I have started to use python for analysis. I would like to do the following:
Get the distribution of dataset
Get the peaks in this distribution
I used gaussian_kde from scipy.stats to make estimation for kernel density function. Does guassian_kde make any assumption about the data ?. I am using data that are changed over time. so if data has one distribution (e.g. Gaussian), it could have another distribution later. Does gaussian_kde have any drawbacks in this scenario?. It was suggested in question to try to fit the data in every distribution in order to get the data distribution. So what's the difference between using gaussian_kde and the answer provided in question. I used the code below, I was wondering also to know is gaussian_kde good way to estimate pdf if the data will be changed over time ?. I know one advantage of gaussian_kde is that it calculate bandwidth automatically by a rule of thumb as in here. Also, how can I get its peaks?
import pandas as pd
import numpy as np
import pylab as pl
import scipy.stats
df = pd.read_csv('D:\dataset.csv')
pdf = scipy.stats.kde.gaussian_kde(df)
x = np.linspace((df.min()-1),(df.max()+1), len(df))
y = pdf(x)
pl.plot(x, y, color = 'r')
pl.hist(data_column, normed= True)
pl.show(block=True)
I think you need to distinguish non-parametric density (the one implemented in scipy.stats.kde) from parametric density (the one in the StackOverflow question you mention). To illustrate the difference between these two, try the following code.
import pandas as pd
import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt
np.random.seed(0)
gaussian1 = -6 + 3 * np.random.randn(1700)
gaussian2 = 4 + 1.5 * np.random.randn(300)
gaussian_mixture = np.hstack([gaussian1, gaussian2])
df = pd.DataFrame(gaussian_mixture, columns=['data'])
# non-parametric pdf
nparam_density = stats.kde.gaussian_kde(df.values.ravel())
x = np.linspace(-20, 10, 200)
nparam_density = nparam_density(x)
# parametric fit: assume normal distribution
loc_param, scale_param = stats.norm.fit(df)
param_density = stats.norm.pdf(x, loc=loc_param, scale=scale_param)
fig, ax = plt.subplots(figsize=(10, 6))
ax.hist(df.values, bins=30, normed=True)
ax.plot(x, nparam_density, 'r-', label='non-parametric density (smoothed by Gaussian kernel)')
ax.plot(x, param_density, 'k--', label='parametric density')
ax.set_ylim([0, 0.15])
ax.legend(loc='best')
From the graph, we see that the non-parametric density is nothing but a smoothed version of histogram. In histogram, for a particular observation x=x0, we use a bar to represent it (put all probability mass on that single point x=x0 and zero elsewhere) whereas in non-parametric density estimation, we use a bell-shaped curve (the gaussian kernel) to represent that point (spreads over its neighbourhood). And the result is a smoothed density curve. This internal gaussian kernel has nothing to do with your distributional assumption on the underlying data x. Its sole purpose is for smoothing.
To get the mode of non-parametric density, we need to do an exhaustive search, as the density is not guaranteed to have uni-mode. As shown in the example above, if you quasi-Newton optimization algo starts between [5,10], it is very likely to end up with a local optimal point rather than the global one.
# get mode: exhastive search
x[np.argsort(nparam_density)[-1]]
Related
I visualize density function (PDF) using two plotting approaches: displot() and plot(). I don't understand why displot() doesn't produce normally distributed plot wheras plot() do this perfectly. Density plots should look alike but they don't. What's wrong with displot() here?
from scipy.stats import norm
import seaborn as sns
import numpy as np
data_x= np.arange(-4, 4, 0.001)
norm_pdf = norm.pdf(data_x)
sns.displot(data = norm_pdf, x = data_x, kind='kde')
from scipy.stats import norm
import matplotlib.pyplot as plt
import numpy as np
data_x= np.arange(-4, 4, 0.001)
plt.plot(data_x, norm.pdf(data_x))
plt.show()
displot (or the underlying kdeplot) creates an approximation of a probability density function (pdf) to resemble the function that might have generated the given random data. As input, you'll need random data. The function will mimic these data as a sum of Gaussian bell shapes (a "kernel density estimation" with a Gaussian kernel).
Here is an example using 8000 random points as input. You'll notice the curve resembles the normal pdf, but is also a bit "bumpier" (that's how randomness looks like).
data_x = norm.rvs(size=8000)
sns.kdeplot(x=data_x)
When you call kdeplot (or displot(..., kind='kde')) with both data= and x=, while x= isn't a columnname in a dataframe, data= gets ignored. So, you are using 8000 evenly distributed values between -4 and 4. The kde of such data looks like a flat line between -4 and 4. But as the kde supposes the underlying function locally resembles a Gaussian, the start and end are smoothed out.
data_x = np.arange(-4, 4, 0.001)
sns.kdeplot(x=data_x)
plt.hist's density argument does not work.
