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preface
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Self-introduction ଘ(੭, ᵕ)੭
Nickname: Haihong
Tag: programmer monkey | C++ contestant | student
Introduction: because of C language to get acquainted with programming, then transferred to the computer major, had the honor to get some national awards, provincial awards… Has been confirmed. Currently learning C++/Linux/Python
Learning experience: solid foundation + more notes + more code + more thinking + learn English well!
Little White stage of learning Python
The article is only used as my own study notes for the establishment of knowledge system and review
The problem is not to learn one more question to understand one more question
Know what is, know why!
1. Recursive relationship – yeast growth model
The key of difference equation modeling is how to get the relationship between the NTH group of data and the NTH +1 group of data
As you can see here, when we grow bacteria in culture, they usually go through these four phases.
- adjustment
- Logarithmic phase
- plateau
Plot based on existing data:
import matplotlib.pyplot as plt
time = [i for i in range(0.19)]
number = [9.6.18.3.29.47.2.71.1.119.1.174.6.257.3.350.7.441.0.513.3.559.7.594.8.629.4.640.8.651.1.655.9.659.6.661.8]
plt.title('Relationship between time and number')# create title
plt.xlabel('time')# X axis labels
plt.ylabel('number')# Y axis labels
plt.plot(time,number)# drawing
plt.show()# show
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There is a law of yeast population growth: when certain resources can only support a maximum population, but cannot support unlimited population growth, the population growth rate will slow down.
- The difference between the values of the two observation points, △ P, represents the growth rate
- △ P is related to the current population, the larger the population, the faster the growth rate
- △ P is also related to the amount of remaining unallocated resources. The more resources, the faster the growth rate
- Then the remaining resources were represented by the difference between the number of limiting groups and the number of existing populations
Model: Δ p = pn + 1 – pn = k (665 – pn) pn \ Delta p = p_ {n + 1} – p_n = k (665 – p_n) p_n Δ p = pn + 1 – pn = k (665 – pn) pn
Then calculate the coefficient k of the model expression (given pn, δ pp_n, \Delta PPN, δ P). When the formula is fitted as a quadratic curve:
import numpy as np
import matplotlib.pylab as plt
p_n = [9.6.18.3.29.47.2.71.1.119.1.174.6.257.3.350.7.441.0.513.3.559.7.594.8.629.4.640.8.651.1.655.9.659.6]
delta_p = [8.7.10.7.18.2.23.9.48.55.5.82.7.93.4.90.3.72.3.46.4.35.1.34.6.11.4.10.3.4.8.3.7.2.2]
plt.plot(p_n,delta_p)
poly = np.polyfit(p_n, delta_p, 2)
z = np.polyval(poly,p_n)
print(poly)
plt.plot(p_n, z)
plt.show()
# [-8.01975671e-04 5.16054679e-01 6.41123361e+00]
# k = -8.01975671e-04
Copy the codeThe results
665 minus p sub n times p sub n is a function of one order
import numpy as np
import matplotlib.pylab as plt
p_n = [9.6.18.3.29.47.2.71.1.119.1.174.6.257.3.350.7.441.0.513.3.559.7.594.8.629.4.640.8.651.1.655.9.659.6]
delta_p = [8.7.10.7.18.2.23.9.48.55.5.82.7.93.4.90.3.72.3.46.4.35.1.34.6.11.4.10.3.4.8.3.7.2.2]
p_n = np.array(p_n)
x= (665 - p_n) * p_n
plt.plot(x,delta_p)
ploy = np.polyfit(x,delta_p,1)
print(ploy)
z = np.polyval(ploy,x)
plt.plot(x,z)
plt.show()
# [0.00081448 0.30791574]
# k = 0.00081448
Copy the codeThe results
Prediction curve:
import matplotlib.pyplot as plt
p0 = 9.6
p_list = []
for i in range(20):
p_list.append(p0)
p0 = 0.00081448* (665-p0)*p0+p0
plt.plot(p_list)
plt.show()
Copy the codeRunning results:
2 shows the differential – heat conduction equation
One-dimensional heat conduction equation is:
Where, k is the heat conduction coefficient, Equation 2 is the initial value condition of the equation, and Equation 3 and 4 are the boundary value condition. The heat conduction equation is as follows:
import numpy as np
import matplotlib.pylab as plt
x = np.linspace(0.1.100)
y = 4 * x * (1 - x)
plt.plot(x,y)
plt.show()
Copy the codeThe results
N = 25
M = 2500
T = 1.0
X = 1.0
xArray = np.linspace(0.1.0.50)
yArray = map(initialCondition, xArray)
starValues = yArray
U = np.zeros((N+1,M+1))
U[:,0]=starValues
dx=X/N
dt=T/N
kappa=1.0
rho=kappa*dt/dx/dx
for k in range(0,N):
for j in range(1,N):
U[j][k+1]=rho*U[j-1][k]+ (1.-2*rho)*U[j][k]+ rho*U[j+1][k]
U[0][k+1] =0.
