I am trying to recreate maximum likelihood distribution fitting, I can already do this in Matlab and R, but now I want to use scipy. In particular, I would like to estimate the Weibull distribution parameters for my data set.
I have tried this:
import scipy.stats as s
import numpy as np
import matplotlib.pyplot as plt
def weib(x,n,a):
return (a / n) * (x / n)**(a - 1) * np.exp(-(x / n)**a)
data = np.loadtxt("stack_data.csv")
(loc, scale) = s.exponweib.fit_loc_scale(data, 1, 1)
print loc, scale
x = np.linspace(data.min(), data.max(), 1000)
plt.plot(x, weib(x, loc, scale))
plt.hist(data, data.max(), density=True)
plt.show()
And get this:
(2.5827280639441961, 3.4955032285727947)
And a distribution that looks like this:
I have been using the exponweib after reading this http://www.johndcook.com/distributions_scipy.html. I have also tried the other Weibull functions in scipy (just in case!).
In Matlab (using the Distribution Fitting Tool - see screenshot) and in R (using both the MASS library function fitdistr and the GAMLSS package) I get a (loc) and b (scale) parameters more like 1.58463497 5.93030013. I believe all three methods use the maximum likelihood method for distribution fitting.
I have posted my data here if you would like to have a go! And for completeness I am using Python 2.7.5, Scipy 0.12.0, R 2.15.2 and Matlab 2012b.
Why am I getting a different result!?
My guess is that you want to estimate the shape parameter and the scale of the Weibull distribution while keeping the location fixed. Fixing loc assumes that the values of your data and of the distribution are positive with lower bound at zero.
floc=0 keeps the location fixed at zero, f0=1 keeps the first shape parameter of the exponential weibull fixed at one.
>>> stats.exponweib.fit(data, floc=0, f0=1)
[1, 1.8553346917584836, 0, 6.8820748596850905]
>>> stats.weibull_min.fit(data, floc=0)
[1.8553346917584836, 0, 6.8820748596850549]
The fit compared to the histogram looks ok, but not very good. The parameter estimates are a bit higher than the ones you mention are from R and matlab.
Update
The closest I can get to the plot that is now available is with unrestricted fit, but using starting values. The plot is still less peaked. Note values in fit that don't have an f in front are used as starting values.
>>> from scipy import stats
>>> import matplotlib.pyplot as plt
>>> plt.plot(data, stats.exponweib.pdf(data, *stats.exponweib.fit(data, 1, 1, scale=02, loc=0)))
>>> _ = plt.hist(data, bins=np.linspace(0, 16, 33), normed=True, alpha=0.5);
>>> plt.show()
It is easy to verify which result is the true MLE, just need a simple function to calculate log likelihood:
>>> def wb2LL(p, x): #log-likelihood
return sum(log(stats.weibull_min.pdf(x, p[1], 0., p[0])))
>>> adata=loadtxt('/home/user/stack_data.csv')
>>> wb2LL(array([6.8820748596850905, 1.8553346917584836]), adata)
-8290.1227946678173
>>> wb2LL(array([5.93030013, 1.57463497]), adata)
-8410.3327470347667
The result from fit method of exponweib and R fitdistr (#Warren) is better and has higher log likelihood. It is more likely to be the true MLE. It is not surprising that the result from GAMLSS is different. It is a complete different statistic model: Generalized Additive Model.
Still not convinced? We can draw a 2D confidence limit plot around MLE, see Meeker and Escobar's book for detail).
Again this verifies that array([6.8820748596850905, 1.8553346917584836]) is the right answer as loglikelihood is lower that any other point in the parameter space. Note:
>>> log(array([6.8820748596850905, 1.8553346917584836]))
array([ 1.92892018, 0.61806511])
BTW1, MLE fit may not appears to fit the distribution histogram tightly. An easy way to think about MLE is that MLE is the parameter estimate most probable given the observed data. It doesn't need to visually fit the histogram well, that will be something minimizing mean square error.
BTW2, your data appears to be leptokurtic and left-skewed, which means Weibull distribution may not fit your data well. Try, e.g. Gompertz-Logistic, which improves log-likelihood by another about 100.
Cheers!
I know it's an old post, but I just faced a similar problem and this thread helped me solve it. Thought my solution might be helpful for others like me:
# Fit Weibull function, some explanation below
params = stats.exponweib.fit(data, floc=0, f0=1)
shape = params[1]
scale = params[3]
print 'shape:',shape
print 'scale:',scale
#### Plotting
# Histogram first
values,bins,hist = plt.hist(data,bins=51,range=(0,25),normed=True)
center = (bins[:-1] + bins[1:]) / 2.
# Using all params and the stats function
plt.plot(center,stats.exponweib.pdf(center,*params),lw=4,label='scipy')
# Using my own Weibull function as a check
def weibull(u,shape,scale):
'''Weibull distribution for wind speed u with shape parameter k and scale parameter A'''
return (shape / scale) * (u / scale)**(shape-1) * np.exp(-(u/scale)**shape)
plt.plot(center,weibull(center,shape,scale),label='Wind analysis',lw=2)
plt.legend()
Some extra info that helped me understand:
Scipy Weibull function can take four input parameters: (a,c),loc and scale.
