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I am currently trying to do a regression of a function calculated via a RK4 method performed on a non-linear Volterra integral of the second kind. The problem I found is that the code is extremely slow, for 1 call of the curve_fit function (fitt), it takes about 30-40 minute to generate a data. Overall, there will be a lot of calls to fitt before the parameters are determined, this takes more than 6 hours to run. Is there anyway to optimize this code? Thanks in advance!
from scipy.special import gamma
from ml_internal import LTInversion
from scipy.optimize import curve_fit , fsolve
from scipy.misc import derivative
from sklearn.metrics import r2_score
from math import comb , factorial
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
sns.set()
# Gets the data
df = pd.read_excel('D:\\CoMat\\Fractional_fit\\optimized\\data_optimized.xlsx')
skipTime = 1
skipIndex = df[df['Time']== skipTime].index.values[0]
xls = pd.read_excel('D:\\CoMat\\Fractional_fit\\optimized\\data_optimized.xlsx',skiprows=np.arange(1,skipIndex+1,1))
timeDF = xls['Time']
tempDF = xls['Temp']
taDF = xls['Ta']
timeDF = timeDF - timeDF[0]
tempDF = tempDF + 273.15
t0 = tempDF[0]
ta = sum(taDF)/len(taDF)
ta = ta + 273.15
###########################################
#Spliting into intervals
h = 0.05
a = 0
b = timeDF[len(timeDF)-1]
N = int(np.round((b-a)/h))
#Each xi
def xidx(index):
return a + h*index
#Function in the image are written here.
def gx(t,lamda,alpha):
return t0 * ml(lamda*(t**alpha),alpha)
gx = np.vectorize(gx)
def kernel(t,s,rad,lamda,alpha,beta):
if t == s:
return 0
return (t-s)**(alpha-1) * ml_(lamda*((t-s)**alpha),alpha,alpha,1) * (beta*(rad**4) - beta*(ta**4) - lamda*ta)
kernel = np.vectorize(kernel)
############################
# The problem is here!!!!!!
def fx(x,n,lamda,alpha,beta):
ans = gx(x,lamda,alpha)
for j in range(n):
ans += (h/6)*(kernel(x,xidx(j),f0[j],lamda,alpha,beta) + 2*kernel(x,xidx(j+1/2),f1[j],lamda,alpha,beta) + 2*kernel(x,xidx(j+1/2),f2[j],lamda,alpha,beta) + kernel(x,xidx(j+1),f3[j],lamda,alpha,beta))
return ans
#########################
f0 = np.zeros(N+1)
f0[0] = t0
f1 = np.zeros(N+1)
f2 = np.zeros(N+1)
f3 = np.zeros(N+1)
F = np.zeros((3,N+1))
def fitt(xvalue,lamda,alpha,beta):
global f0,f1,f2,f3,F
n = int(np.round(xvalue/h))
f1[n] = fx(xidx(n) + 1/2,n,lamda,alpha,beta) + (h/2)*kernel(xidx(n + 1/2),xidx(n),f0[n],lamda,alpha,beta)
f2[n] = fx(xidx(n + 1/2),n,lamda,alpha,beta)
f3[n] = fx(xidx(n+1),n,lamda,alpha,beta) + h*kernel(xidx(n+1),xidx(n+1/2),f2[n],lamda,alpha,beta)
if n+1 <= N:
f0[n+1] = fx(xidx(n+1),n,lamda,alpha,beta) + (h/6)*(kernel(xidx(n+1),xidx(n),f0[n],lamda,alpha,beta) + 2*kernel(xidx(n+1),xidx(n+1/2),f1[n],lamda,alpha,beta) + 2*kernel(xidx(n+1),xidx(n+1/2),f2[n],lamda,alpha,beta) + kernel(xidx(n+1),xidx(n+1),f3[n],lamda,alpha,beta))
if(xvalue == timeDF[len(timeDF) - 1]):
print(f0[n],n)
returnValue = f0[n]
f0 = np.zeros(N+1)
f0[0] = t0
f1 = np.zeros(N+1)
f2 = np.zeros(N+1)
f3 = np.zeros(N+1)
return returnValue
print(f0[n],n)
return f0[n]
fitt = np.vectorize(fitt)
#Fitting, plotting and giving (Adj) R-squared
popt , pcov = curve_fit(fitt,timeDF,tempDF,p0=(-0.1317,0.95,-1e-11),bounds=((-np.inf,0,-np.inf),(0,1,0)))
print(popt)
y_fit = np.array(fitt(timeDF,popt[0],popt[1],popt[2]))
plt.scatter(timeDF,tempDF,color='ORANGE',marker='.',s=0.5)
plt.fill_between(timeDF,tempDF-0.5,tempDF+0.5,color='ORANGE', alpha=0.2)
plt.plot(timeDF,y_fit,color='RED',linewidth=1)
plt.legend(["Experimental data", "Caputo fit"], loc ="upper right")
plt.xlabel("Time (min)")
plt.ylabel("Temperature (Kelvin)")
plt.show()
plt.close()
r2 = r2_score(tempDF,y_fit)
print(r2)
adjr2 = 1 - (1 - r2)*((len(xls)-1)/(len(xls)-3-1))
print(adjr2)
I already tried computing the values f0,f1,f2,f3 all at once, but the thing consuming the most time is Fn(x) which I haven't figured in out how to compute them all at once. If this is possible to compute at once, I think the program will run much faster. PS: ML,ML_ is a function from https://github.com/khinsen/mittag-leffler.
