I am new to Logistic regression. The following is from package mypc example
project. It is still unclear to me for its purpose. More specifically,
variable n is [5, 5, 5, 5], which is used in the mode: pymc.Binomial.
I suppose it should have both 0 and 1 in binomial fitting. n is to represent
'1' cases?
Could you explain the idea of this example? Thanks,
The example is from:
www.map.ox.ac.uk/media/PDF/Patil_et_al_2010.pdf
.........
import pymc
import numpy as np
n = 5*np.ones(4,dtype=int)
x = np.array([-.86,-.3,-.05,.73])
alpha = pymc.Normal('alpha',mu=0,tau=.01)
beta = pymc.Normal('beta',mu=0,tau=.01)
#pymc.deterministic
def theta(a=alpha, b=beta):
"""theta = logit^{-1}(a+b)"""
return pymc.invlogit(a+b*x)
d = pymc.Binomial('d', n=n, p=theta, value=np.array([0.,1.,3.,5.]),\
observed=True)
.......
import pymc
import pymc.Matplot
import mymodel
S = pymc.MCMC(mymodel, db='pickle')
S.sample(iter=10000, burn=5000, thin=2)
pymc.Matplot.plot(S)
import matplotlib.pyplot as plt
plt.show()
Related
I have the following Sympy code that works as expected:
import numpy as np
from sympy.utilities.lambdify import lambdify
from sympy.core import sympify
from sympy import factorial
ex = sympify('-x**2 / cos(x)')
flam = lambdify(['x'], ex, "numpy")
flam(np.array(range(5)))
This returns:
array([ 0. , -1.85081572, 9.61199185, 9.09097799, 24.4781705 ])
Now, what I need to know is how to do the same for factorials, that is, using factorial(x) instead of cos(x). The code:
ex = sympify('-x**2 / factorial(x)')
flam = lambdify(['x'], ex, "numpy")
flam(np.array(range(5)))
raises a NameError
NameError: global name 'factorial' is not defined
What string should I use so that it gets converted to a factorial that can be evaluated after lambdify?
Thanks in advance for any help!
By tinkering I obtain the following code. It seems that the numpy factorial function do not works with ndarrays...
import numpy as np
from sympy.utilities.lambdify import lambdify
flam = lambdify(['x'], '-x**2 / cos(x)', "numpy")
flam(np.array(range(5)))
# >>> array([ 0. , -1.85081572, 9.61199185, 9.09097799, 24.4781705 ])
import scipy.special
flam = lambdify('x', 'factorial(x)', ['numpy', {'factorial':scipy.special.factorial}])
flam(np.array(range(5)))
# >>> array([ 1., 1., 2., 6., 24.])
I'm trying to implement a simple Bayesian Inference using a ODE model. I want to use the NUTS algorithm to sample but it gives me an initialization error. I do not know much about the PyMC3 as I'm new to this. Please take a look and tell me what is wrong.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
import seaborn
import pymc3 as pm
import theano.tensor as T
from theano.compile.ops import as_op
#Actual Solution of the Differential Equation(Used to generate data)
def actual(a,b,x):
Y = np.exp(-b*x)*(a*np.exp(b*x)*(b*x-1)+a+b**2)/b**2
return Y
#Method For Solving the ODE
def lv(xdata, a=5.0, b=0.2):
def dy_dx(y, x):
return a*x - b*y
y0 = 1.0
Y, dict = odeint(dy_dx,y0,xdata,full_output=True)
return Y
#Generating Data for Bayesian Inference
a0, b0 = 5, 0.2
xdata = np.linspace(0, 21, 100)
ydata = actual(a0,b0,xdata)
# Adding some error to the ydata points
yerror = 10*np.random.rand(len(xdata))
ydata += np.random.normal(0.0, np.sqrt(yerror))
ydata = np.ravel(ydata)
#as_op(itypes=[T.dscalar, T.dscalar], otypes=[T.dvector])
def func(al,be):
Q = lv(xdata, a=al, b=be)
return np.ravel(Q)
# Number of Samples and Initial Conditions
nsample = 5000
y0 = 1.0
# Model for Bayesian Inference
model = pm.Model()
with model:
# Priors for unknown model parameters
alpha = pm.Uniform('alpha', lower=a0/2, upper=a0+a0/2)
beta = pm.Uniform('beta', lower=b0/2, upper=b0+b0/2)
# Expected value of outcome
mu = func(alpha,beta)
# Likelihood (sampling distribution) of observations
Y_obs = pm.Normal('Y_obs', mu=mu, sd=yerror, observed=ydata)
trace = pm.sample(nsample, nchains=1)
pm.traceplot(trace)
plt.show()
The error that I get is
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Initializing NUTS failed. Falling back to elementwise auto-assignment.