You want to fix the loc and the first shape parameter (a), this is done with floc=0,f0=1. Fitting will then give you params c and scale, where c corresponds to the shape parameter of the two-parameter Weibull distribution (often used in wind data analysis) and scale corresponds to its scale factor.
From docs:
exponweib.pdf(x, a, c) =
a * c * (1-exp(-x**c))**(a-1) * exp(-x**c)*x**(c-1)
If a is 1, then
exponweib.pdf(x, a, c) =
c * (1-exp(-x**c))**(0) * exp(-x**c)*x**(c-1)
= c * (1) * exp(-x**c)*x**(c-1)
= c * x **(c-1) * exp(-x**c)
From this, the relation to the 'wind analysis' Weibull function should be more clear
I was curious about your question and, despite this is not an answer, it compares the Matlab result with your result and with the result using leastsq, which showed the best correlation with the given data:
The code is as follows:
import scipy.stats as s
import numpy as np
import matplotlib.pyplot as plt
import numpy.random as mtrand
from scipy.integrate import quad
from scipy.optimize import leastsq
## my distribution (Inverse Normal with shape parameter mu=1.0)
def weib(x,n,a):
return (a / n) * (x / n)**(a-1) * np.exp(-(x/n)**a)
def residuals(p,x,y):
integral = quad( weib, 0, 16, args=(p[0],p[1]) )[0]
penalization = abs(1.-integral)*100000
return y - weib(x, p[0],p[1]) + penalization
#
data = np.loadtxt("stack_data.csv")
x = np.linspace(data.min(), data.max(), 100)
n, bins, patches = plt.hist(data,bins=x, normed=True)
binsm = (bins[1:]+bins[:-1])/2
popt, pcov = leastsq(func=residuals, x0=(1.,1.), args=(binsm,n))
loc, scale = 1.58463497, 5.93030013
plt.plot(binsm,n)
plt.plot(x, weib(x, loc, scale),
label='weib matlab, loc=%1.3f, scale=%1.3f' % (loc, scale), lw=4.)
loc, scale = s.exponweib.fit_loc_scale(data, 1, 1)
plt.plot(x, weib(x, loc, scale),
label='weib stack, loc=%1.3f, scale=%1.3f' % (loc, scale), lw=4.)
plt.plot(x, weib(x,*popt),
label='weib leastsq, loc=%1.3f, scale=%1.3f' % tuple(popt), lw=4.)
plt.legend(loc='upper right')
plt.show()
I had the same problem, but found that setting loc=0 in exponweib.fit primed the pump for the optimization. That was all that was needed from #user333700's answer. I couldn't load your data -- your data link points to an image, not data. So I ran a test on my data instead:
import scipy.stats as ss
import matplotlib.pyplot as plt
import numpy as np
N=30
counts, bins = np.histogram(x, bins=N)
bin_width = bins[1]-bins[0]
total_count = float(sum(counts))
f, ax = plt.subplots(1, 1)
f.suptitle(query_uri)
ax.bar(bins[:-1]+bin_width/2., counts, align='center', width=.85*bin_width)
ax.grid('on')
def fit_pdf(x, name='lognorm', color='r'):
dist = getattr(ss, name) # params = shape, loc, scale
# dist = ss.gamma # 3 params
params = dist.fit(x, loc=0) # 1-day lag minimum for shipping
y = dist.pdf(bins, *params)*total_count*bin_width
sqerror_sum = np.log(sum(ci*(yi - ci)**2. for (ci, yi) in zip(counts, y)))
ax.plot(bins, y, color, lw=3, alpha=0.6, label='%s err=%3.2f' % (name, sqerror_sum))
return y
colors = ['r-', 'g-', 'r:', 'g:']
for name, color in zip(['exponweib', 't', 'gamma'], colors): # 'lognorm', 'erlang', 'chi2', 'weibull_min',
y = fit_pdf(x, name=name, color=color)
ax.legend(loc='best', frameon=False)
plt.show()
There have been a few answers to this already here and in other places. likt in Weibull distribution and the data in the same figure (with numpy and scipy)
It still took me a while to come up with a clean toy example so I though it would be useful to post.
from scipy import stats
import matplotlib.pyplot as plt
#input for pseudo data
N = 10000
Kappa_in = 1.8
Lambda_in = 10
a_in = 1
loc_in = 0
#Generate data from given input
data = stats.exponweib.rvs(a=a_in,c=Kappa_in, loc=loc_in, scale=Lambda_in, size = N)
#The a and loc are fixed in the fit since it is standard to assume they are known
a_out, Kappa_out, loc_out, Lambda_out = stats.exponweib.fit(data, f0=a_in,floc=loc_in)
#Plot
bins = range(51)
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.plot(bins, stats.exponweib.pdf(bins, a=a_out,c=Kappa_out,loc=loc_out,scale = Lambda_out))
ax.hist(data, bins = bins , density=True, alpha=0.5)
ax.annotate("Shape: $k = %.2f$ \n Scale: $\lambda = %.2f$"%(Kappa_out,Lambda_out), xy=(0.7, 0.85), xycoords=ax.transAxes)
plt.show()
In the meantime, there is really good package out there: reliability. Here is the documentation: reliability # readthedocs.
