我想编写一个“应用程序”，它将绘制第 n 个试剂的浓度随时间的连锁反应函数：A->B->C->D->...
问题是，c_n(t) 包含 2^n - 1 个指数函数 - 根据我发现的模式嵌套：
c_1(t) = c_0_1 * exp(-k_1 * t)
c_2(t) = c_0_2 * exp(-k_2 * t) + c_0_1 * k_1 * {[exp(-k_1 * t) - exp(-k_2 * t)]/[k_2 - k_1]}
c_3(t) = c_0_3 * exp(-k_3 * t) + c_0_2 * k_2 * {[exp(-k_2 * t) - exp(-k_3 * t)]/[k_3 - k_2]} + c_0_1 * k_1 * k_2 * [1/(k_2-k_1)] * <{[exp(-k_1 * t) - exp(-k_3 * t)]/[k_3 - k_1]} - {[exp(-k_2 * t) - exp(-k_3 * t)]/[k_3 - k_2]}>
如您所见，每个方程都是重复出现的元素的总和。嵌套的数量取决于关系的程度：0 次（A 到 A）- 简单指数函数，1 次（A 到 B，B 到 C 等）- 1 次嵌套，2 次（A 到 C） , B 到 D 等） - 2 个嵌套等。
每个方程可以分为重复出现的部分：
“独立”单位：c_0_n * exp(-k_n * t)，
基本单位：f(a,b) = (exp(- k_n[b - 1] * t) - exp(- k_n[a - 1] * t)) / (k_n[a - 1] - k_n[b - 1]),
基于基本单元的嵌套单元，
每个嵌套单元之前常数（参数）的乘积。
第 n 个方程的每个嵌套单元都源自第 (n-1) 个方程的嵌套单元。方程本身可以通过迭代积分得到。第 n 个试剂的可能方程的数量（基于独立动力学常数 k 的数量）由贝尔数 B(n) 给出。
每个这样的方程都可以从具有第 n 个试剂的 n 个独立动力学常数的方程中获得（都相互独立）。人们只需要找到这样的方程的石灰。例如，如果 k_3 = k_4 和 k_7 = k_2，那么我们正在寻找 lim k_4->k_3 [lim k_7->k_2 (f(t))]。
工作代码：
print
print ("Commands: komendy() - list of commands, test() - sets initial parameters, zakres() - asks for the number of reagents, tabela() - displays the table, stez() - asks for the initial concentrations, kin() - asks for the kinetic constants.")
print
n = 0
import matplotlib.pyplot as plt
import numpy as np
def komendy(): # displays the list of commands
print
print ("Commands: komendy() - list of commands, test() - sets initial parameters, zakres() - asks for the number of reagents, tabela() - displays the table, stez() - asks for the initial concentrations, kin() - asks for the kinetic constants.")
print
return
def zakres(): # number of reagents query
global n, zakres_n, c_0_n, k_n
n = int(raw_input("Define the number of n reagents: "))
zakres_n = range(1, n + 1)
c_0_n = [int(0)] * n
k_n = [int(0)] * n
return
def stez(): # initial concentrations query
while True:
y = int(raw_input("Define the value of c_0_n for n equal to (press 0 to break): "))
if y == 0:
break
x = raw_input("Define the value of c_0_" + str(y) + ": ")
if "." in x:
c_0_n[y - 1] = float(x)
else:
c_0_n[y - 1] = int(x)
return
def kin(): # kinetic constants query
while True:
q = int(raw_input("Define the value of k_n for n equal to (press 0 to break): "))
if q == 0:
break
p = raw_input("Define the value of k_" + str(q) + ": ")
if "." in p:
k_n[q - 1] = float(p)
else:
k_n[q - 1] = int(p)
return
def tabela(): # displays the table with the initial data
if n == 0:
zakres()
print
print "n: ", zakres_n
print "c_0_n: ", c_0_n
print "k_n: ", k_n
print
else:
print
print "n: ", zakres_n
print "c_0_n: ", c_0_n
print "k_n: ", k_n
print
return
def wykres(): # plots the basic unit
global f_t, t_k, t, t_d
a = int(raw_input("a = "))
b = int(raw_input("b = "))
reag = map(int, raw_input("Provide the reagents to plot (separate with spacebar): ").split(" "))
t_k = float(raw_input("Define time range from 0 to: "))
t_d = float(raw_input("Set the precision of the time axis: "))
t = np.arange(0,t_k,t_d)
p = []
def f_t(t):
return (np.exp(- k_n[b - 1] * t) - np.exp(- k_n[a - 1] * t)) / (k_n[a - 1] - k_n[b - 1])
f_t = f_t(t)
for i in reag:
p += plt.plot(t,i*f_t)
并且 [还] 不起作用的代码（唯一的区别是我正在尝试构建的新 wykres() 函数）：
print
print ("Commands: komendy() - list of commands, test() - sets initial parameters, zakres() - asks for the number of reagents, tabela() - displays the table, stez() - asks for the initial concentrations, kin() - asks for the kinetic constants.")