Any help would be really appreciated
Is it possible create custom multivariate distributions in pymc3? In the following, I have tried to create a linear transformation of a Dirichlet distribution. All variants on this have returned numerous errors, perhaps to do with theano data types? Any help would be gratefully appreciated.
import numpy as np
import pymc3 as pymc
import theano.tensor as tt
# data
n = 5
prior_params = np.ones(n - 1) / (n - 1)
mx = np.array([[0.25 , 0.5 , 0.75 , 1. ],
[0.25 , 0.333, 0.25 , 0. ],
[0.25 , 0.167, 0. , 0. ],
[0.25 , 0. , 0. , 0. ]])
# Note that the matrix mx takes the unit simplex into the unit simplex.
# custom log-liklihood
def generate_function(mx, prior_params):
def log_trunc_dir(x):
return pymc.Dirichlet.dist(a=prior_params).logp(mx.dot(x.T)).eval()
return log_trunc_dir
#model
with pymc.Model() as simple_model:
x = pymc.Dirichlet('x', a=np.ones(n - 1))
q = pymc.DensityDist('q', generate_function(mx, prior_params), observed={'x': x})
Thanks to significant help from the PyMC3 development community, I can post the
following working example of a customised Dirichlet prior in PyMC3.
import pymc3 as pm
import numpy as np
import scipy.special as special
import theano.tensor as tt
import matplotlib.pyplot as plt
n = 4
with pm.Model() as model:
prior = np.ones(n) / n
def dirich_logpdf(value=prior):
return -n * special.gammaln(1/n) + (-1 + 1/n) * tt.log(value).sum()
stick = pm.distributions.transforms.StickBreaking()
probs = pm.DensityDist('probs', dirich_logpdf, shape=n,
testval=np.array(prior), transform=stick)
data = np.array([5, 7, 1, 0])
sfs_obs = pm.Multinomial('sfs_obs', n=np.sum(data), p=probs, observed=data)
with model:
step = pm.Metropolis()
trace = pm.sample(100000, tune=10000, step=step)
print('MLE = ', data / np.sum(data))
print(pm.summary(trace))
pm.traceplot(trace, [probs])
plt.show()
I am working on a python code to plot Eddy Kinetic Energy. I am fairly new to python and I'm confused about an error I have been getting. I'm not worried about plotting my data on a map just yet, I just want to see if I can get it to plot. Here is my code and error:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from pylab import *
from netCDF4 import Dataset
from mpl_toolkits.basemap import Basemap
import matplotlib.cm as cm
from mpl_toolkits.basemap import shiftgrid
test = Dataset('p.34331101.atmos_daily.nc', 'r')
lat = test.variables['lat'][:]
lon = test.variables['lon'][:]
level = test.variables['level'][5]
time = test.variables['time'][:]
u = test.variables['ucomp'][:]
v = test.variables['vcomp'][:]
temp = test.variables['temp'][:]
print(lat.shape)
print(u.shape)
#uz = np.reshape(u, (30, 26, 90))
uzm = np.nanmean(u, axis=3)
#vz = np.reshape(v, (30, 26, 90))
vzm = np.nanmean(v, axis=3)
print(uzm.shape)
ustar = u-uzm[:,:,:,np.newaxis]
vstar = v-vzm[:,:,:,np.newaxis]
EKE = np.nanmean(.5*(ustar**2 + vstar**2), axis=3)
EKE1 = np.asarray(EKE)
%matplotlib inline
print(EKE.shape)
levels=[-10, -5, 0, 5, 10]
plt.contour(EKE[1,1,:])
#EKE is time, level, lat and the shape is (30, 26, 90)
TypeError: Input must be a 2D array.
Bret, you would probably get more help if you included a bit more info with your error, did you not get a line number to look at?
I would hazard a guess that your problem is passing a 1D array to contour(). This sometimes seems counter-intuitive but numpy reduces the dimensions 'automatically' when you specify a single value in an index.
i.e. try
print(EKE.shape)
print(EKE[1,1,:].shape)
print(EKE[1:2,1:2,:].shape)
I have this signal :
from math import*
Fs=8000
f=500
sample=16
a=[0]*sample
for n in range(sample):
a[n]=sin(2*pi*f*n/Fs)
How can I plot a graph (this sine wave)?
and create name of xlabel as 'voltage(V)' and ylabel as 'sample(n)'
What code to do this?
