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Histogram function - Python

Looking for someone who can explain this to me:
phase = mod(phase,Nper*2*pi)
cl_phase = arange(0,Nper*2*pi+step,step)
c,p = histogram(phase,cl_phase)
while 0 in c:
step = step*2
cl_phase = arange(0,Nper*2*pi+step,step)
c,p = histogram(phase,cl_phase)
Where phase is the phase of a wave, Nper is the number of periods I'm analysing.
What I want to know is if some one can give me the name/link to an explanation of the histogram function..! Im not even sure from what package it comes from. Maybe numpy? Or maybe it even is a function that comes with python..! Super lost here..!
Any help here would be greatly appreciated!!

histogram() function is from numpy library. It doesn't come as a default function in Python.
You can use it by:
import numpy as np
np.histogram(phase,cl_phase)
In your code, it looks like you are using it as:
from numpy import histogram
histogram(phase,cl_phase)
c,p = histogram(phase, cl_phase) will give you two values as output. c will be the values of the histogram, and p will return the bin edges. You should take a look at the above docs for more info.

Fitting parameters of a data set from a model

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...

lmfit for exponential data returns linear function

I'm working on fitting muon lifetime data to a curve to extract the mean lifetime using the lmfit function. The general process I'm using is to bin the 13,000 data points into 10 bins using the histogram function, calculating the uncertainty with the square root of the counts in each bin (it's an exponential model), then use the lmfit module to determine the best fit along with means and uncertainty. However, graphing the output of the model.fit() method returns this graph, where the red line is the fit (and obviously not the correct fit). Fit result output graph
I've looked online and can't find a solution to this, I'd really appreciate some help figuring out what's going on. Here's the code.
import os
import numpy as np
import matplotlib.pyplot as plt
from numpy import sqrt, pi, exp, linspace
from lmfit import Model
class data():
def __init__(self,file_name):
times_dirty = sorted(np.genfromtxt(file_name, delimiter=' ',unpack=False)[:,0])
self.times = []
for i in range(len(times_dirty)):
if times_dirty[i]<40000:
self.times.append(times_dirty[i])
self.counts = []
self.binBounds = []
self.uncertainties = []
self.means = []
def binData(self,k):
self.counts, self.binBounds = np.histogram(self.times, bins=k)
self.binBounds = self.binBounds[:-1]
def calcStats(self):
if len(self.counts)==0:
print('Run binData function first')
else:
self.uncertainties = sqrt(self.counts)
def plotData(self,fit):
plt.errorbar(self.binBounds, self.counts, yerr=self.uncertainties, fmt='bo')
plt.plot(self.binBounds, fit.init_fit, 'k--')
plt.plot(self.binBounds, fit.best_fit, 'r')
plt.show()
def decay(t, N, lamb, B):
return N * lamb * exp(-lamb * t) +B
def main():
muonEvents = data('C:\Users\Colt\Downloads\muon.data')
muonEvents.binData(10)
muonEvents.calcStats()
mod = Model(decay)
result = mod.fit(muonEvents.counts, t=muonEvents.binBounds, N=1, lamb=1, B = 1)
muonEvents.plotData(result)
print(result.fit_report())
print (len(muonEvents.times))
if __name__ == "__main__":
main()

This might be a simple scaling problem. As a quick test, try dividing all raw data by a factor of 1000 (both X and Y) to see if changing the magnitude of the data has any effect.

Just to build on James Phillips answer, I think the data you show in your graph imply values for N, lamb, and B that are very different from 1, 1, 1. Keep in mind that exp(-lamb*t) is essentially 0 for lamb = 1, and t> 100. So, if the algorithm starts at lamb=1 and varies that by a little bit to find a better value, it won't actually be able to see any difference in how well the model matches the data.
I would suggest trying to start with values that are more reasonable for the data you have, perhaps N=1.e6, lamb=1.e-4, and B=100.
As James suggested, having the variables have values on the order of 1 and putting in scale factors as necessary is often helpful in getting numerically stable solutions.

How to define General deterministic function in PyMC

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.

Why are LASSO in sklearn (python) and matlab statistical package different?

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.