I am using matplotlib.pyplot to create histograms. I'm not actually interested in the plots of these histograms, but interested in the frequencies and bins (I know I can write my own code to do this, but would prefer to use this package).
I know I can do the following,
import numpy as np
import matplotlib.pyplot as plt
x1 = np.random.normal(1.5,1.0)
x2 = np.random.normal(0,1.0)
freq, bins, patches = plt.hist([x1,x1],50,histtype='step')
to create a histogram. All I need is freq[0], freq[1], and bins[0]. The problem occurs when I try and use,
freq, bins, patches = plt.hist([x1,x1],50,histtype='step')
in a function. For example,
def func(x, y, Nbins):
freq, bins, patches = plt.hist([x,y],Nbins,histtype='step') # create histogram
bincenters = 0.5*(bins[1:] + bins[:-1]) # center bins
xf= [float(i) for i in freq[0]] # convert integers to float
xf = [float(i) for i in freq[1]]
p = [ (bincenters[j], (1.0 / (xf[j] + yf[j] )) for j in range(Nbins) if (xf[j] + yf[j]) != 0]
Xt = [j for i,j in p] # separate pairs formed in p
Yt = [i for i,j in p]
Y = np.array(Yt) # convert to arrays for later fitting
X = np.array(Xt)
return X, Y # return arrays X and Y
When I call func(x1,x2,Nbins) and plot or print X and Y, I do not get my expected curve/values. I suspect it something to do with plt.hist, since there is a partial histogram in my plot.
I don't know if I'm understanding your question very well, but here, you have an example of a very simple home-made histogram (in 1D or 2D), each one inside a function, and properly called:
import numpy as np
import matplotlib.pyplot as plt
def func2d(x, y, nbins):
histo, xedges, yedges = np.histogram2d(x,y,nbins)
plt.plot(x,y,'wo',alpha=0.3)
plt.imshow(histo.T,
extent=[xedges.min(),xedges.max(),yedges.min(),yedges.max()],
origin='lower',
interpolation='nearest',
cmap=plt.cm.hot)
plt.show()
def func1d(x, nbins):
histo, bin_edges = np.histogram(x,nbins)
bin_center = 0.5*(bin_edges[1:] + bin_edges[:-1])
plt.step(bin_center,histo,where='mid')
plt.show()
x = np.random.normal(1.5,1.0, (1000,1000))
func1d(x[0],40)
func2d(x[0],x[1],40)
Of course, you may check if the centering of the data is right, but I think that the example shows some useful things about this topic.
My recommendation: Try to avoid any loop in your code! They kill the performance. If you look, In my example there aren't loops. The best practice in numerical problems with python is avoiding loops! Numpy has a lot of C-implemented functions that do all the hard looping work.
You can use np.histogram2d (for 2D histogram) or np.histogram (for 1D histogram):
hst = np.histogram(A, bins)
hst2d = np.histogram2d(X,Y,bins)
Output form will be the same as plt.hist and plt.hist2d, the only difference is there is no plot.
No.
But you can bypass the pyplot:
import matplotlib.pyplot
fig = matplotlib.figure.Figure()
ax = matplotlib.axes.Axes(fig, (0,0,0,0))
numeric_results = ax.hist(data)
del ax, fig
It won't impact active axes and figures, so it would be ok to use it even in the middle of plotting something else.
This is because any usage of plt.draw_something() will put the plot in current axis - which is a global variable.
If you would like to simply compute the histogram (that is, count the number of points in a given bin) and not display it, the np.histogram() function is available
Related
I am using matplotlib.pyplot in python. Consider y-axis is real values called "ranking loss" and x-axis is the number of the iterations(1000). Then I plot the average ranking loss of 2 runs of an algorithm in each iteration.
Does anyone know why do I get this strange thick chart instead of a line?