Your code simply becomes:
from reliability.Fitters import Fit_Weibull_2P
...
wb = Fit_Weibull_2P(failures=data)
plt.show()
Saves a lot of headaches and makes beautiful plots, too.
the order of loc and scale is messed up in the code:
plt.plot(x, weib(x, scale, loc))
the scale parameter should come first.
Related
From https://stackoverflow.com/a/30460089/2202107, we can generate CDF of a normal distribution:
import numpy as np
import matplotlib.pyplot as plt
N = 100
Z = np.random.normal(size = N)
# method 1
H,X1 = np.histogram( Z, bins = 10, normed = True )
dx = X1[1] - X1[0]
F1 = np.cumsum(H)*dx
#method 2
X2 = np.sort(Z)
F2 = np.array(range(N))/float(N)
# plt.plot(X1[1:], F1)
plt.plot(X2, F2)
plt.show()
Question: How do we generate the "original" normal distribution, given only x (eg X2) and y (eg F2) coordinates?
My first thought was plt.plot(x,np.gradient(y)), but gradient of y was all zero (data points are evenly spaced in y, but not in x) These kind of data is often met in percentile calculations. The key is to get the data evenly space in x and not in y, using interpolation:
x=X2
y=F2
num_points=10
xinterp = np.linspace(-2,2,num_points)
yinterp = np.interp(xinterp, x, y)
# for normalizing that sum of all bars equals to 1.0
tot_val=1.0
normalization_factor = tot_val/np.trapz(np.ones(len(xinterp)),yinterp)
plt.bar(xinterp, normalization_factor * np.gradient(yinterp), width=0.2)
plt.show()
output looks good to me:
I put my approach here for examination. Let me know if my logic is flawed.
One issue is: when num_points is large, the plot looks bad, but it's a issue in discretization, not sure how to avoid it.
Related posts:
I failed to understand why the answer was so complicated in https://stats.stackexchange.com/a/6065/131632
I also didn't understand why my approach was different than Generate distribution given percentile ranks
I am trying to find a python package that would give an option to fit natural smoothing splines with user selectable smoothing factor. Is there an implementation for that? If not, how would you use what is available to implement it yourself?
By natural spline I mean that there should be a condition that the second derivative of the fitted function at the endpoints is zero (linear).
By smoothing spline I mean that the spline should not be 'interpolating' (passing through all the datapoints). I would like to decide the correct smoothing factor lambda (see the Wikipedia page for smoothing splines) myself.
What I have found
scipy.interpolate.CubicSpline [link]: Does natural (cubic) spline fitting. Does interpolation, and there is no way to smooth the data.
scipy.interpolate.UnivariateSpline [link]: Does spline fitting with user selectable smoothing factor. However, there is no option to make the splines natural.
After hours of investigation, I did not find any pip installable packages which could fit a natural cubic spline with user-controllable smoothness. However, after deciding to write one myself, while reading about the topic I stumbled upon a blog post by github user madrury. He has written python code capable of producing natural cubic spline models.
The model code is available here (NaturalCubicSpline) with a BSD-licence. He has also written some examples in an IPython notebook.
But since this is the Internet and links tend to die, I will copy the relevant parts of the source code here + a helper function (get_natural_cubic_spline_model) written by me, and show an example of how to use it. The smoothness of the fit can be controlled by using different number of knots. The position of the knots can be also specified by the user.
Example
from matplotlib import pyplot as plt
import numpy as np
def func(x):
return 1/(1+25*x**2)
# make example data
x = np.linspace(-1,1,300)
y = func(x) + np.random.normal(0, 0.2, len(x))
# The number of knots can be used to control the amount of smoothness
model_6 = get_natural_cubic_spline_model(x, y, minval=min(x), maxval=max(x), n_knots=6)
model_15 = get_natural_cubic_spline_model(x, y, minval=min(x), maxval=max(x), n_knots=15)
y_est_6 = model_6.predict(x)
y_est_15 = model_15.predict(x)
plt.plot(x, y, ls='', marker='.', label='originals')
plt.plot(x, y_est_6, marker='.', label='n_knots = 6')
plt.plot(x, y_est_15, marker='.', label='n_knots = 15')
plt.legend(); plt.show()
The source code for get_natural_cubic_spline_model
import numpy as np
import pandas as pd
from sklearn.base import BaseEstimator, TransformerMixin
from sklearn.linear_model import LinearRegression
from sklearn.pipeline import Pipeline
def get_natural_cubic_spline_model(x, y, minval=None, maxval=None, n_knots=None, knots=None):
"""
Get a natural cubic spline model for the data.
For the knots, give (a) `knots` (as an array) or (b) minval, maxval and n_knots.
