I have a third-order ODE with time-varying coefficients of the form
y''' + A1(t)y'' + A2(t)y' + A3(t)y = 0
Where A1, A2, and A3 contain differentiation of a different function
R(t) = a - b* atan (c*t-d)
I have written the code below but I keep getting "can't convert expression to float error" error.
My python skill is not strong.
I would appreciate if I could get an advice on how to approach this.
Thanks
#import math
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from sympy import *
#from math import *
def du_dt(y,x):
return [ y[1], y[2], (- A1*y[2] - A2*y[1] + A3*y[0]) ]
V = 2.5e3
Ra = 7.5e3
C1 = 3.3e-6
RL = 1.10
L = 3.32e-6
C2 = 3.16e-12
Rd = 0.5
w1 = 1/(C1*L)
T1 = Ra*C1
C23 = C1*C2/(C1+C2)
w2 = 1/(L*C23)
R5 = 0.015 - 0.006*atan(x-5)
Tm = L/(RL+R5)
A1 = 1/Tm + 1/T1
A2 = (1/Tm).diff(x)+w2+(1/L)*R5.diff(x)+1/(T1*Tm)
A3 = (1/L)*R5.diff(x,2)+(1/(L*T1))*R5.diff(x)+(w2-w1)/T1
t = np.linspace(0,0.2,1000) # time range
y0 = [3.0425e10,0,0] # initial values
y = odeint(du_dt, y0, t)
# TypeError: can't convert expression to float
Related
Right now the rectangle signal is centre on x = 4, how can I make it centre on x = 0
def rect(n,T):
a = np.zeros(int((n-T)/2,))
b = np.ones((T,))
c= np.zeros(int((n-T)/2,))
a1 = np.append(a,b)
a2 = np.append(a1,c)
return a2
x =rect(11,6)
plt.step(x, 'r')
plt.show()
This is so far that I wrote. Appreciate anyone can give the Idea
A method to center the rectangle at x=0 is to provide x values to plt.step. One way to accomplish this is to use numpy arange and center the x values around 0 by using the length of a2 returned in the rects function
# Changed to y because it will be our y values in plt.step
y = rect(11, 6)
# Add 0.5 so it's centered
x = np.arange(-len(y)/2 + 0.5, len(y)/2 + 0.5)
And then plot it using plt.step and setting where to mid (more info in the plt.step docs):
plt.step(x, y, where='mid', color='r')
Hope this helps. Here is the full code:
import numpy as np
import matplotlib.pyplot as plt
def rect(n, T):
a = np.zeros(int((n-T)/2,))
b = np.ones((T,))
c = np.zeros(int((n-T)/2,))
a1 = np.append(a, b)
a2 = np.append(a1, c)
return a2
y = rect(11, 6)
# Add 0.5 so it's centered
x = np.arange(-len(y)/2 + 0.5, len(y)/2 + 0.5)
plt.step(x, y, where='mid', color='r')
plt.show()
I'm working with 2 imu's. I need to offset all frames with the first frame from the sensor. I have created a fictive scenario, where I precisely know the rotation and the wanted result. I need the two sensors to show the same result when their initial (start) orientation is subtracted.
