I don't understand why the ifft(fft(myFunction)) is not the same as my function. It seems to be the same shape but a factor of 2 out (ignoring the constant y-offset). All the documentation I can see says there is some normalisation that fft doesn't do, but that ifft should take care of that. Here's some example code below - you can see where I've bodged the factor of 2 to give me the right answer. Thanks for any help - its driving me nuts.
import numpy as np
import scipy.fftpack as fftp
import matplotlib.pyplot as plt
import matplotlib.pyplot as plt
def fourier_series(x, y, wn, n=None):
# get FFT
myfft = fftp.fft(y, n)
# kill higher freqs above wavenumber wn
myfft[wn:] = 0
# make new series
y2 = fftp.ifft(myfft).real
# find constant y offset
myfft[1:]=0
c = fftp.ifft(myfft)[0]
# remove c, apply factor of 2 and re apply c
y2 = (y2-c)*2 + c
plt.figure(num=None)
plt.plot(x, y, x, y2)
plt.show()
if __name__=='__main__':
x = np.array([float(i) for i in range(0,360)])
y = np.sin(2*np.pi/360*x) + np.sin(2*2*np.pi/360*x) + 5
fourier_series(x, y, 3, 360)
You're removing half the spectrum when you do myfft[wn:] = 0. The negative frequencies are those in the top half of the array and are required.
You have a second fudge to get your results which is taking the real part to find y2: y2 = fftp.ifft(myfft).real (fftp.ifft(myfft) has a non-negligible imaginary part due to the asymmetry in the spectrum).
Fix it with myfft[wn:-wn] = 0 instead of myfft[wn:] = 0, and remove the fudges. So the fixed code looks something like:
import numpy as np
import scipy.fftpack as fftp
import matplotlib.pyplot as plt
def fourier_series(x, y, wn, n=None):
# get FFT
myfft = fftp.fft(y, n)
# kill higher freqs above wavenumber wn
myfft[wn:-wn] = 0
# make new series
y2 = fftp.ifft(myfft)
plt.figure(num=None)
plt.plot(x, y, x, y2)
plt.show()
if __name__=='__main__':
x = np.array([float(i) for i in range(0,360)])
y = np.sin(2*np.pi/360*x) + np.sin(2*2*np.pi/360*x) + 5
fourier_series(x, y, 3, 360)
It's really worth paying attention to the interim arrays that you are creating when trying to do signal processing. Invariably, there are clues as to what is going wrong that should direct you to the problem. In this case, you taking the real part masked the problem and made your task more difficult.
Just to add another quick point: Sometimes taking the real part of the resultant array is exactly the correct thing to do. It's often the case that you end up with an imaginary part to the signal output which is just down to numerical errors in the input to the inverse FFT. Typically this manifests itself as very small imaginary values, so taking the real part is basically the same array.
You are killing the negative frequencies between 0 and -wn.
I think what you mean to do is to set myfft to 0 for all frequencies outside [-wn, wn].
Change the following line:
myfft[wn:] = 0
to:
myfft[wn:-wn] = 0
Related
I am trying to use a polynomial expression that would fit my function (signal). I am using numpy.polynomial.polynomial.Polynomial.fit function to fit my function(signal) using the coefficients. Now, after generating the coefficients, I want to put those coefficients back into the polynomial equation - get the corresponding y-values - and plot them on the graph. But I am not getting what I want (orange line) . What am I doing wrong here?
Thanks.
import math
def getYValueFromCoeff(f,coeff_list): # low to high order
y_plot_values=[]
for j in range(len(f)):
item_list= []
for i in range(len(coeff_list)):
item= (coeff_list[i])*((f[j])**i)
item_list.append(item)
y_plot_values.append(sum(item_list))
print(len(y_plot_values))
return y_plot_values
from numpy.polynomial import Polynomial as poly
import numpy as np
import matplotlib.pyplot as plt
no_of_coef= 10
#original signal
x = np.linspace(0, 0.01, 10)
period = 0.01
y = np.sin(np.pi * x / period)
#poly fit
test1= poly.fit(x,y,no_of_coef)
coeffs= test1.coef
#print(test1.coef)
coef_y= getYValueFromCoeff(x, test1.coef)
#print(coef_y)
plt.plot(x,y)
plt.plot(x, coef_y)
If you check out the documentation, consider the two properties: poly.domain and poly.window. To avoid numerical issues, the range poly.domain = [x.min(), x.max()] of independent variable (x) that we pass to the fit() is being normalized to poly.window = [-1, 1]. This means the coefficients you get from poly.coef apply to this normalized range. But you can adjust this behaviour (sacrificing numerical stability) accordingly, that is, adjustig the poly.window will make your curves match:
...
test1 = poly.fit(x, y, deg=no_of_coef, window=[x.min(), x.max()])
...
