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+--- Thread: How to plot implicit functions (with two variables) in scipy python using matplotlib? (/thread-18228.html)
How to plot implicit functions (with two variables) in scipy python using matplotlib? - Jay_Nerella - May-09-2019
I have a function with two independent variables x & y and three parameters a,b & c. I done optimization of parameters using least_squares. Now I want to plot the function using the obtained parameters.
xdata = [0.3063191028,-0.0156344344,-0.0155750443,-0.7206687321,-0.7473645659,-0.9174428618,-0.8839320182,-1.0399645639,-0.9997277955,-1.0157928079,-0.9888297188,-0.4533985964,0.0091163748,0.0026577054,0.5926386016,0.5992457462,1.004345373,0.9909529136,1.0392221881,1.0405287695,1.0606471537,1.0283835014,1.0149316519,0.9416591604,0.9685628594,0.9155869223,0.9088016075,0.6344640542,0.6142268898]
ydata = [1.1154790304,0.9978867036,1.0111900779,0.5702040049,0.5903372366,-0.0072010453,-0.0007720708,-0.45206232,-0.4390262009,-1.0375889614,-0.9978570085,-1.0612855969,-0.957932038,-0.904673998,-0.7489532501,-0.7689528542,-0.0266364437,-0.0265473586,0.2857701407,0.5784443763,0.5849624358,0.8579043226,0.8446900334,0.9316519346,0.9580805131,1.091470597,1.071560078,1.1199481327,1.0868233245]
xdata = np.array(xdata)
ydata = np.array(ydata)
def func1(coeff,x,y):
# x = X[0]
# y = X[1]
n = 8
# % A = ydata
# % B = -xdata
# % C = xdata. - ydata
# % H = zdata
g = np.subtract(x,y)
I_0 = np.subtract(x,y) # x-y = C
I_1 = np.multiply(I_0,coeff[2]) # c(x-y) = cC
I_2 = np.multiply(coeff[1],-x) #b(-x) = bB
I_3 = np.multiply(coeff[0],y) # aA
I3_0 = np.subtract(I_1,I_2) # cC-bB
I3_1 = np.subtract(I_3,I_1) # aA-cC
I3_2 = np.subtract(I_2,I_3) # bB-aA
I3_00 = np.multiply(I3_0,I3_1) # (cC-bB)(aA-cC)
I3_01 = np.multiply(I3_00,I3_2) # (cC-bB)(aA-cC)(bB-aA)
I3 = np.divide(I3_01,54) # (cC-bB)(aA-cC)(bB-aA)/54
I2_0 = np.power((I3_1),2) # (aA-cC)^2
I2_1 = np.power((I3_0),2) # (cC-bB)^2
I2_2 = np.power((I3_2),2) # (bB-aA)^2
I2_00 = np.add(I2_0,I2_1) # (aA-cC)^2 + (cC-bB)^2
I2_01 = np.add(I2_00,I2_2) # (aA-cC)^2 + (cC-bB)^2 + (bB-aA)^2
I2 = np.divide(I2_01,54) # ((aA-cC)^2 + (cC-bB)^2 + (bB-aA)^2)/54
th_0 = np.divide(I3,(np.power(I2,(3/2)))) # I3/(I2^(3/2))
# print(th_0)
th = np.arccos(np.clip((th_0),-1,1)) # arccos(I3/(I2^(3/2)))
# print(th)
ans_0 = np.divide(np.add((2*th),(np.pi)),6) # (2*th + pi)/6
ans_1 = np.divide(np.add((2*th),(3*np.pi)),6) # (2*th + 3*pi)/6
ans_2 = np.divide(np.add((2*th),(5*np.pi)),6) # (2*th + 5*pi)/6
ans_00 = np.multiply(np.cos(ans_0),2) # 2*cos((2*th + pi)/6)
ans_11 = np.multiply(np.cos(ans_1),2) # 2*cos((2*th + 3*pi)/6)
ans_22 = np.multiply(np.cos(ans_2),2) # 2*cos((2*th + 5*pi)/6)
ans_000 = np.power(np.absolute(ans_00),n) # (abs(2*cos((2*th + pi)/6)))^n
ans_111 = np.power(np.absolute(ans_11),n) # (abs(2*cos((2*th + 3*pi)/6)))^n
ans_222 = np.power(np.absolute(ans_22),n) # (abs(2*cos((2*th + 5*pi)/6)))^n
ans_0000 = np.add((np.power(np.absolute(ans_00),n)),(np.power(np.absolute(ans_11),n))) # (abs(2*cos((2*th + pi)/6)))^n + (abs(2*cos((2*th + 3*pi)/6)))^n
ans_1111 = np.add((ans_0000),(np.power(np.absolute(ans_22),n))) # (abs(2*cos((2*th + pi)/6)))^n + (abs(2*cos((2*th + 3*pi)/6)))^n + (abs(2*cos((2*th + 5*pi)/6)))^n
sna_0 = np.power(np.multiply(3,I2),(n/2)) # (3*I2)^(n/2) !!
sna_1 = 2*(np.power(1.0167,n)) # 2*(sigma^n) !!
sna_00 = np.multiply(sna_0,ans_1111)
sna_11 = np.subtract(sna_00,sna_1)
return sna_11
x0 = np.array([1.0, 1.0, 1.0])
res_lsq = least_squares(func1, x0,loss='cauchy',f_scale=0.001,args=(xdata, ydata))
res_lsq.x
RE: How to plot implicit functions (with two variables) in scipy python using matplotlib? - scidam - May-11-2019
xmin, xmax, ymin, ymax = xdata.min(), xdata.max(), ydata.min(), ydata.max() g = lambda x, y: func1(res_lsq.x, x, y)**2 X, Y = np.meshgrid(np.linspace(xmin-0.2, xmax+0.2, 1000), np.linspace(ymin-0.2, ymax+0.2, 1000)) Z = g(X, Y) # heavy computations.... # plotting the result plt.contourf(X, Y, np.log(Z), levels=[-10,-8]) plt.gca().scatter(xdata, ydata) plt.colorbar() plt.show()