import numpy as np
# Sensor 0,1 and 2 start orientation in degrees
s0_x = 30
s0_y = 0
s0_z = 0
s1_x = 0
s1_y = 40
s1_z = 0
s2_x = 10
s2_y = 40
s2_z= -10
# Change from start frame 1
x1 = 20
y1 = 10
z1 = 0
# Change from start frame 2
x2 = 60
y2 = 30
z2 = 30
GCS= [[1,0,0],[0,1,0],[0,0,1]]
sensor0 = [[s0_x, s0_y, s0_z], [s0_x, s0_y, s0_z], [s0_x, s0_y, s0_z]]
sensor1 = [[s1_x, s1_y, s1_z], [s1_x + x1, s1_y + y1, s1_z + z1],[s1_x + x1 + x2, s1_y + y1+ y2, s1_z + z1+ z2]]
sensor2 = [[s2_x, s2_y, s2_z], [s2_x + x1, s2_y + y1, s2_z + z1], [s2_x + x1+ x2, s2_y + y1+ y2, s2_z + z1+ z2]]
def Rot_Mat_X(theta):
r = np.array([[1,0,0],[0,np.cos(np.deg2rad(theta)),-np.sin(np.deg2rad(theta))],[0,np.sin(np.deg2rad(theta)),np.cos(np.deg2rad(theta))]])
return r
# rotation the rotation matrix around the Y axis (input in deg)
def Rot_Mat_Y(theta):
r = np.array([[np.cos(np.deg2rad(theta)),0,np.sin(np.deg2rad(theta))],
[0,1,0],
[-np.sin(np.deg2rad(theta)),0,np.cos(np.deg2rad(theta))]])
return r
# rotation the rotation matrix around the Z axis (input in deg)
def Rot_Mat_Z(theta):
r = np.array([[np.cos(np.deg2rad(theta)),-np.sin(np.deg2rad(theta)),0],
[np.sin(np.deg2rad(theta)),np.cos(np.deg2rad(theta)),0],
[0,0,1]])
return r
# Creating the rotation matrices
r_sensor0 = []
r_sensor1= []
r_sensor2= []
for i in range(len(sensor1)):
r_sensor1_z = np.matmul(Rot_Mat_X(sensor1[i][0]),GCS)
r_sensor1_zy = np.matmul(Rot_Mat_Y(sensor1[i][1]),r_sensor1_z)
r_R_Upperarm_medial_zyx = np.matmul(Rot_Mat_Z(sensor1[i][2]),r_sensor1_zy )
r_sensor1.append(r_R_Upperarm_medial_zyx )
r_sensor2_z = np.matmul(Rot_Mat_X(sensor2[i][0]),GCS)
r_sensor2_zy = np.matmul(Rot_Mat_Y(sensor2[i][1]),r_sensor2_z )
r_sensor2_zyx = np.matmul(Rot_Mat_Z(sensor2[i][2]),r_sensor2_zy )
r_sensor2.append(r_sensor2_zyx )
r_start_sensor1 = r_sensor1[0]
r_start_sensor2 = r_sensor2[0]
r_offset_sensor1 = []
r_offset_sensor2 = []
for i in range(len(sensor0)):
r_offset_sensor1.append(np.matmul(np.transpose(r_start_sensor1),r_sensor1[i]))
r_offset_sensor2.append(np.matmul(np.transpose(r_start_sensor2),r_sensor2[i]))
# result:
r_offset_sensor1[0] = [[1,0,0],[0,1,0],[0,0,1]]
r_offset_sensor1[1] = [[0.984,0.059,0.163],[0,0.939,-0.342],[-0.173,0.336,0.925]]
r_offset_sensor1[2] = [[0.748,0.466,0.471],[0.086,0.635,-0.767],[-0.657,0.615,0.434]]
r_offset_sensor2[0] = [[1,0,0],[0,1,0],[0,0,1]]
r_offset_sensor2[1] = [[0.984,0.086,0.150],[-0.03,0.938,-0.344],[-0.171,0.334,0.926]]
r_offset_sensor2[2] = [[0.748,0.541,0.383],[-0.028,0.603,-0.797],[-0.662,0.585,0.466]]
I expect the result of sensors 1 and 2 to be equal for all frames but it doesn't? And they should be:
frame[0] = [1,0,0],[0,1,0],[0,0,1]
frame[1] = [0.984,0,0.173],[0.059,0.939,-0.336],[-0.163,0.342,0.9254]
frame[2] = [0.750,-0.433,0.50],[0.625,0.216,-0.750],[0.216,0.875,0.433]
I'm trying to sequentially sample from a Gaussian Process prior.
The problem is that the samples eventually converge to zero or diverge to infinity.
I'm using the basic conditionals described e.g. here
Note: the kernel(X,X) function returns the squared exponential kernel with isometric noise.