But unless you have a good reason to do that, I'd stick to the default behaviour of fit().
As a side note: Evaluating polynomials or lists of coefficients is already implemented in numpy, e.g. using directly
coef_y = test1(x)
or alternatively using np.polyval.
I always like to see original solutions to problems. I urge you to continue to pursue that as that is the best way to learn how to fit functions programmatically. I also wanted to provide the solution that is much more tailored towards a standard numpy implementation. As for your custom function, you did really well. The only issue is that the coefficients are from high to low order, while you were counting up in powers from 0 to highest power. Simply counting down from highest power to 0, allows your function to give the correct result. Notice how your function overlays perfectly with the numpy polyval.
import numpy as np
import matplotlib.pyplot as plt
def getYValueFromCoeff(f,coeff_list): # low to high order
y_plot_values=[]
for j in range(len(f)):
item_list= []
for i in range(len(coeff_list)):
item= (coeff_list[i])*((f[j])**(len(coeff_list)-i-1))
item_list.append(item)
y_plot_values.append(sum(item_list))
print(len(y_plot_values))
return y_plot_values
no_of_coef = 10
#original signal
x = np.linspace(0, 0.01, 10)
period = 0.01
y = np.sin(np.pi * x / period)
#poly fit
coeffs = np.polyfit(x,y,no_of_coef)
coef_y = np.polyval(coeffs,x)
COEF_Y = getYValueFromCoeff(x,coeffs)
plt.figure()
plt.plot(x,y)
plt.plot(x, coef_y)
plt.plot(x, COEF_Y)
plt.legend(['Original Function', 'Fitted Function', 'Custom Fitting'])
plt.show()
Output
Here's the simple way of doing it if you didn't know that already...
import math
from numpy.polynomial import Polynomial as poly
import numpy as np
import matplotlib.pyplot as plt
no_of_coef= 10
#original signal
x = np.linspace(0, 0.01, 10)
period = 0.01
y = np.sin(np.pi * x / period)
#poly fit
test1= poly.fit(x,y,no_of_coef)
plt.plot(x, y, 'r', label='original y')
x = np.linspace(0, 0.01, 1000)
plt.plot(x, test1(x), 'b', label='y_fit')
plt.legend()
plt.show()
I would like to solve motion first order ODE equations using scipy solve_ivp function. I can see that I'm doing something wrong because this should be an ellipse but I'm plotting only four points. Are you able to spot the mistake?
import math
import matplotlib.pyplot as plt
import numpy as np
import scipy.integrate
gim = 4*(math.pi**2)
x0 = 1 #x-position of the center or h
y0 = 0 #y-position of the center or k
vx0 = 0 #vx position
vy0 = 1.1* 2* math.pi #vy position
initial = [x0, y0, vx0, vy0] #initial state of the system
time = np.arange(0, 1000, 0.01) #period
def motion(t, Z):
dx = Z[2] # vx
dy = Z[3] # vy
dvx = -gim/(x**2+y**2)**(3/2) * x * Z[2]
dvy = -gim/(x**2+y**2)**(3/2) * y * Z[3]
return [dx, dy, dvx, dvy]
sol = scipy.integrate.solve_ivp(motion, t_span=time, y0= initial, method='RK45')
plt.plot(sol.y[0],sol.y[1],"x", label="Scipy RK45 solution")
plt.show()
The code should not be able to run from a fresh workspace. The variables x,y that are used in the gravitation formula are not declared anywhere. So insert the line x,y = Z[:2] or similar.
The gravitation formula usually does not contain the velocity components. Remove Z[2], Z[3].
Check again what the time span and evaluation times arguments expect. The time span takes the first two values from the array. So change to t_span=time[[0,-1]] to build the pair of first and last time value.
The second plot suffers from insufficient evaluation points, the line segments used are too large. With your time array that should not be a problem.
A post gives some code to plot this figure
import scipy.stats as ss
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(-10, 11)
xU, xL = x + 0.5, x - 0.5
prob = ss.norm.cdf(xU, scale = 3) - ss.norm.cdf(xL, scale = 3)
prob = prob / prob.sum() #normalize the probabilities so their sum is 1
nums = np.random.choice(x, size = 10000, p = prob)
plt.hist(nums, bins = len(x))
I modifyied this line
x = np.arange(-10, 11)
to this line
x = np.arange(10, 31)
I got this figure
How to fix that?