This is the function necesssary. Fn is the only one I haven't figured out yet.
There are two typing errors in the cited image. The combination of x_n and 1/2 is always meant to be the midpoint x_{n+1/2} = x_n + h/2. The second error is a duplication of x_{n+1/2} in the formula for f^{(4)}_n in its third term. The first error is probably producing errors that are large enough to make convergence complicated and any limit wrong for the intended problem.
In the Simpson/RK4 step, the 4 fx computations can be reduced to 2.
The F_n implement the left side of the integral equation
F(x) = g(x) + int(s=0 to x of K(x,s,f(s))
where the integral is approximated with the sample sequences f0,...,f3. Due to the structure of problem and algorithm F_n(x_n)=f^0_n = f^4_{n-1}.
Note that K(x,s,f) should be set to zero for s >= x. In the exact version of the equation these values "above the diagonal" are not used.
If an increase in accuracy is needed, for instance to avoid divergence where there is none in the exact solution, you can decrease the step site by a factor of 10 and then sub-sample the f^0_n sequence to produce the numerical guess for the given data. Other factors than 10 are of course also possible.
I'm trying to develop a regression model with constraints on effect from the independent variables. So my model equation is y = a0 + a1x1 + a2x2 with 200 datapoints. What I want to achieve is sum(a1x1) over 200 datapoints should fall in certain range i.e. lb1<sum(a1x1)<ub1. I am using Gekko for the optimization part and a got stuck while applying this condition.
I am using the following code where ubdict is the dictionary for the boundaries:
m = gk.GEKKO(remote=False)
m.options.IMODE=2 #Regression mode
y = np.array(df['y']) #dependant vars for optimization
x = np.array(df[X]) #array of independent vars for optimization
n = x.shape[1] #number of variables
c = m.Array(m.FV, n+1) #array of parameters and intercept
for ci in c:
ci.STATUS = 1 #calculate fixed parameter
xp = [None]*n
#load data
xd = m.Array(m.Param,n)
yd = m.Param(value=y)
for i in range(n):
xd[i].value = x[:,i]
xp[i] = m.Var()
if ubound_dict[i] >= 0:
xp[i] = m.Var(lb=0, ub=ubdict[i])
elif ubound_dict[i] < 0:
xp[i] = m.Var(lb=ubdict[i], ub=0)
m.Equation(xp[i]==c[i]*xd[i])
yp = m.Var()
m.Equation(yp==m.sum([xp[i] for i in range(n)] + [c[n]]))
#Minimize difference between actual and predicted y
m.Minimize((yd-yp)**2)
#APOPT solver
m.options.SOLVER = 1
#Solve
m.solve(disp=True)
#Retrieve parameter values
a = [i.value[0] for i in c]
print(a)
But this is applying the constraint row-wise. What I want is something like
xp[i] = m.Var(lb=0, ub=ubdict[i])
m.Equation(xp[i]==sum(c[i]*xd[i]) over observations)
Any suggestion would be of great help!
Below is a similar problem with sample data.