Here is my code:
n = 32
x_grid = np.linspace(-5,5,n)
x_all = []
y_all = []
for x in x_grid:
x_all = [x] + x_all
X = np.array(x_all).reshape(-1, 1)
# Mean and covariance of the prior
mu = np.zeros((X.shape), np.float)
cov = kernel(X, X)
if len(mu)==1: # first sample is not conditional
y = np.random.randn()*cov + mu
else:
# condition on all previous samples
u1 = mu[0]
u2 = mu[1:]
y2 = np.atleast_2d(np.array(y_all)).T
C11 = cov[:1,:1] # dependent sample
C12 = np.atleast_2d(cov[0,1:])
C21 = np.atleast_2d(cov[1:,0]).T
C22 = np.atleast_2d(cov[1:, 1:])
C22_ = la.inv(C22)
u = u1 + np.dot(C12, np.dot(C22_, (y2 - u2)))
C22_xC21 = np.dot(C22_, C21)
C_minus = np.dot(C12, C22_xC21) # this weirdly becomes larger than C!
C = C11 - C_minus
y = u + np.random.randn()*C
y_all = [y.flatten()[0]] + y_all
Here's an example with 32 samples, where it collapses:
enter image description here
Here's an example with 34 samples, where it explodes:
enter image description here
(for this particular kernel, 34 is the number of samples at which (or more) the samples start to diverge.
I have two dataframes (X & Y). I would like to link them together and to predict the probability that each potential match is correct.
X = pd.DataFrame({'A': ["One", "Two", "Three"]})
Y = pd.DataFrame({'A': ["One", "To", "Free"]})
Method A
I have not yet fully understood the theory but there is an approach presented in:
Sayers, A., Ben-Shlomo, Y., Blom, A.W. and Steele, F., 2015. Probabilistic record linkage. International journal of epidemiology, 45(3), pp.954-964.
Here is my attempt to implementat it in Pandas:
# Probability that Matches are True Matches
m = 0.95
# Probability that non-Matches are True non-Matches
u = min(len(X), len(Y)) / (len(X) * len(Y))
# Priors
M_Pr = u
U_Pr = 1 - M_Pr
O_Pr = M_Pr / U_Pr # Prior odds of a match
# Combine the dataframes
X['key'] = 1
Y['key'] = 1
Z = pd.merge(X, Y, on='key')
Z = Z.drop('key',axis=1)
X = X.drop('key',axis=1)
Y = Y.drop('key',axis=1)
# Levenshtein distance
def Levenshtein_distance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
L_D = np.vectorize(Levenshtein_distance, otypes=[float])
Z["D"] = L_D(Z['A_x'], Z['A_y'])
# Max string length
def Max_string_length(X, Y):
return max(len(X), len(Y))
M_L = np.vectorize(Max_string_length, otypes=[float])
Z["L"] = M_L(Z['A_x'], Z['A_y'])
# Agreement weight
def Agreement_weight(D, L):
return 1 - ( D / L )
A_W = np.vectorize(Agreement_weight, otypes=[float])
Z["C"] = A_W(Z['D'], Z['L'])
# Likelihood ratio
def Likelihood_ratio(C):
return (m/u) - ((m/u) - ((1-m) / (1-u))) * (1-C)
L_R = np.vectorize(Likelihood_ratio, otypes=[float])
Z["G"] = L_R(Z['C'])
# Match weight
def Match_weight(G):
return math.log(G) * math.log(2)
M_W = np.vectorize(Match_weight, otypes=[float])
Z["R"] = M_W(Z['G'])
# Posterior odds
def Posterior_odds(R):
return math.exp( R / math.log(2)) * O_Pr
P_O = np.vectorize(Posterior_odds, otypes=[float])
Z["O"] = P_O(Z['R'])
# Probability
def Probability(O):
return O / (1 + O)
Pro = np.vectorize(Probability, otypes=[float])
Z["P"] = Pro(Z['O'])
I have verified that this gives the same results as in the paper. Here is a sensitivity check on m, showing that it doesn't make a lot of difference:
Method B