Given what you're asking Python to do, there's no error in this plot: it's a histogram of 10,000 samples from the tail (anything that rounds to between 10 and 31) of a normal distribution with mean 0 and standard deviation 3. Since probabilities drop off steeply in the tail of a normal, it happens that none of the 10,000 exceeded 17, which is why you didn't get the full range up to 31.
If you just want the x-axis of the plot to cover your full intended range, you could add plt.xlim(9.5, 31.5) after plt.hist.
If you want a histogram with support over this entire range, then you'll need to adjust the mean and/or variance of the distribution. For instance, if you specify that your normal distribution has mean 20 rather than mean 0 when you obtain prob, i.e.
prob = ss.norm.cdf(xU, loc=20, scale=3) - ss.norm.cdf(xL, loc=20, scale=3)
then you'll recover a similar-looking histogram, just translated to the right by 20.
I need to draw many such rows (for a0 .. a128) in a single window. I've searched in FacetGrid, PairGrid and all over around but couldn't find. Only regplot has similar argument ax but it doesn't plot histograms. My data is 128 real valued features with label column [0, 1]. I need the graphs to be shown from my Python code as a separate application on Linux.
Also, it there a way to scale this histogram to show relative values on Y such that the right curve is not skewed?
g = sns.FacetGrid(df, col="Result")
g.map(plt.hist, "a0", bins=20)
plt.show()
Just a simple example using matplotlib. The code is not optimized (ugly, but simple plot-indexing):
import numpy as np
import matplotlib.pyplot as plt
N = 5
data = np.random.normal(size=(N*N, 1000))
f, axarr = plt.subplots(N, N) # maybe you want sharex=True, sharey=True
pi = [0,0]
for i in range(data.shape[0]):
if pi[1] == N:
pi[0] += 1 # next row
pi[1] = 0 # first column again
axarr[pi[0], pi[1]].hist(data[i], normed=True) # i was wrong with density;
# normed=True should be used
pi[1] += 1
plt.show()
Output:
I have an equation dy/dx = x + y/5 and an initial value, y(0) = -3.
I would like to know how to plot the exact graph of this function using pyplot.
I also have a x = np.linspace(0, interval, steps+1) which I would like to use as the x axis. So I'm only looking for the y axis values.
Thanks in advance.
Just for completeness, this kind of equation can easily be integrated numerically, using scipy.integrate.odeint.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
# function dy/dx = x + y/5.
func = lambda y,x : x + y/5.
# Initial condition
y0 = -3 # at x=0
# values at which to compute the solution (needs to start at x=0)
x = np.linspace(0, 4, 101)
# solution
y = odeint(func, y0, x)
# plot the solution, note that y is a column vector
plt.plot(x, y[:,0])
plt.xlabel('x')
plt.ylabel('y')
plt.show()
Given that you need to solve the d.e. you might prefer doing this algebraically, with sympy. (Or you might not.)
Import the module and define the function and the dependent variable.
>>> from sympy import *
>>> f = Function('f')
>>> var('x')
x
Invoke the solver. Note that all terms of the d.e. must be transposed to the left of the equals sign, and that the y must be replaced by the designator for the function.
>>> dsolve(Derivative(f(x),x)-x-f(x)/5)
Eq(f(x), (C1 + 5*(-x - 5)*exp(-x/5))*exp(x/5))
As you would expect, the solution is given in terms of an arbitrary constant. We must solve for that using the initial value. We define it as a sympy variable.
>>> var('C1')
C1
Now we create an expression to represent this arbitrary constant as the left side of an equation that we can solve. We replace f(0) with its value in the initial condition. Then we substitute the value of x in that condition to get an equation in C1.
>>> expr = -3 - ( (C1 + 5*(-x - 5)*exp(-x/5))*exp(x/5) )
>>> expr.subs(x,0)
-C1 + 22
In other words, C1 = 22. Finally, we can use this value to obtain the particular solution of the differential equation.
>>> ((C1 + 5*(-x - 5)*exp(-x/5))*exp(x/5)).subs(C1,22)
((-5*x - 25)*exp(-x/5) + 22)*exp(x/5)
Because I'm absentminded and ever fearful of making egregious mistakes I check that this function satisfies the initial condition.
>>> (((-5*x - 25)*exp(-x/5) + 22)*exp(x/5)).subs(x,0)
-3
(Usually things are incorrect only when I forget to check them. Such is life.)
And I can plot this in sympy too.
>>> plot(((-5*x - 25)*exp(-x/5) + 22)*exp(x/5),(x,-1,5))
<sympy.plotting.plot.Plot object at 0x0000000008C2F780>