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How to calculate average values of list elements in python?

I'd like to calculate average of each values in a list. To do so, I wrote a function which gets list as parameter and calculate the average and returns the list of average again.
Here is the signal:
random_data = [10 * random.uniform(0,1) for i in range(1000)]
random_peak = [100 * random.uniform(0,1) for i in range(50)] + [0] * 950
random.shuffle(peak)
for i in range(0, len(signal)):
signal = [peak[x] + random_data[x] for x in range(len(random_data))]
And now, I'd like to calculate m as following.
'''
m1 = 1/(number of signal) * x1
m2 = 1/(number of signal) * (x1+x2)
m3 = 1/(number of signal) * (x1+x2+x3)
...
'''
I wrote a following function to calculate m. How would I change the function to return list of m s?
def mean_values(s):
for i in range(len(s)):
m[i] = 1/len(s)*s[i]
return m[i]
mean_values(signal)
#mean_values(np.array(signal)

use m as a float instead of list it make more sense
to get mean
s = 1 / n * Σxi
you can use this to get new mean from a previous one
s' = s + (x1 - s) / n1
where s is the lastest mean, x1 the new value and n1 the new length
However in numpy their is a prebuilt function np.mean() which do that and manage python list too

Python - Multiprocessing with multiple for-loops

I know there are other questions asked concerning this topic so I'm sorry I have to ask it again, but I cannot get it to work since I'm quite new to this topic.
I have four for-loop (nested) in which certain algbraic calculations are done (matrix operations for example). These calculations take too much time to complete, so I was hoping I could speed this up with Multiprocessing.
The code is given below. I simulated the ranges and matrix sizes here, but in my code these ranges are really used (so it's not strange that it takes so long). You should be able to run it directly when copy-paste the code.
import numpy as np
from scipy.linalg import fractional_matrix_power
import math
#Lists for the loop (and one value)
x_list = np.arange(0, 32, 1)
y_list = np.arange(0, 32, 1)
a_list = np.arange(0, 501, 1)
b_list = np.arange(0, 501, 1)
c_list = np.arange(0, 64, 1)
d_number = 32
#Matrices
Y = np.arange(2048).reshape(32, 64)
g = np.asmatrix(np.empty([d_number, 1], dtype=np.complex_))
A = np.empty([len(a_list), len(b_list), len(c_list)], dtype=np.complex_)
A_in = np.empty([len(a_list), len(b_list)], dtype=np.complex_)
for ai in range(len(a_list)):
for bi in range(len(b_list)):
for ci in range(len(c_list)):
f_k_i = c_list[ci]
X_i = np.asmatrix([Y[:, ci]]).T
for di in range(d_number):
r = math.sqrt((x_list[di] - a_list[ai])**2 + (y_list[di] - b_list[bi])**2 + 63**2)
g[di, 0] = np.exp(-2 * np.pi * 1j * f_k_i * (r / 8)) / r #g is a vector
A[-bi, -ai, ci] = ((1 / np.linalg.norm(g)**2) * (((g.conj().T * fractional_matrix_power((X_i * X_i.conj().T), (1/5)) * g) / np.linalg.norm(g)**2)**2)).item(0)
A_in[-bi, -ai] = (1 / len(c_list)) * sum(A[-bi, -ai, :])
What is the best way to approach this? If multiprocessing is the solution, how to implement this for my case (since I couldn't figure that out).
Thanks in advance.

One way to approach it would be to move the two inside loops into a function taking ai and bi as parameters and returning the indexes and the result. Then use multiprocessing.Pool.imap_unordered() to run the function on ai, bi pairs. Something like this (untested):
def partial_calc(index):
"""
This function replaces the inner two loops to calculate the value of
A_in[-bi, -ai]. index is a tuple (ai, bi).
"""
ai, bi = index
for ci in range(len(c_list)):
f_k_i = c_list[ci]
X_i = np.asmatrix([Y[:, ci]]).T
for di in range(d_number):
r = math.sqrt((x_list[di] - a_list[ai])**2 + (y_list[di] - b_list[bi])**2 + 63**2)
g[di, 0] = np.exp(-2 * np.pi * 1j * f_k_i * (r / 8)) / r #g is a vector
A[-bi, -ai, ci] = ((1 / np.linalg.norm(g)**2) * (((g.conj().T * fractional_matrix_power((X_i * X_i.conj().T), (1/5)) * g) / np.linalg.norm(g)**2)**2)).item(0)
return ai, bi, (1 / len(c_list)) * sum(A[-bi, -ai, :])
def main():
with multiprocessing.Pool(None) as p:
# this replaces the outer two loops
indices = itertools.product(range(len(a_list)), range(len(b_list)))
partial_results = p.imap_unordered(partial_calc, indices)
for ai, bi, value in partial_results:
A_in[-bi, -ai] = value
#... do something with A_in ...
if __name__ == "__main__":
main()
Or put the inner three loops into the function and generate one "row" for A_in at a time. Profile it both ways and see which is faster.
The trick will be setting up the lists (a_list, b_list, etc) and the Y matrix. And that depends on their characteristics (constant, quickly/slowly calculated, large/small, etc).