Regression Mode with IMODE=2
Use the m.vsum() object in Gekko with IMODE=2. Gekko lets you write the equations once and then applies the data to each equation. This is more efficient for large-scale data sets.
import numpy as np
from gekko import GEKKO
# load data
x1 = np.array([1,2,5,3,2,5,2])
x2 = np.array([5,6,7,2,1,3,2])
ym = np.array([3,2,3,5,6,7,8])
# model
m = GEKKO()
c = m.Array(m.FV,3)
for ci in c:
ci.STATUS=1
x1 = m.Param(value=x1)
x2 = m.Param(value=x2)
ymeas = m.Param(value=ym)
ypred = m.Var()
m.Equation(ypred == c[0] + c[1]*x1 + c[2]*x2)
# add constraint on sum(c[1]*x1) with vsum
v1 = m.Var(); m.Equation(v1==c[1]*x1)
con = m.Var(lb=0,ub=10); m.Equation(con==m.vsum(v1))
m.Minimize((ypred-ymeas)**2)
m.options.IMODE = 2
m.solve()
print('Final SSE Objective: ' + str(m.options.objfcnval))
print('Solution')
for i,ci in enumerate(c):
print(i,ci.value[0])
# plot solution
import matplotlib.pyplot as plt
plt.figure(figsize=(8,4))
plt.plot(ymeas,ypred,'ro')
plt.plot([0,10],[0,10],'k-')
plt.xlabel('Meas')
plt.ylabel('Pred')
plt.savefig('results.png',dpi=300)
plt.show()
Optimization Mode (IMODE=3)
The optimization mode 3 allows you to write each equation and objective term individually. Both give the same solution.
import numpy as np
from gekko import GEKKO
# load data
x1 = np.array([1,2,5,3,2,5,2])
x2 = np.array([5,6,7,2,1,3,2])
ym = np.array([3,2,3,5,6,7,8])
n = len(ym)
# model
m = GEKKO()
c = m.Array(m.FV,3)
for ci in c:
ci.STATUS=1
yp = m.Array(m.Var,n)
for i in range(n):
m.Equation(yp[i]==c[0]+c[1]*x1[i]+c[2]*x2[i])
m.Minimize((yp[i]-ym[i])**2)
# add constraint on sum(c[1]*x1)
s = m.Var(lb=0,ub=10); m.Equation(s==c[1]*sum(x1))
m.options.IMODE = 3
m.solve()
print('Final SSE Objective: ' + str(m.options.objfcnval))
print('Solution')
for i,ci in enumerate(c):
print(i,ci.value[0])
# plot solution
import matplotlib.pyplot as plt
plt.figure(figsize=(8,4))
ypv = [yp[i].value[0] for i in range(n)]
plt.plot(ym,ypv,'ro')
plt.plot([0,10],[0,10],'k-')
plt.xlabel('Meas')
plt.ylabel('Pred')
plt.savefig('results.png',dpi=300)
plt.show()
For future questions, please create a simple and complete example that demonstrates the issue.
I am trying to numerically solve a rather complex system of ordinary differential equations. Odeint gives error I guess for the excessive computational work. Any idea on alternatives?
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.colors as mcolors
import scipy
from scipy.integrate import odeint
# declare parameters:
k1 = 2
k2 = 1.8E6
k3 = 48
k3r = 2.8e8
k4 = 1.1e6
k5 = 3
k6 = 6.6e9
k6r = 9.4
k7 = 40
k8 = 7.1
k9 = 0.25
k10 = 0.053
H = 1.28
A = 0.42 #BrO3-
C = 0.0012 #CH2(COOH)2
#define tspan and I.C.
t = np.linspace(0,50,num=5000, endpoint=False)
J0=[0,0.11,0.0012,0,0,0,0.19,0]
def adv_oregonator(j, t):
dxdt = k1*A*j[1]*H**2-k2*H*j[1]*j[0]-k3*H*A*j[0]+k3r*j[5]**2+k4*H*j[5]*(C-j[2])-2*k5*j[0]**2;
dydt = -k1*A*j[1]*H**2-k2*H*j[1]*j[0]-k6*j[3]*j[1]*H+k7*j[4]*j[6]+k9*j[7]*j[2];
dzdt = k4*H*j[5]*(C-j[2])-k9*j[7]*j[2]-k10*j[2]*j[6];
dpdt = k1*A*j[1]*H**2+2*k2*H*j[1]*j[0]+k5*j[0]**2-k6*j[3]*j[1]*H-k8*j[3]*j[6];
dudt = k6*j[3]*j[1]*H-k6r*j[4]-k7*j[4]*j[6];
dwdt = 2*k2*H*A*j[0]-2*k3r*j[5]**2-k4*H*j[5]*(C-j[2]);
dmdt = -k7*j[4]*j[6]-k8*j[3]*j[6]-k10*j[2]*j[6];
dbdt = k7*j[4]*j[6]+k8*j[3]*j[6]-k9*j[2]*j[7];
djdt = [dxdt,dydt,dzdt,dpdt,dudt,dwdt,dmdt,dbdt]
return djdt
j = odeint(adv_oregonator,J0,t)
This is the original system.