Thank you very much in advance
And the command is :
fig = plt.figure()
fig.suptitle('Batch-GD', fontsize=20)
plt.xlabel('Iteration', fontsize=18)
plt.ylabel('Avg ranking loss', fontsize=16)
plt.grid(True)
plt.xlim(0, iter)
plt.plot(avg_loss)
fig.savefig('GD_with_ini.jpg')
plt.show()
What's happening here is probably, that your line density is so high that lines overlap in a way that a plain surface is shown instead of the line itself.
If we take e.g. 10000 points and make the plot oscillate at very high frequency, we get a similar behaviour. Zooming in shows that there actually is a line.
Code to reproduce the plot:
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.axes_grid1.inset_locator import inset_axes, mark_inset
x = np.linspace(0,1000,num=10000)
y = np.sin(x*100.)*x/5000.+np.exp(-x/60.)+np.sin(x/50.)*0.016
plt.plot(x,y)
###### show inset ####
ax = plt.gca()
axins = inset_axes(ax, 2,2, loc=1)
axins.plot(x,y)
axins.set_xlim(400, 410)
axins.set_ylim(-0.1, 0.17)
mark_inset(ax, axins, loc1=2, loc2=4, fc="none", ec="0.5")
plt.show()
A solution can then be to calculate some kind of rolling average. E.g.:
def running_mean(x, N):
cumsum = np.cumsum(np.insert(x, 0, 0))
return (cumsum[N:] - cumsum[:-N]) / N
N=300
cumsum = running_mean(y, N)
ax.plot(x[N//2:-N//2+1], cumsum, c="orange")
axins.plot(x[N//2:-N//2+1], cumsum, c="orange")
How do I fit a curve on a barplot?
I have an equation, the diffusion equation, which has some unknown parameters, these parameters make the curve larger, taller, etc. On the other hand I have a barplot coming from a simulation. I would like to fit the curve on the barplot, and find the best parameters for the curve, how can I do that?
This is what I obtained by 'manual fitting', so basically I changed manually all the parameters for hours. However is there a way to do this with python?
To make it simple, imagine I have the following code:
import matplotlib.pyplot as plt
list1 = []
for i in range(-5,6):
list1.append(i)
width = 1/1.5
list2 = [0,0.2,0.6,3.5,8,10,8,3.5,0.6,0.2,0]
plt.bar(list1,list2,width)
plt.show()
T = 0.13
xx = np.arange(-6,6,0.01)
yy = 5*np.sqrt(np.pi)*np.exp(-((xx)**2)/(4*T))*scipy.special.erfc((xx)/(2*np.sqrt(T))) + np.exp(-((xx)**2)/(4*T))
plt.plot(xx,yy)
plt.show()
Clearly the fitting here would be pretty hard, but anyway, is there any function or such that allows me to find the best coefficients for the equation: (where T is known)
y = A*np.sqrt(np.pi*D)*np.exp(-((x-E)**2)/(4*D*T))*scipy.special.erfc((x-E)/(2*np.sqrt(D*T))) + 300*np.exp(-((x-E)**2)/(4*D*T))
EDIT: This is different from the already asked question and from the scipy documentation example. In the latter the 'xdata' is the same, while in my case it might and might not be. Furthermore I would also be able to plot this curve fitting, which isn't shown on the documentation. The height of the bars is not a function of the x's! So my xdata is not a function of my ydata, this is different from what is in the documentation.
To see what I mean try to change the code in the documentation a little bit, to fall into my example, try this:
def func(x,a,b,c):
return a * np.exp(-b * x) + c
xdata = np.linspace(0,4,50)
y = func(xdata, 2.5, 1.3, 0.5)
ydata = [1,6,3,4,6,7,8,5,7,0,9,8,2,3,4,5]
popt, pcov = curve_fit(func,xdata,ydata)
if you run this, it doesn't work. The reason is that I have 16 elements for the ydata and 50 for the function. This happens because y takes values from xdata, while ydata takes values from another set of x values, which is here unknown.