If the knots are not directly specified, the resulting knots are equally
space within the *interior* of (max, min). That is, the endpoints are
*not* included as knots.
Parameters
----------
x: np.array of float
The input data
y: np.array of float
The outpur data
minval: float
Minimum of interval containing the knots.
maxval: float
Maximum of the interval containing the knots.
n_knots: positive integer
The number of knots to create.
knots: array or list of floats
The knots.
Returns
--------
model: a model object
The returned model will have following method:
- predict(x):
x is a numpy array. This will return the predicted y-values.
"""
if knots:
spline = NaturalCubicSpline(knots=knots)
else:
spline = NaturalCubicSpline(max=maxval, min=minval, n_knots=n_knots)
p = Pipeline([
('nat_cubic', spline),
('regression', LinearRegression(fit_intercept=True))
])
p.fit(x, y)
return p
class AbstractSpline(BaseEstimator, TransformerMixin):
"""Base class for all spline basis expansions."""
def __init__(self, max=None, min=None, n_knots=None, n_params=None, knots=None):
if knots is None:
if not n_knots:
n_knots = self._compute_n_knots(n_params)
knots = np.linspace(min, max, num=(n_knots + 2))[1:-1]
max, min = np.max(knots), np.min(knots)
self.knots = np.asarray(knots)
#property
def n_knots(self):
return len(self.knots)
def fit(self, *args, **kwargs):
return self
class NaturalCubicSpline(AbstractSpline):
"""Apply a natural cubic basis expansion to an array.
The features created with this basis expansion can be used to fit a
piecewise cubic function under the constraint that the fitted curve is
linear *outside* the range of the knots.. The fitted curve is continuously
differentiable to the second order at all of the knots.
This transformer can be created in two ways:
- By specifying the maximum, minimum, and number of knots.
- By specifying the cutpoints directly.
If the knots are not directly specified, the resulting knots are equally
space within the *interior* of (max, min). That is, the endpoints are
*not* included as knots.
Parameters
----------
min: float
Minimum of interval containing the knots.
max: float
Maximum of the interval containing the knots.
n_knots: positive integer
The number of knots to create.
knots: array or list of floats
The knots.
"""
def _compute_n_knots(self, n_params):
return n_params
#property
def n_params(self):
return self.n_knots - 1
def transform(self, X, **transform_params):
X_spl = self._transform_array(X)
if isinstance(X, pd.Series):
col_names = self._make_names(X)
X_spl = pd.DataFrame(X_spl, columns=col_names, index=X.index)
return X_spl
def _make_names(self, X):
first_name = "{}_spline_linear".format(X.name)
rest_names = ["{}_spline_{}".format(X.name, idx)
for idx in range(self.n_knots - 2)]
return [first_name] + rest_names
def _transform_array(self, X, **transform_params):
X = X.squeeze()
try:
X_spl = np.zeros((X.shape[0], self.n_knots - 1))
except IndexError: # For arrays with only one element
X_spl = np.zeros((1, self.n_knots - 1))
X_spl[:, 0] = X.squeeze()
def d(knot_idx, x):
def ppart(t): return np.maximum(0, t)
def cube(t): return t*t*t
numerator = (cube(ppart(x - self.knots[knot_idx]))
- cube(ppart(x - self.knots[self.n_knots - 1])))
denominator = self.knots[self.n_knots - 1] - self.knots[knot_idx]
return numerator / denominator
for i in range(0, self.n_knots - 2):
X_spl[:, i+1] = (d(i, X) - d(self.n_knots - 2, X)).squeeze()
return X_spl
You could use this numpy/scipy implementation of natural cubic smoothing spline for univariate/multivariate data smoothing. Smoothing parameter should be in range [0.0, 1.0]. If we use smoothing parameter equal to 1.0 we get natural cubic spline interpolant without data smoothing. Also the implementation supports vectorization for univariate data.
Univariate example:
import numpy as np
import matplotlib.pyplot as plt
import csaps
np.random.seed(1234)
x = np.linspace(-5., 5., 25)
y = np.exp(-(x/2.5)**2) + (np.random.rand(25) - 0.2) * 0.3
sp = csaps.UnivariateCubicSmoothingSpline(x, y, smooth=0.85)
xs = np.linspace(x[0], x[-1], 150)
ys = sp(xs)
plt.plot(x, y, 'o', xs, ys, '-')
plt.show()
Bivariate example:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import csaps
xdata = [np.linspace(-3, 3, 61), np.linspace(-3.5, 3.5, 51)]
i, j = np.meshgrid(*xdata, indexing='ij')
ydata = (3 * (1 - j)**2. * np.exp(-(j**2) - (i + 1)**2)
- 10 * (j / 5 - j**3 - i**5) * np.exp(-j**2 - i**2)
- 1 / 3 * np.exp(-(j + 1)**2 - i**2))
np.random.seed(12345)
noisy = ydata + (np.random.randn(*ydata.shape) * 0.75)
sp = csaps.MultivariateCubicSmoothingSpline(xdata, noisy, smooth=0.988)
ysmth = sp(xdata)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(j, i, noisy, linewidths=0.5, color='r')
ax.scatter(j, i, noisy, s=5, c='r')
ax.plot_surface(j, i, ysmth, linewidth=0, alpha=1.0)
plt.show()
The python package patsy has functions for generating spline bases, including a natural cubic spline basis. Described in the documentation.