U[N][k+1] =0.
pylab.figure(figsize=(12.6))
pylab.plot(xArray, U[:,0])
pylab.plot(xArray, U[:,int(0.10/dt)])
pylab.plot(xArray, U[:,int(0.20/dt)])
pylab.plot(xArray, U[:,int(0.50/ dt)]) pylab. Xlabel ($x $, fontsize =15) pylab. Ylabel (' $r U $' (\ dot, \ tau), fontsize =15Pylab.title (u 'One-dimensional heat conduction equation', fontProperties =font) Pylab.legend ([r '$\tau=0.$', 'r' $\ tau =0.10$', 'r' $\ tau =0.20$', 'r' $\ tau =0.50Fontsize = $']15)
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tArray = np.linspace(0.0.2.int(0.2/dt)+1)
xGride, tGride = np.meshgrid(xArray, tArray)
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
fig = pylab.figure(figsize=(16.10))
ax = fig.add_subplot(1.1.1,projection= '3d') surface = ax.plot_surface(xGride, tGride, U[:,:int(0.2/dt)+1]. J T, cmap = cm. Coolwarm) ax. Set_xlabel (" $$" x, fontdict = {" size ":18"$}) ax. Set_ylabel (r \ tau $", fontdict = {" size" :18}) ax. Set_zlabel (r $U $", "fontdict = {" size" :18}) ax.set_title(u "Heat conduction equation $u_\\tau = u_{xx}$", fontproperties=font) colorbar(surface, shrink=0.75)
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3. Markov Chain – Election voting prediction
A Markov chain is a process consisting of a series of events with the following properties:
- An event has finitely many outcomes, called states, and the process is always one of these states;
- At each stage or period of the process, a particular result may be transferred from its present state to any state, or remain unchanged;
- The probability of moving from one state to another at each stage is represented by a transition matrix with elements in each row between 0 and 1, and the sum of each row is 1.
Taking the American election as an example, the historical data of the past ten elections were obtained first, and then the transfer matrix of voter intentions was obtained according to the historical data.
import matplotlib.pyplot as plt
RLIST = [0.33333]
DLIST = [0.33333]
ILIST = [0.33333]
for i in range(40):
R = RLIST[i]*0.75+DLIST[i]*0.20+ILIST[i]*0.40
RLIST.append(R)
D = RLIST[i]*0.05+DLIST[i]*0.60+ILIST[i]*0.20
DLIST.append(D)
I = RLIST[i]*0.20+DLIST[i]*0.20+ILIST[i]*0.40
ILIST.append(I)
plt.plot(RLIST)
plt.plot(DLIST)
plt.plot(ILIST)
plt.xlabel('Time')
plt.ylabel('Voting percent')
plt.annotate('DemocraticParty',xy = (5.0.2))
plt.annotate('RepublicanParty',xy = (5.0.5))
plt.annotate('IndependentCandidate',xy = (5.0.25))
plt.show()
print(RLIST,DLIST,ILIST)
Copy the codeThe results
- Fifty-six percent voted Republican
- Nineteen percent voted Democratic
- Twenty-five percent voted for independent candidates
This problem can also be solved by using the C-K equation
import numpy as np
a = np.array([[0.75.0.05.0.20], [0.20.0.60.0.20], [0.40.0.20.0.40]])
p = np.mat(a)
for i in range(40):
p = p*p
print(p)
Copy the codeRunning results:
conclusion
Source of learning: STATION B and its classroom PPT, the code is reproduced
The essay is just a study note, recording a process from 0 to 1
Hope to help you, if there is a mistake welcome small partners to correct ~