Thank you
I stand by my thinking that this question is a duplicate. Here is a brief example of the typical workflow using curve_fit. Let me know if you still think that your situation is different.
import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
# bar plot data
list1 = range(-5, 6)
list2 = [0, 0.2, 0.6, 3.5, 8, 10,
8, 3.5, 0.6, 0.2, 0]
width = 1/1.5
plt.bar(list1, list2, width, alpha=0.75)
# fit bar plot data using curve_fit
def func(x, a, b, c):
# a Gaussian distribution
return a * np.exp(-(x-b)**2/(2*c**2))
popt, pcov = curve_fit(func, list1, list2)
x = np.linspace(-5, 5, 100)
y = func(x, *popt)
plt.plot(x + width/2, y, c='g')
it is quite common to calculate the probability density of a value within a probability density function (PDF). Imagine we have a gaussian distribution with mean = 40, a standard deviation of 5 and now would like to get the probability density of value 32. We'd go like:
In [1]: import scipy.stats as stats
In [2]: print stats.norm.pdf(32, loc=40, scale=5)
Out [2]: 0.022
--> The probability density is 2.2%.
But now, let's consider the inverse problem. I have the mean value, I have the value at probabilty density of 0.05 and I would like to get the standard deviation (i.e. the scale parameter).
What I could implement is a numerical approach: create stats.norm.pdf several times with the scale-parameter increased stepwise and take that one with the result getting as closest as possible.
In my case, I specify the value 30 as the 5% mark. So I need to solve this "equation":
stats.norm.pdf(30, loc=40, scale=X) = 0.05
There is a scipy function called "ppf" which is the inverse of the PDF, so it will return the value for a specific probability density, but I haven't found a function to return the scale parameter.
Implementing an iteration would take too much time (both creating and calculating). My script is going to be huge, so I should save computation time. Could the lambda-function help in this case? I roughly know what it's doing, but I haven't used it so far. Any ideas on this?
Thank you!
The normal probability density function, f is given by
Given f and x we wish to solve for 𝞼. Let's ask sympy if it can solve the equation:
import sympy as sy
from sympy.abc import x, y, sigma
expr = (1/(sy.sqrt(2*sy.pi)*sigma) * sy.exp(-x**2/(2*sigma**2))) - y
ans = sy.solve(expr, sigma)[0]
print(ans)
# sqrt(2)*exp(LambertW(-2*pi*x**2*y**2)/2)/(2*sqrt(pi)*y)
So it appears there is a closed-formed solution in terms of the LambertW function, W, which satisfies
z = W(z) * exp(W(z))
for all complex-valued z.
We could use sympy to also find the numerical result for given x and y, but
perhaps it would be faster to do the numerical work with
scipy.special.lambertw:
import numpy as np
import scipy.special as special
def sigma_func(x, y):
results = set([np.real_if_close(
np.sqrt(2)*np.exp(special.lambertw(-2*np.pi*x**2*y**2, k=k)/2)
/(2*np.sqrt(np.pi)*y)).item() for k in (0, -1)])
results = [s for s in results if np.isreal(s)]
return results
In general, the LambertW function returns complex values, but we are only
interested in real-valued solutions for sigma. Per the
docs,
special.lambertw has two partially-real branches, when k=0 and k=1. So the
code above checks if the returned value (for those two branches) is real, and
returns a list of any real solutions if they exist. If no real solution exists,
then an empty list is returned. That happens if the pdf value y is not
attained for any real value of sigma (for the given value of x).
You can use it like this:
x = 30.0
loc = 40.0
y = 0.02
s = sigma_func(loc-x, y)
print(s)
# [16.65817044316178, 6.830458938511113]
import scipy.stats as stats
for si in s:
assert np.allclose(stats.norm.pdf(x, loc=loc, scale=si), y)
In the example you gave, with y = 0.025, there is no solution for sigma:
import numpy as np
import scipy.stats as stats
import matplotlib.pyplot as plt
x = 30.0
loc = 40.0
y = 0.025
s = np.linspace(5, 20, 100)
plt.plot(s, stats.norm.pdf(x, loc=loc, scale=s))
plt.hlines(y, 4, 20, color='red') # the horizontal line y = 0.025
plt.ylabel('pdf')
plt.xlabel('sigma')
plt.show()
and so sigma_func(40-30, 0.025) returns an empty list:
In [93]: sigma_func(40-30, 0.025)
Out [93]: []
The plot above is typical in the sense that when y is too large there are zero
solutions, at the maximum of the curve (let's call it y_max) there is one
solution
In [199]: y_max = np.nextafter(np.sqrt(1/(np.exp(1)*2*np.pi*(10)**2)), -np.inf)
In [200]: y_max
Out[200]: 0.024197072451914336
In [201]: sigma_func(40-30, y_max)
Out[201]: [9.9999999776424]
and for y smaller than the y_max there are two solutions.