Any library can then be used for fitting a model, e.g. scikit-learn or statsmodels.
The df parameter for cr() can be used to control the "smoothness"
Note that too low df can result to underfit (see below).
A simple example using scikit-learn.
import numpy as np
from sklearn.linear_model import LinearRegression
from patsy import cr
import matplotlib.pyplot as plt
n_obs = 600
np.random.seed(0)
x = np.linspace(-3, 3, n_obs)
y = 1 / (x ** 2 + 1) * np.cos(np.pi * x) + np.random.normal(0, 0.2, size=n_obs)
def plot_smoothed(df=5):
# Generate spline basis with different degrees of freedom
x_basis = cr(x, df=df, constraints="center")
# Fit model to the data
model = LinearRegression().fit(x_basis, y)
# Get estimates
y_hat = model.predict(x_basis)
plt.plot(x, y_hat, label=f"df={df}")
plt.scatter(x, y, s=4, color="tab:blue")
for df in (5, 7, 10, 25):
plot_smoothed(df)
plt.legend()
plt.title(f"Natural cubic spline with varying degrees of freedom")
plt.show()
For a project of mine, I needed to create intervals for time-series modeling, and to make the procedure more efficient I created tsmoothie: A python library for time-series smoothing and outlier detection in a vectorized way.
It provides different smoothing algorithms together with the possibility to computes intervals.
In the case of SplineSmoother of natural cubic type:
import numpy as np
import matplotlib.pyplot as plt
from tsmoothie.smoother import *
def func(x):
return 1/(1+25*x**2)
# make example data
x = np.linspace(-1,1,300)
y = func(x) + np.random.normal(0, 0.2, len(x))
# operate smoothing
smoother = SplineSmoother(n_knots=10, spline_type='natural_cubic_spline')
smoother.smooth(y)
# generate intervals
low, up = smoother.get_intervals('prediction_interval', confidence=0.05)
# plot the first smoothed timeseries with intervals
plt.figure(figsize=(11,6))
plt.plot(smoother.smooth_data[0], linewidth=3, color='blue')
plt.plot(smoother.data[0], '.k')
plt.fill_between(range(len(smoother.data[0])), low[0], up[0], alpha=0.3)
I point out also that tsmoothie can carry out the smoothing of multiple time-series in a vectorized way
The programming language R offers a very good implementation of natural cubic smoothing splines. You can use R functions in Python with rpy2:
import rpy2.robjects as robjects
r_y = robjects.FloatVector(y_train)
r_x = robjects.FloatVector(x_train)
r_smooth_spline = robjects.r['smooth.spline'] #extract R function# run smoothing function
spline1 = r_smooth_spline(x=r_x, y=r_y, spar=0.7)
ySpline=np.array(robjects.r['predict'](spline1,robjects.FloatVector(x_smooth)).rx2('y'))
plt.plot(x_smooth,ySpline)
If you want to directly set lambda: spline1 = r_smooth_spline(x=r_x, y=r_y, lambda=42) doesn't work, because lambda has already another meaning in Python, but there is a solution: How to use the lambda argument of smooth.spline in RPy WITHOUT Python interprating it as lambda.
To get the code running you first need to define the data x_train and y_train and you can define x_smooth=np.array(np.linspace(-3,5,1920)). if you want to plot it between -3 and 5 in Full-HD-resolution.
Note that this code is not fully compatible with Jupyter-notebooks for the latest versions of rpy2. You can fix this by using !pip install -Iv rpy2==3.4.2 as described in NotImplementedError: Conversion 'rpy2py' not defined for objects of type '<class 'rpy2.rinterface.SexpClosure'>' only after I run the code twice
So, I'm trying to fit a set of data with a power law of the following kind:
def f(x,N,a): # Power law fit
if a >0:
return N*x**(-a)
else:
return 10.**300
par,cov = scipy.optimize.curve_fit(f,data,time,array([10**(-7),1.2]))
where the else condition is just to force a to be positive. Using scipy.optimize.curve_fit yields an awful fit (green line), returning values of 1.2e+04 and 1.9e0-7 for N and a, respectively, with absolutely no intersection with the data. From fits I've put in manually, the values should land around 1e-07 and 1.2 for N and a, respectively, though putting those into curve_fit as initial parameters doesn't change the result. Removing the condition for a to be positive results in a worse fit, as it chooses a negative, which leads to a fit with the wrong sign slope.
I can't figure out how to get a believable, let alone reliable, fit out of this routine, but I can't find any other good Python curve fitting routines. Do I need to write my own least-squares algorithm or is there something I'm doing wrong here?
UPDATE
In the original post, I showed a solution that uses lmfit which allows to assign bounds to your parameters. Starting with version 0.17, scipy also allows to assign bounds to your parameters directly (see documentation). Please find this solution below after the EDIT which can hopefully serve as a minimal example on how to use scipy's curve_fit with parameter bounds.