I tried to use the density argument in the plt.hist function to normalize stock returns in my plot, but it didn't work.
The following code worked fine for me and give me the probability density function which I desired.
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(19680801)
# example data
mu = 100 # mean of distribution
sigma = 15 # standard deviation of distribution
x = mu + sigma * np.random.randn(437)
num_bins = 50
plt.hist(x, num_bins, density=1)
plt.show()
But when I tried it with stock data, it simply didn't work. The result gave the unnormalized data. I didn't find any abnormal data in my data array.
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
plt.hist(returns, 50,density = True)
plt.show()
# "returns" is a np array consisting of 360 days of stock returns
This is a known issue in Matplotlib.
As stated in Bug Report: The density flag in pyplot.hist() does not work correctly
When density = False, the histogram plot would have counts on the Y-axis. But when density = True, the Y-axis does not mean anything useful. I think a better implementation would plot the PDF as the histogram when density = True.
The developers view this as a feature not a bug since it maintains compatibility with numpy. They have closed several the bug reports about it already with since it is working as intended. Creating even more confusion the example on the matplotlib site appears to show this feature working with the y-axis being assigned a meaningful value.
What you want to do with matplotlib is reasonable but matplotlib will not let you do it that way.
It is not a bug.
Area of the bars equal to 1.
Numbers only seem strange because your bin sizes are small
Since this isn't resolved; based on #user14518925's response which is actually correct, this is treating bin width as an actual valid number whereas from my understanding you want each bin to have a width of 1 such that the sum of frequencies is 1. More succinctly, what you're seeing is:
\sum_{i}y_{i}\times\text{bin size} =1
Whereas what you want is:
\sum_{i}y_{i} =1
therefore, all you really need to change is the tick labels on the y-axis. One way to this is to disable the density option :
density = false
and instead divide by the total sample size as such (shown in your example):
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(19680801)
# example data
mu = 0 # mean of distribution
sigma = 0.0000625 # standard deviation of distribution
x = mu + sigma * np.random.randn(437)
fig = plt.figure()
plt.hist(x, 50, density=False)
locs, _ = plt.yticks()
print(locs)
plt.yticks(locs,np.round(locs/len(x),3))
plt.show()
Another approach, besides that of tvbc, is to change the yticks on the plot.
import matplotlib.pyplot as plt
import numpy as np
steps = 10
bins = np.arange(0, 101, steps)
data = np.random.random(100000) * 100
plt.hist(data, bins=bins, density=True)
yticks = plt.gca().get_yticks()
plt.yticks(yticks, np.round(yticks * steps, 2))
plt.show()
The gaussian_kde function in scipy.stats has a function evaluate that can returns the value of the PDF of an input point. I'm trying to use gaussian_kde to estimate the inverse CDF. The motivation is for generating Monte Carlo realizations of some input data whose statistical distribution is numerically estimated using KDE. Is there a method bound to gaussian_kde that serves this purpose?
The example below shows how this should work for the case of a Gaussian distribution. First I show how to do the PDF calculation to set up the specific API I'm trying to achieve:
import numpy as np
from scipy.stats import norm, gaussian_kde
npts_kde = int(5e3)
n = np.random.normal(loc=0, scale=1, size=npts_kde)
kde = gaussian_kde(n)
npts_sample = int(1e3)
x = np.linspace(-3, 3, npts_sample)
kde_pdf = kde.evaluate(x)
norm_pdf = norm.pdf(x)
Is there an analogously simple way to compute the inverse CDF? The norm function has a very handy isf function that does exactly this:
cdf_value = np.sort(np.random.rand(npts_sample))
cdf_inv = norm.isf(1 - cdf_value)
Does such a function exist for kde_gaussian? Or is it straightforward to construct such a function from the already implemented methods?
The method integrate_box_1d can be used to compute the CDF, but it is not vectorized; you'll need to loop over points. If memory is not an issue, rewriting its source code (which is essentially just a call to special.ndtr) in vector form may speed things up.
from scipy.special import ndtr
stdev = np.sqrt(kde.covariance)[0, 0]
pde_cdf = ndtr(np.subtract.outer(x, n)).mean(axis=1)
plot(x, pde_cdf)
The plot of the inverse function would be plot(pde_cdf, x). If the goal is to compute the inverse function at a specific point, consider using the inverse of interpolating spline, interpolating the computed values of the CDF.
You can use some python tricks for fast and memory-effective estimation of the CDF (based on this answer):
from scipy.special import ndtr
cdf = tuple(ndtr(np.ravel(item - kde.dataset) / kde.factor).mean()
for item in x)
It works as fast as this answer, but has linear (len(kde.dataset)) space complexity instead of the quadratic (actually, len(kde.dataset) * len(x)) one.
All you have to do next is to use inverse approximation, for instance, from statsmodels.
