I have a 2D numpy-array as input for a basic sigmoid-classifier.
I would like the classifier to return an array with the probabilities.
import numpy as np
def sigmoid(x):
sigm = 1 / (1 + np.exp(-x))
return sigm
def p(D, w, b):
prob=sigmoid(np.dot(D[:][7],w)+b)
return prob
How can I change p() so that it returns a 1D numpy array with the probabilities listed in order of the input data ?
Atm "prob" is an array of length 14, however the input array "D" is over 400 in size, so there is an error somewhere in the logic.
EDIT: The issue is the incorrect slicing of the array. One has to use D[:,7] instead of D[:][7] to extract a column.
Perhaps like this:
import numpy as np
def sigmoid(x):
sigm = 1 / (1 + np.exp(-x))
return sigm
def p(D, w, b):
prob=sigmoid(np.dot(D[:][7],w)+b)
return np.ravel(prob)
Related
I need help to compute a mathematical expression using only numpy operations. The expression I want to compute is the following :
Where : x is an (N, S) array and f is a numpy function (that can work with broadcastable arrays e.g np.maximum, np.sum, np.prod, ...). If that is of importance, in my case f is a symetric function.
So far my code looks like this:
s = 0
for xp in x: # Loop over N...
s += np.sum(np.prod(f(xp, x), axis=1))
And still has loop that I'd like to get rid of.
Typically N is "large" (around 30k) but S is small (less than 20) so if anyone can find a trick to only loop over S this would still be a major improvement.
I belive the problem is easy by N-plicating the array but one of size (32768, 32768, 20) requires 150Go of RAM that I don't have. However, (32768, 32768) fits in memory though I would appreciate a solution that does not allocate such array.
Maybe a use of np.einsum with well-chosen arrays is possible?
Thanks for your replies. If any information is missing let me know!
Have a nice day !
Edit 1 :
Form of f I'm interested in includes (for now) : f(x, y) = |x - y|, f(x, y) = |x - y|^2, f(x, y) = 2 - max(x, y).
Your loop is very efficient. Some possible ways are
Method-1 (looping over S)
import numpy as np
def f(x,y):
return np.abs(x-y)
N = 200
S = 20
x_data = random.rand(N,S) #(i,s)
y_data = random.rand(N,S) #(i',s)
product = f(broadcast_to(x_data[:,0][...,None],(N,N)) ,broadcast_to(y_data[:,0][...,None],(N,N)).T)
for i in range(1,S):
product *= f(broadcast_to(x_data[:,i][...,None],(N,N)) ,broadcast_to(y_data[:,i][...,None],(N,N)).T)
sum = np.sum(product)
Method-2 (dispatching S number of blocks)
import numpy as np
def f(x,y):
x1 = np.broadcast_to(x[:,None,...],(x.shape[0],y.shape[0],x.shape[1]))
y1 = np.broadcast_to(y[None,...],(x.shape[0],y.shape[0],x.shape[1]))
return np.abs(x1-y1)
def f1(x1,y1):
return np.abs(x1-y1)
N = 5000
S = 20
x_data = np.random.rand(N,S) #(i,s)
y_data = np.random.rand(N,S) #(i',s)
def fun_new(x_data1,y_data1):
s = 0
pp =np.split(x_data1,S,axis=0)
for xp in pp:
s += np.sum(np.prod(f(xp, y_data1), axis=2))
return s
def fun_op(x_data1,y_data1):
s = 0
for xp in x_data1: # Loop over N...
s += np.sum(np.prod(f1(xp, y_data1), axis=1))
return s
fun_new(x_data,y_data)
I have the following np.array which describes a rectangular wave. I would like to transform it into a callable function with a continuous argument. The np.array is:
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import seed
from numpy.random import rand
def piecewise_control( off_times,amp_inj, period_inj ):
def select(T):
return lambda t: (-T/2 <= t) & (t < T/2)
def pulse_train(t, at, shape):
return np.sum(shape(t - at[:,np.newaxis]), axis=0)
for i in range(1,len(off_times)):
off_times[i] += off_times[i-1] + period_inj
return amp_inj*pulse_train(t,off_times,shape=select(period_inj))
t=np.linspace(0,100,10000)
off_times = 10*rand(10)
period_inj = 1
amp_inj = 1
control = piecewise_control( off_times,amp_inj, period_inj )
plt.plot(t,control)
plt.show()
This answer inspired me.
The plot is the following:
The question is: can we transform the array control into a function with a continuous argument?