Original post
As suggested by #Warren Weckesser, you could use lmfit to get this task done, which allows you to assign bounds to your parameters and avoids this 'ugly' if-clause.
Since you do not provide any data, I created some which are shown here:
They follow the law f(x) = 10.5 * x ** (-0.08)
I fit them - as suggested by #roadrunner66 - by transforming the power law in a linear function:
y = N * x ** a
ln(y) = ln(N * x ** a)
ln(y) = a * ln(x) + ln(N)
So I first use np.log on the original data and then do the fit. When I now use lmfit, I get the following output:
[[Variables]]
lN: 2.35450302 +/- 0.019531 (0.83%) (init= 1.704748)
a: -0.08035342 +/- 0.005158 (6.42%) (init=-0.5)
So a is pretty close to the original value and np.exp(2.35450302) gives 10.53 which is also very close to the original value.
The plot then looks as follows; as you can see the fit describes the data very well:
Here is the entire code with a couple of inline comments:
import numpy as np
import matplotlib.pyplot as plt
from lmfit import minimize, Parameters, Parameter, report_fit
# generate some data with noise
xData = np.linspace(0.01, 100., 50.)
aOrg = 0.08
Norg = 10.5
yData = Norg * xData ** (-aOrg) + np.random.normal(0, 0.5, len(xData))
plt.plot(xData, yData, 'bo')
plt.show()
# transform data so that we can use a linear fit
lx = np.log(xData)
ly = np.log(yData)
plt.plot(lx, ly, 'bo')
plt.show()
def decay(params, x, data):
lN = params['lN'].value
a = params['a'].value
# our linear model
model = a * x + lN
return model - data # that's what you want to minimize
# create a set of Parameters
params = Parameters()
params.add('lN', value=np.log(5.5), min=0.01, max=100) # value is the initial value
params.add('a', value=-0.5, min=-1, max=-0.001) # min, max define parameter bounds
# do fit, here with leastsq model
result = minimize(decay, params, args=(lx, ly))
# write error report
report_fit(params)
# plot data
xnew = np.linspace(0., 100., 5000.)
# plot the data
plt.plot(xData, yData, 'bo')
plt.plot(xnew, np.exp(result.values['lN']) * xnew ** (result.values['a']), 'r')
plt.show()
EDIT
Assuming that you have scipy 0.17 installed, you can also do the following using curve_fit. I show it for your original definition of the power law (red line in the plot below) as well as for the logarithmic data (black line in the plot below). The data is generated in the same way as above. The plot the looks as follows:
As you can see, the data is described very well. If you print popt and popt_log, you obtain array([ 10.47463426, 0.07914812]) and array([ 2.35158653, -0.08045776]), respectively (note: for the letter one you will have to take the exponantial of the first argument - np.exp(popt_log[0]) = 10.502 which is close to the original data).
Here is the entire code:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
# generate some data with noise
xData = np.linspace(0.01, 100., 50)
aOrg = 0.08
Norg = 10.5
yData = Norg * xData ** (-aOrg) + np.random.normal(0, 0.5, len(xData))
# get logarithmic data
lx = np.log(xData)
ly = np.log(yData)
def f(x, N, a):
return N * x ** (-a)
def f_log(x, lN, a):
return a * x + lN
# optimize using the appropriate bounds
popt, pcov = curve_fit(f, xData, yData, bounds=(0, [30., 20.]))
popt_log, pcov_log = curve_fit(f_log, lx, ly, bounds=([0, -10], [30., 20.]))
xnew = np.linspace(0.01, 100., 5000)
# plot the data
plt.plot(xData, yData, 'bo')
plt.plot(xnew, f(xnew, *popt), 'r')
plt.plot(xnew, f(xnew, np.exp(popt_log[0]), -popt_log[1]), 'k')
plt.show()
I want to generate a Gaussian distribution in Python with the x and y dimensions denoting position and the z dimension denoting the magnitude of a certain quantity.
The distribution has a maximum value of 2e6 and a standard deviation sigma=0.025.
In MATLAB I can do this with:
x1 = linspace(-1,1,30);
x2 = linspace(-1,1,30);
mu = [0,0];
Sigma = [.025,.025];
[X1,X2] = meshgrid(x1,x2);
F = mvnpdf([X1(:) X2(:)],mu,Sigma);
F = 314159.153*reshape(F,length(x2),length(x1));
surf(x1,x2,F);
In Python, what I have so far is:
x = np.linspace(-1,1,30)
y = np.linspace(-1,1,30)
mu = (np.median(x),np.median(y))
sigma = (.025,.025)
There is a Numpy function numpy.random.multivariate_normal what can supposedly do the same as MATLAB's mvnpdf, but I am struggling to undestand the documentation. Especially in obtaining the covariance matrix needed by numpy.random.multivariate_normal.