The question has been answered in the other answers but it took me a while to wrap my mind around everything. Here is a complete example of the final solution:
import numpy as np
from scipy import interpolate
from scipy.special import ndtr
import matplotlib.pyplot as plt
from scipy.stats import norm, gaussian_kde
# create kde
npts_kde = int(5e3)
n = np.random.normal(loc=0, scale=1, size=npts_kde)
kde = gaussian_kde(n)
# grid for plotting
npts_sample = int(1e3)
x = np.linspace(-3, 3, npts_sample)
# evaluate pdfs
kde_pdf = kde.evaluate(x)
norm_pdf = norm.pdf(x)
# cdf and inv cdf are available directly from scipy
norm_cdf = norm.cdf(x)
norm_inv = norm.ppf(x)
# estimate cdf
cdf = tuple(ndtr(np.ravel(item - kde.dataset) / kde.factor).mean()
for item in x)
# estimate inv cdf
inversefunction = interpolate.interp1d(cdf, x, kind='cubic', bounds_error=False)
fig, ax = plt.subplots(1, 3, figsize=(6, 3))
ax[0].plot(x, norm_pdf, c='k')
ax[0].plot(x, kde_pdf, c='r', ls='--')
ax[0].set_title('PDF')
ax[1].plot(x, norm_cdf, c='k')
ax[1].plot(x, cdf, c='r', ls='--')
ax[1].set_title('CDF')
ax[2].plot(x, norm_inv, c='k')
ax[2].plot(x, inversefunction(x), c='r', ls='--')
ax[2].set_title("Inverse CDF")
How would I calculate the confidence intervals for a LOWESS regression in Python? I would like to add these as a shaded region to the LOESS plot created with the following code (other packages than statsmodels are fine as well).
import numpy as np
import pylab as plt
import statsmodels.api as sm
x = np.linspace(0,2*np.pi,100)
y = np.sin(x) + np.random.random(100) * 0.2
lowess = sm.nonparametric.lowess(y, x, frac=0.1)
plt.plot(x, y, '+')
plt.plot(lowess[:, 0], lowess[:, 1])
plt.show()
I've added an example plot with confidence interval below from the webblog Serious Stats (it is created using ggplot in R).
LOESS doesn't have an explicit concept for standard error. It just doesn't mean anything in this context. Since that's out, your stuck with the brute-force approach.
Bootstrap your data. Your going to fit a LOESS curve to the bootstrapped data. See the middle of this page to find a pretty picture of what your doing. http://statweb.stanford.edu/~susan/courses/s208/node20.html
Once you have your large number of different LOESS curves, you can find the top and bottom Xth percentile.
This is a very old question but it's one of the first that pops up on google search. You can do this using the loess() function from scikit-misc. Here's an example (I tried to keep your original variable names, but I bumped up the noise a bit to make it more visible)
import numpy as np
import pylab as plt
from skmisc.loess import loess
x = np.linspace(0,2*np.pi,100)
y = np.sin(x) + np.random.random(100) * 0.4
l = loess(x,y)
l.fit()
pred = l.predict(x, stderror=True)
conf = pred.confidence()
lowess = pred.values
ll = conf.lower
ul = conf.upper
plt.plot(x, y, '+')
plt.plot(x, lowess)
plt.fill_between(x,ll,ul,alpha=.33)
plt.show()
result:
For a project of mine, I need 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 LowessSmoother:
import numpy as np
import matplotlib.pyplot as plt
from tsmoothie.smoother import *
from tsmoothie.utils_func import sim_randomwalk
# generate 10 randomwalks of length 200
np.random.seed(33)
data = sim_randomwalk(n_series=10, timesteps=200,
process_noise=10, measure_noise=30)
# operate smoothing
smoother = LowessSmoother(smooth_fraction=0.1, iterations=1)
smoother.smooth(data)
# 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. Hope this can help someone
How can I make a figure like the following one but with flat curve using matlibplot in Python?
Instead of using a histogram to bin your data have a look at using a KDE for a continuous estimate of the probability distribution. There is an implementation using a gaussian kernel in scipy.stats.gaussian_kde.
As an example:
import numpy as np
from scipy.stats import gaussian_kde
import matplotlib.pyplot as plt
data = np.random.normal(0.0, 1.0, 10000) #Generate some data
kde = gaussian_kde(data)
xplot = np.linspace(-5,5,1000)
plt.plot( xplot, kde(xplot), label='KDE' )
plt.hist( data, bins=50, histtype='step', normed=True, label='histogram' )
plt.legend()
plt.show()
Will produce the plot:
Note that when using KDEs the bandwidth of the kernel that you choose can have a very big impact on the representation of the data that gets produced, this is similar to the effect that the bin size would have when making a histogram. Both the scipy documentation that I linked to and the wikipedia page have good writeups on how to make this selection in a well motivated way.