I understand that there are a lot of answers on this topic but I have scrutinized all of them and did not find something suitable for me.
I'm sure that error is childish but still can not find a solution.
I want to take some element from numpy.linspace.
import numpy
#Porosity range
phi = numpy.linspace(0.1, 1, num=10)
mu = [1, 10, 100, 1000]
Example for how it looks like but not in loop and it works:
mu_total3 = mu[0]*phi[2]+ mu[1]*(1 - phi[2])
print(mu_total3)
7.3
What I want in following:
for x in phi:
mu_total = mu[0]*phi[x]+ mu[1]*(1 - phi[x])
print(mu_total)
Numpy is specialised at doing vector operations. That is taking an one or two arrays and applying an operation to all its elements. For python lists you might write:
zs = []
for x, y in zip(xs, ys):
z = x + 2*y
zs.append(z)
print(zs)
Wheras with a numpy array you can write:
zs = xs + 2*ys
print(zs)
Applied to your code that becomes:
mu_totals = mu[0]*phi + mu[1]*(1 - phi)
Related
I am attempting to write a program which constructs a matrix and performs a singular value decomposition on it. I am evaluating the function ax^2 +bx + 1 on a grid. I then make a uniform meshgrid of a and b. The rows of the matrix correspond to different quadratic coefficients, while each column corresponds to a grid point at which the function is evaluated.
The matlab code is here:
% Collect data
x = linspace(-1,1,100);
[a,b] = meshgrid(0:0.1:1,0:0.1:1);
D=zeros(numel(x),numel(a));
sz = size(D)
% Build “Dose” matrix
for i=1:numel(a)
D(:,i) = a(i)*x.^2+b(i)*x+1;
end
% Do the SVD:
[U,S,V]=svd(D,'econ');
D_reconstructed = U*S*V';
plot(diag(S))
scatter3(a(:),b(:),V(:,1))
This is my attempt at a solution:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(-1, 1, 100)
def f(x, a, b):
return a*x*x + b*x + 1
a, b = np.mgrid[0:1:0.1,0:1:0.1]
#a = b = np.arange(0,1,0.01)
D = np.zeros((x.size, a.size))
for i in range(a.size):
D[i] = a[i]*x*x +b[i]*x +1
U, S, V = np.linalg.svd(D)
plt.plot(np.diag(S))
fig = plt.figure()
ax = plt.axes(projection="3d")
ax.scatter(a, b, V[0])
but I always get broadcasting errors which I am not sure how to fix.
Firstly, in MATLAB you're assigning to D(:,i), but in python you're assigning to D[i]. The latter is equivalent to D[i, ...] which is in your case D[i, :]. Instead you seem to need D[:, i].
Secondly, in MATLAB using a linear index into a 2d array (namely a and b) will give you flattened views. If you do that with numpy you get slices of an array instead, just as I mentioned with D[i].
You can do away with the loop with broadcasting and getting your desired 2d array by .ravelling (or reshaping) your a and b arrays:
x = np.linspace(-1, 1, 100)[:, None] # inject trailing singleton for broadcasting
a, b = np.mgrid[0:1:0.1, 0:1:0.1]
D = a.ravel() * x**2 + b.ravel() * x + 1
The way this works is that x has shape (100, 1) after we inject a trailing singleton (in MATLAB trailing singletons are implied, in numpy leading ones), and both a.ravel() and b.ravel() have shape (10*10,) which is compatible with (1, 10*10), making broadcasting possible into shape (100, 10*10). You could also replace the calls to ravel with
a, b = np.mgrid[...].reshape(2, -1)
which is a trick I sometimes use, but this is harder to read if you're unfamiliar with the pattern.
Side note: it's better to use example data where dimensions end up being of different size so that you notice if something ends up being transposed.
Write a function that normalizes a vector (finds the unit vector). A vector can be normalized by dividing each individual component of the vector by its magnitude. Your input for this function will be a vector i.e. 1 dimensional list containing 3 integers.