As of scipy 0.14, you can use scipy.stats.multivariate_normal.pdf()
import numpy as np
from scipy.stats import multivariate_normal
x, y = np.mgrid[-1.0:1.0:30j, -1.0:1.0:30j]
# Need an (N, 2) array of (x, y) pairs.
xy = np.column_stack([x.flat, y.flat])
mu = np.array([0.0, 0.0])
sigma = np.array([.025, .025])
covariance = np.diag(sigma**2)
z = multivariate_normal.pdf(xy, mean=mu, cov=covariance)
# Reshape back to a (30, 30) grid.
z = z.reshape(x.shape)
I am working on a scikit called scikit-guess that contains some fast estimation routines for non-linear fits. It has a function skg.ngauss.model (also accessible as skg.ngauss_fit.model or skg.ngauss.ngauss_fit.model) which does exactly what you want. The nice thing is that it's not a PDF, so you set the amplitude out of the box:
import numpy as np
import skg.ngauss
a = 2e6
mu = 0, 0
sigma = 0.025, 0.025
x = y = np.linspace(-1, 1, 31)
cov = np.diag(sigma)**2
X = np.meshgrid(x, y)
data = skg.ngauss.model(X, a, mu, cov, axis=0)
You need to tell it axis=0 because it automatically stacks your arrays for you. To avoid passing in that argument, you could write
X = np.stack(np.meshgrid(x, y), axis=-1)
You can plot the result:
from matplotlib import pyplot as plt
plt.imshow(data)
plt.show()
This is not a very exciting distribution because the spread is so small that you end up with a value of ~2e-5 just one pixel away. You may want to up your sampling space to get any sort of meaningful resolution.
Note: At time of writing, the fitting function (ngauss_fit) is still buggy, but the model has been tested successfully, just not in the scikit.
Disclaimer: In case it wasn't obvious from the above, I am the author of scikit-guess.
I am trying to show that economies follow a relatively sinusoidal growth pattern. I am building a python simulation to show that even when we let some degree of randomness take hold, we can still produce something relatively sinusoidal.
I am happy with the data I'm producing, but now I'd like to find some way to get a sine graph that pretty closely matches the data. I know you can do polynomial fit, but can you do sine fit?
Here is a parameter-free fitting function fit_sin() that does not require manual guess of frequency:
import numpy, scipy.optimize
def fit_sin(tt, yy):
'''Fit sin to the input time sequence, and return fitting parameters "amp", "omega", "phase", "offset", "freq", "period" and "fitfunc"'''
tt = numpy.array(tt)
yy = numpy.array(yy)
ff = numpy.fft.fftfreq(len(tt), (tt[1]-tt[0])) # assume uniform spacing
Fyy = abs(numpy.fft.fft(yy))
guess_freq = abs(ff[numpy.argmax(Fyy[1:])+1]) # excluding the zero frequency "peak", which is related to offset
guess_amp = numpy.std(yy) * 2.**0.5
guess_offset = numpy.mean(yy)
guess = numpy.array([guess_amp, 2.*numpy.pi*guess_freq, 0., guess_offset])
def sinfunc(t, A, w, p, c): return A * numpy.sin(w*t + p) + c
popt, pcov = scipy.optimize.curve_fit(sinfunc, tt, yy, p0=guess)
A, w, p, c = popt
f = w/(2.*numpy.pi)
fitfunc = lambda t: A * numpy.sin(w*t + p) + c
return {"amp": A, "omega": w, "phase": p, "offset": c, "freq": f, "period": 1./f, "fitfunc": fitfunc, "maxcov": numpy.max(pcov), "rawres": (guess,popt,pcov)}
The initial frequency guess is given by the peak frequency in the frequency domain using FFT. The fitting result is almost perfect assuming there is only one dominant frequency (other than the zero frequency peak).
import pylab as plt
N, amp, omega, phase, offset, noise = 500, 1., 2., .5, 4., 3
#N, amp, omega, phase, offset, noise = 50, 1., .4, .5, 4., .2
#N, amp, omega, phase, offset, noise = 200, 1., 20, .5, 4., 1
tt = numpy.linspace(0, 10, N)
tt2 = numpy.linspace(0, 10, 10*N)
yy = amp*numpy.sin(omega*tt + phase) + offset
yynoise = yy + noise*(numpy.random.random(len(tt))-0.5)
res = fit_sin(tt, yynoise)
print( "Amplitude=%(amp)s, Angular freq.=%(omega)s, phase=%(phase)s, offset=%(offset)s, Max. Cov.=%(maxcov)s" % res )
plt.plot(tt, yy, "-k", label="y", linewidth=2)
plt.plot(tt, yynoise, "ok", label="y with noise")
plt.plot(tt2, res["fitfunc"](tt2), "r-", label="y fit curve", linewidth=2)
plt.legend(loc="best")
plt.show()
The result is good even with high noise:
Amplitude=1.00660540618, Angular freq.=2.03370472482, phase=0.360276844224, offset=3.95747467506, Max. Cov.=0.0122923578658
You can use the least-square optimization function in scipy to fit any arbitrary function to another. In case of fitting a sin function, the 3 parameters to fit are the offset ('a'), amplitude ('b') and the phase ('c').