Convergence tests of Leapfrog method for vectorial wave equation in Python

Considering the following Leapfrog scheme used to discretize a vectorial wave equation with given initial conditions and periodic boundary conditions. I have implemented the scheme and now I want to make numerical convergence tests to show that the scheme is of second order in space and time.
I'm mainly struggling with two points here:
I'm not 100% sure if I implemented the scheme correctly. I really wanted to use slicing because it is so much faster than using loops.
I don't really know how to get the right error plot, because I'm not sure which norm to use. In the examples I have found (they were in 1D) we've always used the L2-Norm.
import numpy as np
import matplotlib.pyplot as plt
# Initial conditions
def p0(x):
return np.cos(2 * np.pi * x)
def u0(x):
return -np.cos(2 * np.pi * x)
# exact solution
def p_exact(x, t):
# return np.cos(2 * np.pi * (x + t))
return p0(x + t)
def u_exact(x, t):
# return -np.cos(2 * np.pi * (x + t))
return u0(x + t)
# function for doing one time step, considering the periodic boundary conditions
def leapfrog_step(p, u):
p[1:] += CFL * (u[:-1] - u[1:])
p[0] = p[-1]
u[:-1] += CFL * (p[:-1] - p[1:])
u[-1] = u[0]
return p, u
# Parameters
CFL = 0.3
LX = 1 # space length
NX = 100 # number of space steps
T = 2 # end time
NN = np.array(range(50, 1000, 50)) # list of discretizations
Ep = []
Eu = []
for NX in NN:
print(NX)
errorsp = []
errorsu = []
x = np.linspace(0, LX, NX) # space grid
dx = x[1] - x[0] # spatial step
dt = CFL * dx # time step
t = np.arange(0, T, dt) # time grid
# TEST
# time loop
for time in t:
if time == 0:
p = p0(x)
u = u0(x)
else:
p, u = leapfrog_step(p, u)
errorsp.append(np.linalg.norm((p - p_exact(x, time)), 2))
errorsu.append(np.linalg.norm((u - u_exact(x, time)), 2))
errorsp = np.array(errorsp) * dx ** (1 / 2)
errorsu = np.array(errorsu) * dx ** (1 / 2)
Ep.append(errorsp[-1])
Eu.append(errorsu[-1])
# plot the error
plt.figure(figsize=(8, 5))
plt.xlabel("$Nx$")
plt.ylabel(r'$\Vert p-\bar{p}\Vert_{L_2}$')
plt.loglog(NN, 15 / NN ** 2, "green", label=r'$O(\Delta x^{2})$')
plt.loglog(NN, Ep, "o", label=r'$E_p$')
plt.loglog(NN, Eu, "o", label=r'$E_u$')
plt.legend()
plt.show()
I would really appreciate it if someone could quickly check the implementation of the scheme and an indication on how to get the error plot.

Apart from the initialization, I see no errors in your code.
As to the initialization, consider the first step. There you should compute, per the method description, approximations for p(dt,j*dx) from the values of p(0,j*dx) and u(0.5*dt, (j+0.5)*dx). This means that you need to initialize at time==0
u = u_exact(x+0.5*dx, 0.5*dt).
and also need to compare the then obtained solution against u_exact(x+0.5*dx, time+0.5*dt).
That you obtained the correct order is IMO more an artefact of the test problem than an accidentially still correct algorithm.
If no exact solution is known, or if you want to use a more realistic algorithm in the test, you would need to compute the initial u values from p(0,x) and u(0,x) via Taylor expansions
u(t,x) = u(0,x) + t*u_t(0,x) + 0.5*t^2*u_tt(0,x) + ...
u(0.5*dt,x) = u(0,x) - 0.5*dt*p_x(0,x) + 0.125*dt^2*u_xx(0,x) + ...
= u(0,x) - 0.5*CFL*(p(0,x+0.5*dx)-p(0,x-0.5*dx))
+ 0.125*CFL^2*(u(0,x+dx)-2*u(0,x)+u(0,x-dx)) + ...
It might be sufficient to take just the linear expansion,
u[j] = u0(x[j]+0.5*dx) - 0.5*CFL*(p0(x[j]+dx)-p0(x[j])
or with array operations
p = p0(x)
u = u0(x+0.5*dx)
u[:-1] -= 0.5*CFL*(p[1:]-p[:-1])
u[-1]=u[0]
as then the second order error in the initial data just adds to the general second order error.
You might want to change the space grid to x = np.linspace(0, LX, NX+1) to have dx = LX/NX.
I would define the exact solution and the initial condition the other way around, as that allows more flexibility in the test problems.
# components of the solution
def f(x): return np.cos(2 * np.pi * x)
def g(x): return 2*np.sin(6 * np.pi * x)
# exact solution
def u_exact(x,t): return f(x+t)+g(x-t)
def p_exact(x,t): return -f(x+t)+g(x-t)
# Initial conditions
def u0(x): return u_exact(x,0)
def p0(x): return p_exact(x,0)