Of course if we did:
def ccontrol(t, control):
return control[t]
unfortunately we would get a function which only depends on integers.
You can subclass numpy.ndarray and implement the __call__ method:
import numpy as np
class MyArray(np.ndarray):
def __call__(self, idx):
return self[idx]
control = np.random.rand(100)
control_view = control.view(MyArray)
print(control_view(5), control[5])
For interpolation you can use scipy.interpolate. In fact, interpolation routines can return functions that you can call with any input, not necessarily integers.
I'm getting an import error for "norm". What am I not doing correct??
I'm open to constructive feedback on improving the code, however I have to keep the parameters as they are!
Thanks!!!
Code is below:
import numpy as np
from numpy import norm, inalg, array, zeros, diag, diagflat, dot, linalg
"""Test Case Data"""
A = np.matrix([[4,-1,-1],[-2,6,1],[-1,1,7]])
b = np.matrix([[3],[9],[-6]])
x = np.matrix([[0],[0],[0]])
"""Main Function"""
def jacobi(A, b, x, Tolerance, Iterations):
V = np.diag(A)
D = np.diag(V)
R = D-A
D_I = D.I
D = np.asmatrix(D)
Counter_1 = 1
tol_gauge = 100
while Counter_1 <= Iterations:
# I considered using the "dot" function in NUMPY but I was wary of mixed results
iterative_approach_form = D_I * ((R*x)+b)
tol_gauge = np.linalg.norm(iterative_approach_form-x)
x = iterative_approach_form
if initial_tol <= Tolerance:
return("The Solution x = {},y={}, z={} ".format(x[0], x[1], x[2]))
return("The Solution was found in %s interation(s)" %(Counter_1))
else:
pass
Counter_1 +=1
return("The Solution was not found in {} iteration(s)".format(Iterations))
You need to specify which numpy module you are importing from. The following works if you want to use a function only by its name:
from numpy import linalg
from numpy.linalg import norm
from numpy import zeros, array, diag, diagflat, dot
Looking at you code however, you don't need the second import line, because in the rest of the code the numpy functions are specified according to the accepted norm. For example, norm is already present in your code as np.linalg.norm.
There are three more issues with your code: 1) initial_tol is not assigned a value; 2) tol_gauge is assigned but not used in the code; 3) the last return statement is not indented properly (perhaps only here) and the same is very likely for the block in your while loop.
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'd like to multiply two vectors, one column (i.e., (N+1)x1), one row (i.e., 1x(N+1)) to give a (N+1)x(N+1) matrix. I'm fairly new to Numpy but have some experience with MATLAB, this is the equivalent code in MATLAB to what I want in Numpy:
n = 0:N;
xx = cos(pi*n/N)';
T = cos(acos(xx)*n');
in Numpy I've tried:
import numpy as np
n = range(0,N+1)
pi = np.pi
xx = np.cos(np.multiply(pi / float(N), n))
xxa = np.asarray(xx)
na = np.asarray(n)
nd = np.transpose(na)
T = np.cos(np.multiply(np.arccos(xxa),nd))
I added the asarray line after I noticed that without it Numpy seemed to be treating xx and n as lists. np.shape(n), np.shape(xx), np.shape(na) and np.shape(xxa) gives the same result: (100001L,)
np.multiply only does element by element multiplication. You want an outer product. Use np.outer:
np.outer(np.arccos(xxa), nd)
If you want to use NumPy similar to MATLAB, you have to make sure that your arrays have the right shape. You can check the shape of any NumPy array with arrayname.shape and because your array na has shape (4,) instead of (4,1), the transpose method is effectless and multiply calculates the dot product. Use arrayname.reshape(N+1,1) resp. arrayname.reshape(1,N+1) to transform your arrays:
import numpy as np
n = range(0,N+1)
pi = np.pi
xx = np.cos(np.multiply(pi / float(N), n))
xxa = np.asarray(xx).reshape(N+1,1)
na = np.asarray(n).reshape(N+1,1)
nd = np.transpose(na)
T = np.cos(np.multiply(np.arccos(xxa),nd))
Since Python 3.5, you can use the # operator for matrix multiplication. So it's a walkover to get code that's very similar to MATLAB:
import numpy as np
n = np.arange(N + 1).reshape(N + 1, 1)
xx = np.cos(np.pi * n / N)
T = np.cos(np.arccos(xx) # n.T)
Here n.T denotes the transpose of n.