I am fairly new to python, and I am trying to plot a contour plot of water surface over a 2d mesh.
At the moment the code is running but I am not getting the right solution. I have checked the formula carefully and I am fairly confident that the issue is with my loops.
I want the code to run for each point on my mesh based on their x and y coordinates.
The mesh is 100 x 100 resulting in 10000 nodes. I have posted my code below, I believe the problem is with the integrated for loops. Any advice on what I might be able to try would be great.
Apologies for the length of code...
import numpy as np
import matplotlib.pyplot as plt
import math
import sys
from math import sqrt
import decimal
t=0
n=5
l=100000
d=100
g=9.81
nx, ny = (100,100)
x5 = np.linspace(-100000,100000,nx)
y5 = np.linspace(-100000,100000,ny)
xv,yv = np.meshgrid(x5,y5)
x = np.arange(-100000,100000,2000)
y = np.arange(-100000,100000,2000)
c=np.arange(len(x))
x2=np.arange(len(x))
y2=np.arange(len(x))
t59=np.arange (1,10001,1)
h=np.arange(len(t59))
om2=1.458*(10**-4.0)
phi=52
phirad=phi*(math.pi/180)
f=om2*math.sin(phirad)
A=(((d+n)**2.0)-(d**2.0))/(((d+n)**2.0)+(d**2.0))
w=(((8*g*d)/(l**2))+(f**2))**0.5
a=((1-(A**2.0))**0.5)/(1-(A*math.cos(w*t)))
b=(((1-(A**2.0))/(1-(A*math.cos(w*t)))**2.0)-1)
l2=l**2.0
for i in range (len(x)):
for j in range (len(y)):
h[i]=d*(a-1-((((x[i]**2.0)+(y[j]**2.0))/l2)*b))
h5=np.reshape(h,(100,100))
plt.figure(1)
plt.contourf(x5,y5,h5)
plt.colorbar()
plt.show()
Ok apologies I didn't make myself very clear. So I'm hoping to achieve a parabolic basin output with h values varying between roughly -10 and 10. Instead I am getting enormous values and the completely wrong shape. I thought the for loop needed to be more like:
for i in range (len(x)):
for j in range (len(y)):
h[i][j]=d*(a-1-((((x[i][j]**2.0)+(y[i][j]**2.0))/l2)*b))
Is that clearer? Let me know if not.
The first thing is that the complete loop is not necessary.
h = d * (a - 1 - (x[None,:]**2 + y[:,None]**2) / 12 * b)
Here the magic comes with the None in indexing. x[None, :] means "x as a row vector copied to as many rows as needed and y[:, None] means "y as a column vector copied to as many columns as needed`.
This might be easiest to understand with an example:
import numpy as np
x = np.arange(5)
y = np.arange(0,50,10)
print x, y, x[None,:] + y[:, None]
The one-liner above gives:
Some manual calculations show this should be rather ok.
d = 100
a = 1.05
b = 0.1025
For a corner point at (1e5, 1e5), we have 2e10 in the addition, so the values do not look badly off.
Related
I got a simple 2D array of values like this :
[simple array]
and I want to add reverb to it (I don't know how to call it other way) in order for it to look like this, basicly with a damping/smooth effect on y values but only on +x :
[with reverb]
I tried to check with scipy as i'm already using it to smooth values but didn't found out how to do it.
does anybody has an idea ?
You could try a Finite impulse response filter, though it's not clear if it's exactly what you need.
This was produced by the script below.
I've assumed, given your figures, that your data is actually 1-dimensional (a "line" of numbers, not a "rectangle").
import numpy as np
import matplotlib.pyplot as plt
from scipy import signal
npts = 50
# FIR with falling sawtooth impulse response
b = np.linspace(1,0,npts,endpoint=False)
u = np.zeros(3 * npts)
u[0] = 1
u[npts + 10] = 1
u[npts + 10 + npts//2] = 1
y = signal.lfilter(b, [1], u)
fig, ax = plt.subplots(2)
ax[0].stem(u)
ax[0].set_ylabel('input')
ax[1].stem(y)
ax[1].set_ylabel('output')
plt.show()
In my work I have the task to read in a CSV file and do calculations with it. The CSV file consists of 9 different columns and about 150 lines with different values acquired from sensors. First the horizontal acceleration was determined, from which the distance was derived by double integration. This represents the lower plot of the two plots in the picture. The upper plot represents the so-called force data. The orange graph shows the plot over the 9th column of the CSV file and the blue graph shows the plot over the 7th column of the CSV file.