Normalise all vectors on the z axis

I have written code that calculates the angle between two vectors. However the way in which is does this is to start with two vectors, rotate each according to some euler angles calculated in a separate program, then calculate the angle between the vectors.
Up until now I have been working with a use case that means both starting vectors are (0,0,1) that makes life super easy. I could just take one set of euler angles away from the other and then calculate the angle between 0,0,1 and the vector that had been rotated by the difference. It meant I could plot nice distribution plots and vector diagrams because everything was normalised to 0,0,1. (I have 1000s of these vectors for the record).
No I am trying to write in a function that would allow for a use case where the two starting vectors are not on 0,0,1. I figured the easiest way to do this would be to calculate direction of the vector relative to 0,0,1 and after calculating the position of the vector just rotate by the precalculated offsets. (this might be a stupid way to do it, if it is please tell me).
MY current code works for a case where a vector is 0,1,0 but then breaks down if i start entering random numbers.
import numpy as np
import math
def RotationMatrix(axis, rotang):
"""
This uses Euler-Rodrigues formula.
"""
#Input taken in degrees, here we change it to radians
theta = rotang * 0.0174532925
axis = np.asarray(axis)
#Ensure axis is a unit vector
axis = axis/math.sqrt(np.dot(axis, axis))
#calclating a, b, c and d according to euler-rodrigues forumla requirments
a = math.cos(theta/2)
b, c, d = axis*math.sin(theta/2)
a2, b2, c2, d2 = a*a, b*b, c*c, d*d
bc, ad, ac, ab, bd, cd = b*c, a*d, a*c, a*b, b*d, c*d
#Return the rotation matrix
return np.array([[a2+b2-c2-d2, 2*(bc-ad), 2*(bd+ac)],
[2*(bc+ad), a2+c2-b2-d2, 2*(cd-ab)],
[2*(bd-ac), 2*(cd+ab), a2+d2-b2-c2]])
def ApplyRotationMatrix(vector, rotationmatrix):
"""
This function take the output from the RotationMatrix function and
uses that to apply the rotation to an input vector
"""
a1 = (vector[0] * rotationmatrix[0, 0]) + (vector[1] * rotationmatrix[0, 1]) + (vector[2] * rotationmatrix[0, 2])
b1 = (vector[0] * rotationmatrix[1, 0]) + (vector[1] * rotationmatrix[1, 1]) + (vector[2] * rotationmatrix[1, 2])
c1 = (vector[0] * rotationmatrix[2, 0]) + (vector[1] * rotationmatrix[2, 1]) + (vector[2] * rotationmatrix[2, 2])
return np.array((a1, b1, c1)
'''
Functions for Calculating the angles of 3D vectors relative to one another
'''
def CalculateAngleBetweenVector(vector, vector2):
"""
Does what it says on the tin, outputs an angle in degrees between two input vectors.
"""
dp = np.dot(vector, vector2)
maga = math.sqrt((vector[0] ** 2) + (vector[1] ** 2) + (vector[2] ** 2))
magb = math.sqrt((vector2[0] ** 2) + (vector2[1] ** 2) + (vector2[2] ** 2))
magc = maga * magb
dpmag = dp / magc
#These if statements deal with rounding errors of floating point operations
if dpmag > 1:
error = dpmag - 1
print('error = {}, do not worry if this number is very small'.format(error))
dpmag = 1
elif dpmag < -1:
error = 1 + dpmag
print('error = {}, do not worry if this number is very small'.format(error))
dpmag = -1
angleindeg = ((math.acos(dpmag)) * 180) / math.pi
return angleindeg
def CalculateAngleAroundZ(Vector):
X,Y,Z = Vector[0], Vector[1], Vector[2]
AngleAroundZ = math.atan2(Y, X)
AngleAroundZdeg = (AngleAroundZ*180)/math.pi
return AngleAroundZdeg
def CalculateAngleAroundX(Vector):
X,Y,Z = Vector[0], Vector[1], Vector[2]
AngleAroundZ = math.atan2(Y, Z)
AngleAroundZdeg = (AngleAroundZ*180)/math.pi
return AngleAroundZdeg
def CalculateAngleAroundY(Vector):
X,Y,Z = Vector[0], Vector[1], Vector[2]
AngleAroundZ = math.atan2(X, Z)
AngleAroundZdeg = (AngleAroundZ*180)/math.pi
return AngleAroundZdeg
V1 = (0,0,1)
V2 = (3,5,4)
Xoffset = (CalculateAngleAroundX(V2))
Yoffset = (CalculateAngleAroundY(V2))
Zoffset = (CalculateAngleAroundZ(V2))
XRM = RotationMatrix((1,0,0), (Xoffset * 1))
YRM = RotationMatrix((0,1,0), (Yoffset * 1))
ZRM = RotationMatrix((0,0,1), (Zoffset * 1))
V2 = V2 / np.linalg.norm(V2)
V2X = ApplyRotationMatrix(V2, XRM)
V2XY = ApplyRotationMatrix(V2X, YRM)
V2XYZ = ApplyRotationMatrix(V2XY, ZRM)
print(V2XYZ)
print(CalculateAngleBetweenVector(V1, V2XYZ))
Any advice to fix this problem will be much appreciated.