As long as you provide a reasonable first guess of the parameters, the optimization should converge well.Fortunately for a sine function, first estimates of 2 of these are easy: the offset can be estimated by taking the mean of the data and the amplitude via the RMS (3*standard deviation/sqrt(2)).
Note: as a later edit, frequency fitting has also been added. This does not work very well (can lead to extremely poor fits). Thus, use at your discretion, my advise would be to not use frequency fitting unless frequency error is smaller than a few percent.
This leads to the following code:
import numpy as np
from scipy.optimize import leastsq
import pylab as plt
N = 1000 # number of data points
t = np.linspace(0, 4*np.pi, N)
f = 1.15247 # Optional!! Advised not to use
data = 3.0*np.sin(f*t+0.001) + 0.5 + np.random.randn(N) # create artificial data with noise
guess_mean = np.mean(data)
guess_std = 3*np.std(data)/(2**0.5)/(2**0.5)
guess_phase = 0
guess_freq = 1
guess_amp = 1
# we'll use this to plot our first estimate. This might already be good enough for you
data_first_guess = guess_std*np.sin(t+guess_phase) + guess_mean
# Define the function to optimize, in this case, we want to minimize the difference
# between the actual data and our "guessed" parameters
optimize_func = lambda x: x[0]*np.sin(x[1]*t+x[2]) + x[3] - data
est_amp, est_freq, est_phase, est_mean = leastsq(optimize_func, [guess_amp, guess_freq, guess_phase, guess_mean])[0]
# recreate the fitted curve using the optimized parameters
data_fit = est_amp*np.sin(est_freq*t+est_phase) + est_mean
# recreate the fitted curve using the optimized parameters
fine_t = np.arange(0,max(t),0.1)
data_fit=est_amp*np.sin(est_freq*fine_t+est_phase)+est_mean
plt.plot(t, data, '.')
plt.plot(t, data_first_guess, label='first guess')
plt.plot(fine_t, data_fit, label='after fitting')
plt.legend()
plt.show()
Edit: I assumed that you know the number of periods in the sine-wave. If you don't, it's somewhat trickier to fit. You can try and guess the number of periods by manual plotting and try and optimize it as your 6th parameter.
More userfriendly to us is the function curvefit. Here an example:
import numpy as np
from scipy.optimize import curve_fit
import pylab as plt
N = 1000 # number of data points
t = np.linspace(0, 4*np.pi, N)
data = 3.0*np.sin(t+0.001) + 0.5 + np.random.randn(N) # create artificial data with noise
guess_freq = 1
guess_amplitude = 3*np.std(data)/(2**0.5)
guess_phase = 0
guess_offset = np.mean(data)
p0=[guess_freq, guess_amplitude,
guess_phase, guess_offset]
# create the function we want to fit
def my_sin(x, freq, amplitude, phase, offset):
return np.sin(x * freq + phase) * amplitude + offset
# now do the fit
fit = curve_fit(my_sin, t, data, p0=p0)
# we'll use this to plot our first estimate. This might already be good enough for you
data_first_guess = my_sin(t, *p0)
# recreate the fitted curve using the optimized parameters
data_fit = my_sin(t, *fit[0])
plt.plot(data, '.')
plt.plot(data_fit, label='after fitting')
plt.plot(data_first_guess, label='first guess')
plt.legend()
plt.show()
The current methods to fit a sin curve to a given data set require a first guess of the parameters, followed by an interative process. This is a non-linear regression problem.
A different method consists in transforming the non-linear regression to a linear regression thanks to a convenient integral equation. Then, there is no need for initial guess and no need for iterative process : the fitting is directly obtained.
In case of the function y = a + r*sin(w*x+phi) or y=a+b*sin(w*x)+c*cos(w*x), see pages 35-36 of the paper "Régression sinusoidale" published on Scribd
In case of the function y = a + p*x + r*sin(w*x+phi) : pages 49-51 of the chapter "Mixed linear and sinusoidal regressions".
In case of more complicated functions, the general process is explained in the chapter "Generalized sinusoidal regression" pages 54-61, followed by a numerical example y = r*sin(w*x+phi)+(b/x)+c*ln(x), pages 62-63
All the above answers are based on curve fitting, and most use an iterative method - they all work very nicely, but I wanted to add a different approach using an FFT. Here, we transform the data, set all but the peak frequency to zero and then do the inverse transform. Note, that you probably want to remove the data mean (and detrend) before doing the FFT and then you can add those back in after.
import numpy as np
import pylab as plt
# fake data
N = 1000 # number of data points
t = np.linspace(0, 4*np.pi, N)
f = 1.05
data = 3.0*np.sin(f*t+0.001) + np.random.randn(N) # create artificial data with noise
# FFT...
mfft=np.fft.fft(data)
imax=np.argmax(np.absolute(mfft))
mask=np.zeros_like(mfft)
mask[[imax]]=1
mfft*=mask
fdata=np.fft.ifft(mfft)
plt.plot(t, data, '.')
plt.plot(t, fdata,'.', label='FFT')
plt.legend()
plt.show()