According to the solution devised, I have considered a predefined list of 3 elements. But if I want to apply loops, then please explain me that how I could deduce the solution using loops. I tried working on the problem. This is my solution so far:
from math import sqrt
def vector_normalization(my_vector):
result = 0
for x in my_vector:
result = result + (x ** 2)
magnitude = sqrt(result)
nx_vector = my_vector[0] / magnitude
ny_vector = my_vector[1] / magnitude
nz_vector = my_vector[2] / magnitude
n_vector = [nx_vector, ny_vector, nz_vector]
return n_vector
Now, after I calculate the magnitude using for loop of some random list, according to my program I will get only three elements in the list as the output. But I want all the elements in the random list to be normalized. Please suggest me the way to achieve the same.
Also, you can use high order functions in Python like map:
vec = [1,2,3]
magnitude = sqrt(sum(map(lambda x: x**2, vec)))
normalized_vec = list(map(lambda x: x/magnitude, vec))
So normalized_vec becomes:
[0.2672612419124244, 0.5345224838248488, 0.8017837257372732]
Or using Numpy:
import numpy as np
arr = np.array([1,2,3])
arr_normalized = arr/sqrt(sum(arr**2))
arr_normalized results in:
array([ 0.26726124, 0.53452248, 0.80178373])
Please try the following code,
vector = [1,2,4]
y=0
for x in vector:
y+=x**2
y = y**0.5
unit_vector = []
for x in vector:
unit_vector.append(x/y)
Hope this helps.
def vector_normalization(vec):
result = 0
for x in vec:
result = result + (x**2)
magnitude = (result)**0.5
x = vec[0]/magnitude
y = vec[1]/magnitude
z = vec[2]/magnitude
vec = [x,y,z]
return vec
I have a python (NumPy) function which creates a uniform random quaternion. I would like to get two quaternion multiplication as 2-dimensional returned array from the same or an another function. The formula of quaternion multiplication in my recent case is Q1*Q2 and Q2*Q1. Here, Q1=(w0, x0, y0, z0) and Q2=(w1, x1, y1, z1) are two quaternions. The expected two quaternion multiplication output (as 2-d returned array) should be
return([-x1*x0 - y1*y0 - z1*z0 + w1*w0, x1*w0 + y1*z0 - z1*y0 +
w1*x0, -x1*z0 + y1*w0 + z1*x0 + w1*y0, x1*y0 - y1*x0 + z1*w0 +
w1*z0])
Can anyone help me please? My codes are here:
def randQ(N):
#Generates a uniform random quaternion
#James J. Kuffner 2004
#A random array 3xN
s = random.rand(3,N)
sigma1 = sqrt(1.0 - s[0])
sigma2 = sqrt(s[0])
theta1 = 2*pi*s[1]
theta2 = 2*pi*s[2]
w = cos(theta2)*sigma2
x = sin(theta1)*sigma1
y = cos(theta1)*sigma1
z = sin(theta2)*sigma2
return array([w, x, y, z])
I know that the question is old but as I found it interesting, for future reference I herewith write an answer: if no special data type for quaternions is desirable, then a quaternion can be written as a tuple of a real number and a normal vector as an array of floats. Thus, mathematically, based on the process mentioned here, the Hamilton product of two quaternions $\hat{q}_1=(w_1,\mathbf{v}_1k$ and $\hat{q}_2=(w_2,\mathbf{v}_2)$ would be:
$$\hat{q}_1 \hat{q}_2=(w_1 w_2-\mathbf{v}^T_1\mathbf{v}_2, w_1 \mathbf{v}_2+w_2 \mathbf{v}_1+\mathbf{v}_1\times \mathbf{v}_2)$$
Sorry for the math notation that cannot be rendered in Stack Overflow.
Thus in numpy:
import numpy as np
q1=(w1,v1)
q2=(w2,v2)
q1q2=(w1*w2-np.matmul(v1.T,v2),w1*v2+w2*v1+np.cross(v1,v2))
A simple rendition of your request would be:
In [70]: def multQ(Q1,Q2):
...: w0,x0,y0,z0 = Q1 # unpack
...: w1,x1,y1,z1 = Q2
...: return([-x1*x0 - y1*y0 - z1*z0 + w1*w0, x1*w0 + y1*z0 - z1*y0 +
...: w1*x0, -x1*z0 + y1*w0 + z1*x0 + w1*y0, x1*y0 - y1*x0 + z1*w0 +
...: w1*z0])
...:
In [72]: multQ(randQ(1),randQ(2))
Out[72]:
[array([-0.37695449, 0.79178506]),
array([-0.38447116, 0.22030199]),
array([ 0.44019022, 0.56496059]),
array([ 0.71855397, 0.07323243])]
The result is a list of 4 arrays. Just wrap it in np.array() to get a 2d array:
In [73]: M=np.array(_)
In [74]: M
Out[74]:
array([[-0.37695449, 0.79178506],
[-0.38447116, 0.22030199],
[ 0.44019022, 0.56496059],
[ 0.71855397, 0.07323243]])
I haven't tried to understand or clean up your description - just rendering it as working code.