I'm not sure to fully understand what you need but if it is to compute the angle between two vectors in space you can use the formula:
where a.b is the scalar product and theta is the angle between vectors.
thus your function CalculateAngleBetweenVector becomes:
def CalculateAngleBetweenVector(vector, vector2):
return math.acos(np.dot(vector,vector2)/(np.linalg.norm(vector)* np.linalg.norm(vector2))) * 180 /math.pi
You can also simplify your ApplyRotationMatrix function:
def ApplyRotationMatrix(vector, rotationmatrix):
"""
This function take the output from the RotationMatrix function and
uses that to apply the rotation to an input vector
"""
return rotationmatrix # vector
the # symbol is the matrix product
Hope this will help you. Feel free to precise your request if this is not helpfull.

Im an idiot I just needed to do the cross product and the dot product and rotate by the dot product *-1 around the cross product.

python code to calculate angle between three point using their 3D coordinates

I have write down a code to calculate angle between three points using their 3D coordinates.
import numpy as np
a = np.array([32.49, -39.96,-3.86])
b = np.array([31.39, -39.28, -4.66])
c = np.array([31.14, -38.09,-4.49])
f = a-b # normalization of vectors
e = b-c # normalization of vectors
angle = dot(f, e) # calculates dot product
print degrees(cos(angle)) # calculated angle in radians to degree
output of the code:
degree 33.4118214995
but when i used one of the software to calculate the same it gives output bit different 120 degree. please help
reference i have used to write the program:
(How to calculate bond angle in protein db file?)

Your original code is pretty close. Adomas.m's answer is not very idiomatic numpy:
import numpy as np
a = np.array([32.49, -39.96,-3.86])
b = np.array([31.39, -39.28, -4.66])
c = np.array([31.14, -38.09,-4.49])
ba = a - b
bc = c - b
cosine_angle = np.dot(ba, bc) / (np.linalg.norm(ba) * np.linalg.norm(bc))
angle = np.arccos(cosine_angle)
print np.degrees(angle)

I guess numpy is quite enough:
from numpy import *
from numpy.linalg import norm
a = array([32.49, -39.96,-3.86])
b = array([31.39, -39.28, -4.66])
c = array([31.14, -38.09,-4.49])
f = b-a
e = b-c
abVec = norm(f)
bcVec = norm(e)
abNorm = f / abVec;
bcNorm = e / bcVec;
res = abNorm[0] * bcNorm[0] + abNorm[1] * bcNorm[1] + abNorm[2] * bcNorm[2];
angle = arccos(res)*180.0/ pi
print angle
also the res can be calculated with dot:
res = abNorm[0] * bcNorm[0] + abNorm[1] * bcNorm[1] + abNorm[2] * bcNorm[2];
res = dot(abNorm, bcNorm)

For 2D, you can use this method using the math library.
import math
def getAngle(a, b, c):
ang = math.degrees(math.atan2(c[1]-b[1], c[0]-b[0]) - math.atan2(a[1]-b[1], a[0]-b[0]))
return ang + 360 if ang < 0 else ang
print(getAngle((5, 0), (0, 0), (0, 5)))
Credits: https://manivannan-ai.medium.com/find-the-angle-between-three-points-from-2d-using-python-348c513e2cd

In case you have a big list of (x,y,z) coordinates, this works:
import numpy
def compute_angle_between_3d_points(a,b,c):
ba = a - b
bc = c - b
cosine_numerator = np.sum(ba*bc, axis=1)
cosine_denominator_1 = np.linalg.norm(ba, axis=1)
cosine_denominator_2 = np.linalg.norm(bc, axis=1)
cosine_angle = cosine_numerator / (cosine_denominator_1 * cosine_denominator_2)
angles = np.arccos(cosine_angle)
degree_angles = np.rad2deg(angles)
return degree_angles
Above, a,b,c are presumed to be of shape (N_Points, 3). Something in TensorFlow or Torch would surely be faster, but there you go.