The will be two solutions, because normal PDF is symmetric around the mean.
As it stands, you have a single-variable equation to solve. It won't have a closed-form solution, so you can use e.g. scipy.optimize.fsolve to solve it.
EDIT: see #unutbu's answer for the closed form solution in terms of Lambert W function.
I am trying to use the package pynfft in python 2.7 to do the non-uniform fast Fourier transform (nfft). I have learnt python for only two months, so I have some difficulties.
This is my code:
import numpy as np
from pynfft.nfft import NFFT
#loading data, 104 lines
t_diff, x_diff = np.loadtxt('data/analysis/amplitudes.dat', unpack = True)
N = [13,8]
M = 52
#fourier coefficients
f_hat = np.fft.fft(x_diff)/(2*M)
#instantiation
plan = NFFT(N,M)
#precomputation
x = t_diff
plan.x = x
plan.precompute()
# vector of non uniform samples
f = x_diff[0:M]
#execution
plan.f = f
plan.f_hat = f_hat
f = plan.trafo()
I am basically following the instructions I found in the pynfft tutorial (http://pythonhosted.org/pyNFFT/tutorial.html).
I need the nfft because the time intervals in which my data are taken are not constant (I mean, the first measure is taken at t, the second after dt, the third after dt+dt' with dt' different from dt and so on).
The pynfft package wants the vector of the fourier coefficients ("f_hat") before execution, so I calculated it using numpy.fft, but I am not sure this procedure is correct. Is there another way to do it (maybe with the nfft)?
I would like also to calculate the frquencies; I know that with numpy.fft there is a command: is ther anything like that also for pynfft? I did not find anything in the tutorial.
Thank you for any advice you can give me.
Here is a working example, taken from here:
First we define the function we want to reconstruct, which is the sum of four harmonics:
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(12345)
%pylab inline --no-import-all
# function we want to reconstruct
k=[1,5,10,30] # modulating coefficients
def myf(x,k):
return sum(np.sin(x*k0*(2*np.pi)) for k0 in k)
x=np.linspace(-0.5,0.5,1000) # 'continuous' time/spatial domain; -0.5<x<+0.5
y=myf(x,k) # 'true' underlying trigonometric function
fig=plt.figure(1,(20,5))
ax =fig.add_subplot(111)
ax.plot(x,y,'red')
ax.plot(x,y,'r.')
# we should sample at a rate of >2*~max(k)
M=256 # number of nodes
N=128 # number of Fourier coefficients
nodes =np.random.rand(M)-0.5 # non-uniform oversampling
values=myf(nodes,k) # nodes&values will be used below to reconstruct
# original function using the Solver
ax.plot(nodes,values,'bo')
ax.set_xlim(-0.5,+0.5)
The we initialize and run the Solver:
from pynfft import NFFT, Solver
f = np.empty(M, dtype=np.complex128)
f_hat = np.empty([N,N], dtype=np.complex128)
this_nfft = NFFT(N=[N,N], M=M)
this_nfft.x = np.array([[node_i,0.] for node_i in nodes])
this_nfft.precompute()
this_nfft.f = f
ret2=this_nfft.adjoint()
print this_nfft.M # number of nodes, complex typed
print this_nfft.N # number of Fourier coefficients, complex typed
#print this_nfft.x # nodes in [-0.5, 0.5), float typed
this_solver = Solver(this_nfft)
this_solver.y = values # '''right hand side, samples.'''
