong domain (negative numbers) and then it fails. Here I find a=9.14464745425e-06 b=-1.75179880756 c=0.00066720486385which is pretty close to data.
In my model, I need to obtain the value of my deterministic variable from a set of parent variables using a complicated python function.
Is it possible to do that?
Following is a pyMC3 code which shows what I am trying to do in a simplified case.
import numpy as np
import pymc as pm
#Predefine values on two parameter Grid (x,w) for a set of i values (1,2,3)
idata = np.array([1,2,3])
size= 20
gridlength = size*size
Grid = np.empty((gridlength,2+len(idata)))
for x in range(size):
for w in range(size):
# A silly version of my real model evaluated on grid.
Grid[x*size+w,:]= np.array([x,w]+[(x**i + w**i) for i in idata])
# A function to find the nearest value in Grid and return its product with third variable z
def FindFromGrid(x,w,z):
return Grid[int(x)*size+int(w),2:] * z
#Generate fake Y data with error
yerror = np.random.normal(loc=0.0, scale=9.0, size=len(idata))
ydata = Grid[16*size+12,2:]*3.6 + yerror # ie. True x= 16, w= 12 and z= 3.6
with pm.Model() as model:
#Priors
x = pm.Uniform('x',lower=0,upper= size)
w = pm.Uniform('w',lower=0,upper =size)
z = pm.Uniform('z',lower=-5,upper =10)
#Expected value
y_hat = pm.Deterministic('y_hat',FindFromGrid(x,w,z))
#Data likelihood
ysigmas = np.ones(len(idata))*9.0
y_like = pm.Normal('y_like',mu= y_hat, sd=ysigmas, observed=ydata)
# Inference...
start = pm.find_MAP() # Find starting value by optimization
step = pm.NUTS(state=start) # Instantiate MCMC sampling algorithm
trace = pm.sample(1000, step, start=start, progressbar=False) # draw 1000 posterior samples using NUTS sampling
print('The trace plot')
fig = pm.traceplot(trace, lines={'x': 16, 'w': 12, 'z':3.6})
fig.show()
When I run this code, I get error at the y_hat stage, because the int() function inside the FindFromGrid(x,w,z) function needs integer not FreeRV.
Finding y_hat from a pre calculated grid is important because my real model for y_hat does not have an analytical form to express.
I have earlier tried to use OpenBUGS, but I found out here it is not possible to do this in OpenBUGS. Is it possible in PyMC ?
Update
Based on an example in pyMC github page, I found I need to add the following decorator to my FindFromGrid(x,w,z) function.
#pm.theano.compile.ops.as_op(itypes=[t.dscalar, t.dscalar, t.dscalar],otypes=[t.dvector])
This seems to solve the above mentioned issue. But I cannot use NUTS sampler anymore since it needs gradient.
Metropolis seems to be not converging.
Which step method should I use in a scenario like this?
You found the correct solution with as_op.
Regarding the convergence: Are you using pm.Metropolis() instead of pm.NUTS() by any chance? One reason this could not converge is that Metropolis() by default samples in the joint space while often Gibbs within Metropolis is more effective (and this was the default in pymc2). Having said that, I just merged this: https://github.com/pymc-devs/pymc/pull/587 which changes the default behavior of the Metropolis and Slice sampler to be non-blocked by default (so within Gibbs). Other samplers like NUTS that are primarily designed to sample the joint space still default to blocked. You can always explicitly set this with the kwarg blocked=True.
Anyway, update pymc with the most recent master and see if convergence improves. If not, try the Slice sampler.
I am using LaasoCV from sklearn to select the best model is selected by cross-validation. I found that the cross validation gives different result if I use sklearn or matlab statistical toolbox.