These assumptions won't apply to all applications but in some cases each row of X should match a row of Y. In that case:
The probabilities should sum to 1
If there are many credible candidates to match to then that should reduce the probability of getting the right one
then:
X["I"] = X.index
# Combine the dataframes
X['key'] = 1
Y['key'] = 1
Z = pd.merge(X, Y, on='key')
Z = Z.drop('key',axis=1)
X = X.drop('key',axis=1)
Y = Y.drop('key',axis=1)
# Levenshtein distance
def Levenshtein_distance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
L_D = np.vectorize(Levenshtein_distance, otypes=[float])
Z["D"] = L_D(Z['A_x'], Z['A_y'])
# Max string length
def Max_string_length(X, Y):
return max(len(X), len(Y))
M_L = np.vectorize(Max_string_length, otypes=[float])
Z["L"] = M_L(Z['A_x'], Z['A_y'])
# Agreement weight
def Agreement_weight(D, L):
return 1 - ( D / L )
A_W = np.vectorize(Agreement_weight, otypes=[float])
Z["C"] = A_W(Z['D'], Z['L'])
# Normalised Agreement Weight
T = Z .groupby('I') .agg({'C' : sum})
D = pd.DataFrame(T)
D.columns = ['T']
J = Z.set_index('I').join(D)
J['P1'] = J['C'] / J['T']
Comparing it against Method A:
Method C
This combines method A with method B:
# Normalised Probability
U = Z .groupby('I') .agg({'P' : sum})
E = pd.DataFrame(U)
E.columns = ['U']
K = Z.set_index('I').join(E)
K['P1'] = J['P1']
K['P2'] = K['P'] / K['U']
We can see that method B (P1) doesn't take account of uncertainty whereas method C (P2) does.
Below is a forward pass and partly implemented backward pass of back propagation of a neural network :
import numpy as np
def sigmoid(z):
return 1 / (1 + np.exp(-z))
X_train = np.asarray([[1,1], [0,0]]).T
Y_train = np.asarray([[1], [0]]).T
hidden_size = 2
output_size = 1
learning_rate = 0.1
forward propagation
w1 = np.random.randn(hidden_size, 2) * 0.1
b1 = np.zeros((hidden_size, 1))
w2 = np.random.randn(output_size, hidden_size) * 0.1
b2 = np.zeros((output_size, 1))
Z1 = np.dot(w1, X_train) + b1
A1 = sigmoid(Z1)
Z2 = np.dot(w2, A1) + b2
A2 = sigmoid(Z2)
derivativeA2 = A2 * (1 - A2)
derivativeA1 = A1 * (1 - A1)
first steps of back propagation
error = (A2 - Y_train)
dA2 = error / derivativeA2
dZ2 = np.multiply(dA2, derivativeA2)
What is the intuition behind :
error = (A2 - Y_train)
dA2 = error / derivativeA2
dZ2 = np.multiply(dA2, derivativeA2)
I understand error is the difference between the current prediction A2 and actual values Y_train.
But why divide this error by the derivative of A2 and then multiply the result of error / derivativeA2 by derivativeA2 ? What is intuition behind this ?
These expressions are indeed confusing:
derivativeA2 = A2 * (1 - A2)
error = (A2 - Y_train)
dA2 = error / derivativeA2
... because error doesn't have a meaning on its own. At this point, the goal is compute the derivative of the cross-entropy loss, which has this formula:
dA2 = (A2 - Y_train) / (A2 * (1 - A2))
See these lecture notes (formula 6) for the derivation. It just happens that the previous operation is sigmoid and its derivative is A2 * (1 - A2). That's why this expression is used again to compute dZ2 (formula 7).
But if you had a different loss function (say, L2) or a different squeeze layer, then A2 * (1 - A2) wouldn't be reused. These are different nodes in the computational graph.