As you can see I have drawn two vertical lines in the lower plot in the picture. These lines represent the x-value, which in the upper plot is the global minimum of the orange function and the intersection with the blue function. Now I want to do the following, but I need some help: While I want the intersection point between the first vertical line and the graph to be (0,0), i.e. the function has to be moved down. How do I achieve this? Furthermore, the piece of the function before this first intersection point (shown in purple) should be omitted, so that the function really only starts at this point. How can I do this?
In the following picture I try to demonstrate how I would like to do that:
If you need my code, here you can see it:
import numpy as np
import matplotlib.pyplot as plt
import math as m
import loaddataa as ld
import scipy.integrate as inte
from scipy.signal import find_peaks
import pandas as pd
import os
# Loading of the values
print(os.path.realpath(__file__))
a,b = os.path.split(os.path.realpath(__file__))
print(os.chdir(a))
print(os.chdir('..'))
print(os.chdir('..'))
path=os.getcwd()
path=path+"\\Data\\1 Fabienne\\Test1\\left foot\\50cm"
print(path)
dataListStride = ld.loadData(path)
indexStrideData = 0
strideData = dataListStride[indexStrideData]
#%%Calculation of the horizontal acceleration
def horizontal(yAngle, yAcceleration, xAcceleration):
a = ((m.cos(m.radians(yAngle)))*yAcceleration)-((m.sin(m.radians(yAngle)))*xAcceleration)
return a
resultsHorizontal = list()
for i in range (len(strideData)):
strideData_yAngle = strideData.to_numpy()[i, 2]
strideData_xAcceleration = strideData.to_numpy()[i, 4]
strideData_yAcceleration = strideData.to_numpy()[i, 5]
resultsHorizontal.append(horizontal(strideData_yAngle, strideData_yAcceleration, strideData_xAcceleration))
resultsHorizontal.insert(0, 0)
#plt.plot(x_values, resultsHorizontal)
#%%
#x-axis "convert" into time: 100 Hertz makes 0.01 seconds
scale_factor = 0.01
x_values = np.arange(len(resultsHorizontal)) * scale_factor
#Calculation of the global high and low points
heel_one=pd.Series(strideData.iloc[:,7])
plt.scatter(heel_one.idxmax()*scale_factor,heel_one.max(), color='red')
plt.scatter(heel_one.idxmin()*scale_factor,heel_one.min(), color='blue')
heel_two=pd.Series(strideData.iloc[:,9])
plt.scatter(heel_two.idxmax()*scale_factor,heel_two.max(), color='orange')
plt.scatter(heel_two.idxmin()*scale_factor,heel_two.min(), color='green')#!
#Plot of force data
plt.plot(x_values[:-1],strideData.iloc[:,7]) #force heel
plt.plot(x_values[:-1],strideData.iloc[:,9]) #force toe
# while - loop to calculate the point of intersection with the blue function
i = heel_one.idxmax()
while strideData.iloc[i,7] > strideData.iloc[i,9]:
i = i-1
# Length calculation between global minimum orange function and intersection with blue function
laenge=(i-heel_two.idxmin())*scale_factor
print(laenge)
#%% Integration of horizontal acceleration
velocity = inte.cumtrapz(resultsHorizontal,x_values)
plt.plot(x_values[:-1], velocity)
#%% Integration of the velocity
s = inte.cumtrapz(velocity, x_values[:-1])
plt.plot(x_values[:-2],s)
I hope it's clear what I want to do. Thanks for helping me!
I didn't dig all the way through your code, but the following tricks may be useful.
Say you have x and y values:
x = np.linspace(0,3,100)
y = x**2
Now, you only want the values corresponding to, say, .5 < x < 1.5. First, create a boolean mask for the arrays as follows:
mask = np.logical_and(.5 < x, x < 1.5)
(If this seems magical, then run x < 1.5 in your interpreter and observe the results).
Then use this mask to select your desired x and y values:
x_masked = x[mask]
y_masked = y[mask]
Then, you can translate all these values so that the first x,y pair is at the origin:
x_translated = x_masked - x_masked[0]
y_translated = y_masked - y_masked[0]
Is this the type of thing you were looking for?
I am trying to do a rectangular shape function for a signal with ten seconds with values for 1 between 1 and for 4 seconds a 0 for the rest, I looked other problems but they only seemed to cover for repeating pulses while I just want this single pulse. I already tried the code below, but since I am very new to programming I can not seem to get it to work. I also saw this question but since it only gives the absolute values it does not work for me rectangular pulse train in python
y=np.zeros(10)
def rect(x):
x = np.linspace(0, 10, 100)
if 1<=x<=4:
y=1
else:
y=0
return rect(x)
f1=rect(y)
plt.plot(y,f1)
There are two ways to do: Long way using your functional approach and a short vectorized way. I present both:
Longway: Call the function within a for loop and append the values of y to a list.