print
n = 0
import matplotlib.pyplot as plt
import numpy as np
def komendy(): # displays the list of commands
print
print ("Commands: komendy() - list of commands, test() - sets initial parameters, zakres() - asks for the number of reagents, tabela() - displays the table, stez() - asks for the initial concentrations, kin() - asks for the kinetic constants.")
print
return
def zakres(): # number of reagents query
global n, zakres_n, c_0_n, k_n
n = int(raw_input("Define the number of n reagents: "))
zakres_n = range(1, n + 1)
c_0_n = [int(0)] * n
k_n = [int(0)] * n
return
def stez(): # initial concentrations query
while True:
y = int(raw_input("Define the value of c_0_n for n equal to (press 0 to break): "))
if y == 0:
break
x = raw_input("Define the value of c_0_" + str(y) + ": ")
if "." in x:
c_0_n[y - 1] = float(x)
else:
c_0_n[y - 1] = int(x)
return
def kin(): # kinetic constants query
while True:
q = int(raw_input("Define the value of k_n for n equal to (press 0 to break): "))
if q == 0:
break
p = raw_input("Define the value of k_" + str(q) + ": ")
if "." in p:
k_n[q - 1] = float(p)
else:
k_n[q - 1] = int(p)
return
def tabela(): # displays the table with the initial data
if n == 0:
zakres()
print
print "n: ", zakres_n
print "c_0_n: ", c_0_n
print "k_n: ", k_n
print
else:
print
print "n: ", zakres_n
print "c_0_n: ", c_0_n
print "k_n: ", k_n
print
return
def wykres(): # plots the requested functions
global t_k, t, t_d, f, constans
reag = map(int, raw_input("Provide the reagents to plot (separate with spacebar): ").split(" "))
t_k = float(raw_input("Define the time range from 0 to: "))
t_d = float(raw_input("Define the precision of the time axis: "))
t = np.arange(0,t_k,t_d)
p = []
def f(a,b): # basic unit
return (np.exp(- k_n[b - 1] * t) - np.exp(- k_n[a - 1] * t)) / (k_n[a - 1] - k_n[b - 1])
def const(l,r): # products appearing before the nested parts
const = 1
constans = 1
for h in range(l,r):
const = const * k_n[h]
constans = c_0_n[l] * const
return
def czlonF(g): # nested part
czlonF = 0
for u in range(g):
czlonF = czlonF + npoch(f(a,b),g)
if g == 1:
czlonF(g) = 0
return
def npoch(f(a,b),n):
f = f(a,b)
for x in range(b+1, n+1):
f = npoch(f(a,b),x)
return
def c(j): # final result, concentration in time function
return
def czlon0(m): # 'independent' part
return (c_0_n[m - 1] * np.exp(- k_n[m - 1] * t))
for i in reag: # the actual plot command
p += plt.plot(t,c(i))
plt.show()
return
def test():
global n, zakres_n, k_n, c_0_n
n = 5
zakres_n = range(1, n + 1)
k_n = [1,2,3,4,5]
c_0_n = [2,3,4,5,6]
return
plt.show()
return
def test():
global n, zakres_n, k_n, c_0_n
n = 5
zakres_n = range(1, n + 1)
k_n = [1,2,3,4,5]
c_0_n = [2,3,4,5,6]
return
如何修复 wykres() 函数以绘制 c(n)？我如何构建它以便可以绘制它？我希望 Python 自动为我想要的任何 n 构建 c_n(t) 并绘制所有这些。