#this_solver.f_hat_iter = f_hat # assign arbitrary initial solution guess, default is 0
this_solver.before_loop() # initialize solver internals
while not np.all(this_solver.r_iter < 1e-2):
this_solver.loop_one_step()
Finally, we display the frequencies:
import matplotlib.pyplot as plt
fig=plt.figure(1,(20,5))
ax =fig.add_subplot(111)
foo=[ np.abs( this_solver.f_hat_iter[i][0])**2 for i in range(len(this_solver.f_hat_iter) ) ]
ax.plot(np.abs(np.arange(-N/2,+N/2,1)),foo)
cheers
I can use np.polyfit to fit a line in my scatter plot as shown bellow
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
m1, b1 = np.polyfit(a, b, 1)
corr1 =a1.plot(a, m1*a+b1, '-', color='black')
a1.scatter(a, b)
Is there any way to fit a line using polyfit this time taking the errors for my points as shown bellow?
ae = np.empty(10)
ae.fill(0.15)
be = ae
sca1=a1.errorbar(a, b, ae, be, capsize=0, ls='none', color='black', elinewidth=1)
If you want to compute the fit and plot the (fixed at the given value) error bars over the fit points, like this:
Then this code will do the job:
import numpy as np
import matplotlib.pyplot as mp
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
ae = np.empty(10)
ae.fill(0.15)
be = ae
m1, b1 = np.polyfit(a, b, 1)
mp.figure()
corr1 =mp.errorbar(a,m1*a+b1,ae,be, '-', color='black')
mp.scatter(a, b)
mp.show()
If you want to get the covariance of the fit, and use the standard deviation to set the error bars, instead use the code
import numpy as np
import matplotlib.pyplot as mp
import math
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
coeff,covar = np.polyfit(a, b, 1,cov=True)
m1= coeff[0]
b1= coeff[1]
xe = math.sqrt(covar[0][0])
ye = math.sqrt(covar[1][2])
mp.figure()
corr1 =mp.errorbar(a,m1*a+b1,xe,ye, '-', color='black')
mp.scatter(a, b)
mp.show()
which gives a plot like this:
If you want to do a weighted fit, you can supply a weight vector to polyfit with the syntax
m2, b2 = np.polyfit(a, b, 1,w=weightvector)
According to the documentation the weightvector should contain 1 over the standard deviation of the data points.
If you want to do a least squares fit weighted by errors in BOTH x and y, I don't think polyfit does this - it will accept a weight vector for one dimension.
To supply errors in both dimensions as weights you would have to use scipy.optimize.leastsq.
There is a documentation page at this link of the Scipy documentation about doing fits with scipy.optimize.leastsq. The example talks about a fit to power law, but clearly a straight line could be done as well.
For errors in one dimension (Y) here, an example using leastsq is:
import numpy as np
import matplotlib.pyplot as mp
import math
from scipy import optimize
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
aerr = np.empty(10)
aerr.fill(0.15)
berr=aerr
# fit a straight line with scipy scipy.optimize.leastsq
# define our (line) fitting function
fitfunc = lambda p, x: p[0] + p[1] * x
errfunc = lambda p, x, y, err: (y - fitfunc(p, x)) / err
pinit = [1.0, -1.0]
out = optimize.leastsq(errfunc, pinit,args=(a, b, aerr), full_output=1)
coeff = out[0]
covar = out[1]
print 'coeff', coeff
print 'covar', covar
m1= coeff[1]
b1= coeff[0]
xe = math.sqrt(covar[0][0])
ye = math.sqrt(covar[1][1])
# plot results
mp.figure()
corr2 =mp.errorbar(a,m1*a+b1,xe,ye, '-', color='red')
mp.scatter(a, b)
mp.show()
To take into account errors in both X and Y, you would have to change the definition of errfunc to reflect the specific technique you are using to do that. If a lambda isn't convenient you can instead define a function that will do that. I can't comment further on this without knowing what technique is being used to weight by X and Y errors, there are several in the literature.