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0, 10, 100)
def rect(i):
if 1<=i<=4:
y=1
else:
y=0
return y
f1 = []
for i in x:
f1.append(rect(i))
plt.plot(x,f1)
plt.show()
Short way: Create a conditional mask and apply it to your y-array to fill it with 0 and 1 depending on the condition.
x = np.linspace(0, 10, 100)
mask = (x>=1) & (x <=4)
y = np.where(mask, 1, 0)
plt.plot(x,y)
plt.show()
Good evening.
I'm currently trying to solve the 1D Schrödinger eq. (time independent) with the Numerov method. The derivation of the method is clear to me but I have some problems with the implementation. I tried to look for solutions on google, and there are some (like this one or this one), but I don't really understand what they are doing in their codes...
The Problem:
With some math you can get the equation to this form:
where . For the beginning I'd like to look at the potential V(x)=1 if -a<x<a.
Since I don't have values for the energy or the first values of Psi (which are needed to start the algorithm) I just guessed some...
The code looks like this:
import numpy as np
import matplotlib.pyplot as plt
from scipy.constants import hbar
m= 1e-27
E= 0.5
def numerov_step(psi_1,psi_2,k1,k2,k3,h):
#k1=k_(n-1), k2=k_n, k3=k_(n+1)
#psi_1 = psi_(n-1) and psi_2=psi_n
m = 2*(1-5/12. * h**2 * k2**2)*psi_2
n = (1+1/12.*h**2*k1**2)*psi_1
o = 1 + 1/12. *h**2 *k3**2
return (m-n)/o
def numerov(N,x0,xE,a):
x,dx = np.linspace(x0,xE,N+1,retstep=True)
def V(x,a):
if (np.abs(x)<a):
return 1
else:
return 0
k = np.zeros(N+1)
for i in range(len(k)):
k[i] = 2*m*(E-V(x[i],a))/hbar**2
psi= np.zeros(N+1)
psi[0]=0
psi[1]=0.1
for j in np.arange(2,N):
psi[j+1]= numerov_step(psi[j],psi[j+1],k[j-1],k[j],k[j+1],dx)
return psi
x0 =-10
xE = 10
N =1000
psi=numerov(N,x0,xE,3)
x = np.linspace(x0,xE,N+1)
plt.figure()
plt.plot(x,psi)
plt.show()
Since the plot doesn't look like a wavefunction at all something has to be wrong, but I'm having trobule to find out what it is.. Would be nice if someone could help a little.
Thanks Sito
Unfortunately I don't quite remember the quantum physics so I don't understand some details. Still I see some bugs in your code:
Why inside numerov_step you square k1, k2 and k3?
In your main cycle
for j in np.arange(2,N):
psi[j+1]= numerov_step(psi[j],psi[j+1],k[j-1],k[j],k[j+1],dx)
you messed up with indices. It looks like this line should be
for j in np.arange(2, N):
psi[j] = numerov_step(psi[j - 2], psi[j - 1], k[j - 2], k[j - 1], k[j], dx)
This is the part I don't really understand. Looking into animation at your first link it looks like this equation has good solutions only for certain combinations of V(x) and E and in other cases it quickly goes wild. It looks like both your V(x) and proportion of E to hbar and V(x) are quite different from the referenced articles and this might be one more reason why the solution goes wild.
I want to draw a Bifurcation diagram of quadratic map in python.
Basically its a plot of x_{n+1}=x_n^2-c and it should look like http://static.sewanee.edu/Physics/PHYSICS123/image99.gif
But I am newbie so I am not sure do I make it right.
My code
import numpy as n
import scipy as s
import pylab as p
xa=0.252
xb=1.99
C=n.linspace(xa,0.001,xb)
iter=100
Y=n.zeros((len(X),iteracje))
i=1
Y0=1
for Y0 in iter:
Y(i+1)=Y0^2-C
for Y0 in iter:
Y(i+1)=Y0^2-C
p.plot(C,Y)
p.show()
My problem is that I don't know how properly write these for loop properly.
Here is some modified code (partial explanation below)
import numpy as n
import scipy as s
import pylab as p
xa=0.252
xb=1.99
C=n.linspace(xa,xb,100)
print C
iter=1000
Y = n.ones(len(C))
for x in xrange(iter):
Y = Y**2 - C #get rid of early transients
for x in xrange(iter):
Y = Y**2 - C
p.plot(C,Y, '.', color = 'k', markersize = 2)
p.show()
First, the linspace command had the wrong format. help(s.linspace) will give you insight into the syntax. The first two arguments are start and stop. The third is how many values. I then made Y a numpy array of the same length as C, but whose values were all 1. Your Y0 was simply the number 1, and it never changed. Then I did some iteration to get past the initial conditions. Then did more iteration plotting each value.
To really understand what I've done, you'll have to look at how numpy handles calculations with arrays.