I am so thanksful for help ^_^
Setting the x-axis with np.arange(0, 1, 0.001) gives an array from 0 to 1 in 0.001 increments.
x = np.arange(0, 1, 0.001) returns an array of 1000 points from 0 to 1, and y = np.sin(2*np.pi*x) you will get the sin wave from 0 to 1 sampled 1000 times
I hope this will help:
import matplotlib.pyplot as plt
import numpy as np
Fs = 8000
f = 5
sample = 8000
x = np.arange(sample)
y = np.sin(2 * np.pi * f * x / Fs)
plt.plot(x, y)
plt.xlabel('sample(n)')
plt.ylabel('voltage(V)')
plt.show()
P.S.: For comfortable work you can use The Jupyter Notebook.
import matplotlib.pyplot as plt # For ploting
import numpy as np # to work with numerical data efficiently
fs = 100 # sample rate
f = 2 # the frequency of the signal
x = np.arange(fs) # the points on the x axis for plotting
# compute the value (amplitude) of the sin wave at the for each sample
y = np.sin(2*np.pi*f * (x/fs))
#this instruction can only be used with IPython Notbook.
% matplotlib inline
# showing the exact location of the smaples
plt.stem(x,y, 'r', )
plt.plot(x,y)
import numpy as np
import matplotlib.pyplot as plt
F = 5.e2 # No. of cycles per second, F = 500 Hz
T = 2.e-3 # Time period, T = 2 ms
Fs = 50.e3 # No. of samples per second, Fs = 50 kHz
Ts = 1./Fs # Sampling interval, Ts = 20 us
N = int(T/Ts) # No. of samples for 2 ms, N = 100
t = np.linspace(0, T, N)
signal = np.sin(2*np.pi*F*t)
plt.plot(t, signal)
plt.xlabel('Time (s)')
plt.ylabel('Voltage (V)')
plt.show()
import math
import turtle
ws = turtle.Screen()
ws.bgcolor("lightblue")
fred = turtle.Turtle()
for angle in range(360):
y = math.sin(math.radians(angle))
fred.goto(angle, y * 80)
ws.exitonclick()
The window of usefulness has likely come and gone, but I was working at a similar problem. Here is my attempt at plotting sine using the turtle module.
from turtle import *
from math import *
#init turtle
T=Turtle()
#sample size
T.screen.setworldcoordinates(-1,-1,1,1)
#speed up the turtle
T.speed(-1)
#range of hundredths from -1 to 1
xcoords=map(lambda x: x/100.0,xrange(-100,101))
#setup the origin
T.pu();T.goto(-1,0);T.pd()
#move turtle
for x in xcoords:
T.goto(x,sin(xcoords.index(x)))
A simple way to plot sine wave in python using matplotlib.
import numpy as np
import matplotlib.pyplot as plt
x=np.arange(0,3*np.pi,0.1)
y=np.sin(x)
plt.plot(x,y)
plt.title("SINE WAVE")
plt.show()
import matplotlib.pyplot as plt
import numpy as np
#%matplotlib inline
x=list(range(10))
def fun(k):
return np.sin(k)
y=list(map(fun,x))
plt.plot(x,y,'-.')
#print(x)
#print(y)
plt.show()
This is another option
#!/usr/bin/env python
import numpy as np
import matplotlib
matplotlib.use('TKAgg') #use matplotlib backend TkAgg (optional)
import matplotlib.pyplot as plt
sample_rate = 200 # sampling frequency in Hz (atleast 2 times f)
t = np.linspace(0,5,sample_rate) #time axis
f = 100 #Signal frequency in Hz
sig = np.sin(2*np.pi*f*(t/sample_rate))
plt.plot(t,sig)
plt.xlabel("Time")
plt.ylabel("Amplitude")
plt.tight_layout()
plt.show()
Yet another way to plot the sine wave.
import numpy as np
import matplotlib
matplotlib.use('TKAgg') #use matplotlib backend TKAgg (optional)
import matplotlib.pyplot as plt
t = np.linspace(0.0, 5.0, 50000) # time axis
sig = np.sin(t)
plt.plot(t,sig)
from math import *
Fs = 8000
f = 500
sample = 16
a = [0] * sample
for n in range(sample):
a[n] = sin(2*pi*f*n/Fs)
creating the x coordinates
Sample = [i for i in range(sample)]
importing matplotlib for plotting
import matplotlib.pyplot as plt
adding labels and plotting
plt.xlabel('Voltage(V)')
plt.ylabel('Sample(n)')
plt.plot(Sample, a)
plt.show()