A 2-Dimensional Array is an array like this: foo[0][1]
You don't need to do that. Multiplying two quaternions yields one single quaternion. I don't see why you would need a two-dimensional array, or how you would even use one.
Just have a function that takes two arrays as arguments:
def multQuat(q1, q2):
then return the relevant array.
return array([-q2[1] * q1[1], ...])
I know the post is pretty old but would like to add a function using the pyquaternion library to calculate quaternion multiplication. The quaternion multiplication mentioned in the question is called the Hamilton product. You can use it like below...
from pyquaternion import Quaternion
q1 = Quaternion()
q2 = Quaternion()
q1_q2 = q1*q2
You can find more about this library here http://kieranwynn.github.io/pyquaternion/
There is a Python module that adds a quaternion dtype to NumPy.
Please check out the documentation for the quaternion module here.
Here is an example from the documentation. It looks native to the usage of NumPy.
>>> import numpy as np
>>> import quaternion
>>> np.quaternion(1,0,0,0)
quaternion(1, 0, 0, 0)
>>> q1 = np.quaternion(1,2,3,4)
>>> q2 = np.quaternion(5,6,7,8)
>>> q1 * q2
quaternion(-60, 12, 30, 24)
>>> a = np.array([q1, q2])
>>> a
array([quaternion(1, 2, 3, 4), quaternion(5, 6, 7, 8)], dtype=quaternion)
>>> np.exp(a)
array([quaternion(1.69392, -0.78956, -1.18434, -1.57912),
quaternion(138.909, -25.6861, -29.9671, -34.2481)], dtype=quaternion)
I want to interpolate some 3-d data using the scipy LinearNDInterpolator function (Python 2.7). I can't quite figure out how to use it, though: below is my attempt. I'm getting the error ValueError: different number of values and points. This leads me to believe that the shape of "coords" is not appropriate for these data, but it looks in the documentation like the shape is okay.
Note that in the data I really want to use (instead of this example) the spacing of my grid is irregular, so something like RegularGridInterpolator will not do the trick.
Thanks very much for your help!
def f(x,y,z):
return 2 * x**3 + 3 * y**2 - z
x = np.linspace(1,2,2)
y = np.linspace(1,2,2)
z = np.linspace(1,2,2)
data = f(*np.meshgrid(x, y, z, indexing='ij', sparse=True))
coords = np.zeros((len(x),len(y),len(z),3))
coords[...,0] = x.reshape((len(x),1,1))
coords[...,1] = y.reshape((1,len(y),1))
coords[...,2] = z.reshape((1,1,len(z)))
coords = coords.reshape(data.size,3)
my_interpolating_function = LinearNDInterpolator(coords,data)
pts = np.array([[2.1, 6.2, 8.3], [3.3, 5.2, 7.1]])
print(my_interpolating_function(pts))
I have two arrays, lets say x and y that contain a few thousand datapoints.
Plotting a scatterplot gives a beautiful representation of them. Now I'd like to select all points within a certain radius. For example r=10
I tried this, but it does not work, as it's not a grid.
x = [1,2,4,5,7,8,....]
y = [-1,4,8,-1,11,17,....]
RAdeccircle = x**2+y**2
r = 10
regstars = np.where(RAdeccircle < r**2)
This is not the same as an nxn array, and RAdeccircle = x**2+y**2 does not seem to work as it does not try all permutations.
You can only perform ** on a numpy array, But in your case you are using lists, and using ** on a list returns an error,so you first need to convert the list to numpy array using np.array()
import numpy as np
x = np.array([1,2,4,5,7,8])
y = np.array([-1,4,8,-1,11,17])
RAdeccircle = x**2+y**2
print RAdeccircle
r = 10
regstars = np.where(RAdeccircle < r**2)
print regstars
>>> [ 2 20 80 26 170 353]
>>> (array([0, 1, 2, 3], dtype=int64),)