And these are the weird plots that I get when I run the script:
I would like to know what formula is being used in statsmodels ARIMA predict/forecast. For a simple AR(1) model I thought that it would be y_t = a1 * y_t-1. However, I am not able to recreate the results produced by forecast or predict.
Here's what I am trying to do:
from statsmodels.tsa.arima.model import ARIMA
import numpy as np
def ar_series(n):
# generate the series y_t = a1 y_t-1 + eps
np.random.seed(1)
y0 = np.random.rand()
y = [y0]
a1 = 0.7 # the AR coefficient
for i in range(1, n):
y.append(a1 * y[i - 1] + 0.3 * np.random.rand())
return np.array(y)
series = ar_series(10)
model = ARIMA(series, order=(1, 0, 0))
fit = model.fit()
#print(fit.summary())
# const = 0.3441; ar.L1 = 0.6518
print(fit.predict())
y_pred = [0.3441]
for i in range(1, 10):
y_pred.append( 0.6518 * series[i-1])
y_pred = np.array(y_pred)
print(y_pred)
The two series don't match and I have no idea how the in-sample predictions are being calculated?
Found the answer here. I think what I was trying to do is valid only if the process mean is zero.
https://faculty.washington.edu/ezivot/econ584/notes/forecast.pdf
In python 2.7, I want to compare 2 image to the same, How to do this? please show me step by step. Thanks!
There are many ways to do. By using some opensource Library, like OpenCV, Scikit Learn, TensorFlow.
To compare two images, you can do something like Template Matching in OpenCV
import cv2
import numpy as np
from matplotlib import pyplot as plt
img = cv2.imread('img.jpg', 0)
img2 = img.copy()
template = cv2.imread('img2.jpg', 0)
w, h = template.shape[::-1]
methods = ['cv2.TM_CCOEFF', 'cv2.TM_CCOEFF_NORMED']
for meth in methods:
img = img2.copy()
method = eval(meth)
res = cv2.matchTemplate(img, template, method)
min_val, max_val, min_loc, max_loc = cv2.minMaxLoc(res)
if method in [cv2.TM_SQDIFF or cv2. TM_SQDIFF_NORMED]:
top_left = min_loc
else:
top_left = max_loc
bottom_right = (top_left[0] + w, top_left[1] + h)
cv2.rectangle(img, top_left, bottom_right, 255,2)
plt.subplot(121), plt.imshow(res)
plt.title('Matching Result'), plt.xticks([]), plt.yticks([])
plt.subplot(122),plt.imshow(img,cmap = 'gray')
plt.title('Detected Point'), plt.xticks([]), plt.yticks([])
plt.suptitle(meth)
plt.show()
or Histogram comparison
import cv2
import numpy as np
base = cv2.imread('test4.jpg')
test1 = cv2.imread('test3.jpg')
test2 = cv2.imread('test5.jpg')
rows,cols = base.shape[:2]
basehsv = cv2.cvtColor(base,cv2.COLOR_BGR2HSV)
test1hsv = cv2.cvtColor(test1,cv2.COLOR_BGR2HSV)
test2hsv = cv2.cvtColor(test2,cv2.COLOR_BGR2HSV)
halfhsv = basehsv[rows/2:rows-1,cols/2:cols-1].copy() # Take lower half of the base image for testing
hbins = 180
sbins = 255
hrange = [0,180]
srange = [0,256]
ranges = hrange+srange # ranges = [0,180,0,256]
ranges=None
histbase = cv2.calcHist(basehsv,[0,1],None,[180,256],ranges)
cv2.normalize(histbase,histbase,0,255,cv2.NORM_MINMAX)
histhalf = cv2.calcHist(halfhsv,[0,1],None,[180,256],ranges)
cv2.normalize(histhalf,histhalf,0,255,cv2.NORM_MINMAX)
histtest1 = cv2.calcHist(test1hsv,[0,1],None,[180,256],ranges)
cv2.normalize(histtest1,histtest1,0,255,cv2.NORM_MINMAX)
histtest2 = cv2.calcHist(test2hsv,[0,1],None,[180,256],ranges)
cv2.normalize(histtest2,histtest2,0,255,cv2.NORM_MINMAX)
for i in xrange(5):
base_base = cv2.compareHist(histbase,histbase,i)
base_half = cv2.compareHist(histbase,histhalf,i)
base_test1 = cv2.compareHist(histbase,histtest1,i)
base_test2 = cv2.compareHist(histbase,histtest2,i)
print "Method: {0} -- base-base: {1} , base-test1: {2}, base_test2: {3}".format(i,base_base,base_test1,base_test2)