I used matlab and replicate the example given in
http://www.mathworks.se/help/stats/lasso-and-elastic-net.html
to get a figure like this
Then I saved the matlab data, and tried to replicate the figure with laaso_path from sklearn, I got
Although there are some similarity between these two figures, there are also certain differences. As far as I understand parameter lambda in matlab and alpha in sklearn are same, however in this figure it seems that there are some differences. Can somebody point out which is the correct one or am I missing something? Further the coefficient obtained are also different (which is my main concern).
Matlab Code:
rng(3,'twister') % for reproducibility
X = zeros(200,5);
for ii = 1:5
X(:,ii) = exprnd(ii,200,1);
end
r = [0;2;0;-3;0];
Y = X*r + randn(200,1)*.1;
save randomData.mat % To be used in python code
[b fitinfo] = lasso(X,Y,'cv',10);
lassoPlot(b,fitinfo,'plottype','lambda','xscale','log');
disp('Lambda with min MSE')
fitinfo.LambdaMinMSE
disp('Lambda with 1SE')
fitinfo.Lambda1SE
disp('Quality of Fit')
lambdaindex = fitinfo.Index1SE;
fitinfo.MSE(lambdaindex)
disp('Number of non zero predictos')
fitinfo.DF(lambdaindex)
disp('Coefficient of fit at that lambda')
b(:,lambdaindex)
Python Code:
import scipy.io
import numpy as np
import pylab as pl
from sklearn.linear_model import lasso_path, LassoCV
data=scipy.io.loadmat('randomData.mat')
X=data['X']
Y=data['Y'].flatten()
model = LassoCV(cv=10,max_iter=1000).fit(X, Y)
print 'alpha', model.alpha_
print 'coef', model.coef_
eps = 1e-2 # the smaller it is the longer is the path
models = lasso_path(X, Y, eps=eps)
alphas_lasso = np.array([model.alpha for model in models])
coefs_lasso = np.array([model.coef_ for model in models])
pl.figure(1)
ax = pl.gca()
ax.set_color_cycle(2 * ['b', 'r', 'g', 'c', 'k'])
l1 = pl.semilogx(alphas_lasso,coefs_lasso)
pl.gca().invert_xaxis()
pl.xlabel('alpha')
pl.show()
I do not have matlab but be careful that the value obtained with the cross--validation can be unstable. This is because it influenced by the way you subdivide the samples.
Even if you run 2 times the cross-validation in python you can obtain 2 different results.
consider this example :
kf=sklearn.cross_validation.KFold(len(y),n_folds=10,shuffle=True)
cv=sklearn.linear_model.LassoCV(cv=kf,normalize=True).fit(x,y)
print cv.alpha_
kf=sklearn.cross_validation.KFold(len(y),n_folds=10,shuffle=True)
cv=sklearn.linear_model.LassoCV(cv=kf,normalize=True).fit(x,y)
print cv.alpha_
0.00645093258722
0.00691712356467
it's possible that alpha = lambda / n_samples
where n_samples = X.shape[0] in scikit-learn
another remark is that your path is not very piecewise linear as it could/should be. Consider reducing the tol and increasing max_iter.
hope this helps
I know this is an old thread, but:
I'm actually working on piping over to LassoCV from glmnet (in R), and I found that LassoCV doesn't do too well with normalizing the X matrix first (even if you specify the parameter normalize = True).
Try normalizing the X matrix first when using LassoCV.
If it is a pandas object,
(X - X.mean())/X.std()
It seems you also need to multiple alpha by 2
Though I am unable to figure out what is causing the problem, there is a logical direction in which to continue.
These are the facts:
Mathworks have selected an example and decided to include it in their documentation
Your matlab code produces exactly the result as the example.
The alternative does not match the result, and has provided inaccurate results in the past
This is my assumption:
The chance that mathworks have chosen to put an incorrect example in their documentation is neglectable compared to the chance that a reproduction of this example in an alternate way does not give the correct result.
The logical conclusion: Your matlab implementation of this example is reliable and the other is not.
This might be a problem in the code, or maybe in how you use it, but either way the only logical conclusion would be that you should continue with Matlab to select your model.