Question

我是python的初学者，而我完全了解算法。我需要你的帮助，我希望你帮助我解决以下问题。我只能说，请帮助我。对我来说，对这个问题给出消极或积极的印记并不重要，只需帮助我PLZ。

在下面的代码中我遇到了一个问题：

让我解释一下：使用m=(ss-h)的代码中间的datafile.txt一词最初是：m=(u[i]-h)我将u[i]替换为ss }。现在，如果我想将z[i]替换为aa以删除for循环并避免data_file.txt导入，我该怎么办？

如果我只是将此代码中的z[i]替换为aa，则会出现错误： ValueError: setting an array element with a sequence.

我搜索过这个错误，我发现ODE需要一维数组。据我所知，aa是二维数组。现在我谈到了我的问题。我想要的一件事是调用m[1]，代码会自动aa[1]和ss[1]，m[1]=(ss[1]-M-h)和aa[1]进入h }以及其他m。

有些人请帮忙。数据文件如下：

0.01    32.95388698
0.012   33.87900347
0.014   33.84214074
0.016   34.11856704
0.019   34.59344598
0.023   34.93902653
0.026   35.25209633
0.031   35.74850165
0.037   36.06978761
0.043   36.43457047

代码是：

from math import *
import numpy as np
from scipy.integrate import quad
from scipy.integrate import odeint

z,u=np.genfromtxt('Datafile.txt',unpack=True)

aa=np.array([0.01,0.012,0.014,0.016,0.019,0.023,0.026,0.031,0.037,0.043])
ss=np.array([32.95388698,33.87900347,33.84214074,34.11856704,34.59344598,
        34.93902653,35.25209633,35.74850165,36.06978761,36.43457047])
Cov= [[21282,-10840,1918,451,946,614,785,686,581,233],
     [0,28155,-2217,1702,74,322,380,273,424,487],
     [0,0,6162,-1593,1463,419,715,580,664,465],
     [0,0,0,5235,-722,776,588,591,583,403],
     [0,0,0,0,7303,-508,1026,514,596,315],
     [0,0,0,0,0,3150,-249,800,431,358],
     [0,0,0,0,0,0,3729,-88,730,321],
     [0,0,0,0,0,0,0,3222,-143,568],
     [0,0,0,0,0,0,0,0,3225,-508],
     [0,0,0,0,0,0,0,0,0,5646]]

def ant(z,O_m,O_D):           
   return 1/sqrt(((1+z)**2)*(1+O_m*z)-z*(2+z)*O_D)

def HDE(y,z):
    a=1/(1+z)
    p=log(alpha+beta*a)
    pd=beta*a/(alpha+beta*a)    
    dydz = -((C**2-y)*((pd/p)*(y-2)-(pd**3)/(p**2)+(pd**2)/p+1.5*(y*(b+1)-1))/(1-(pd**2)/(p**2)+2*pd/p-0.5*(C**2+y)))*(1/(1+z))
    return dydz

b=0.01
M=4
C=0.5
O_m=0.23
alpha=1
beta=2

for i in range (len(z)):   
    y0=1-O_m
    z0=0

    z1=z[i]

    y = odeint(HDE,y0,[z0,z1])                       
    O_D=y[-1,0]
    q=quad(ant,0,z[i],args=(O_m,O_D))[0]     
    h=5*log10((1+z[i])*(299000/70)*q)+25     # Here is the main problem of my code
                                             # h and q must work with aa instead of z[i]  

    fl=(ss-h)


    l1=[[fl[0],fl[1],fl[2],fl[3],fl[4],fl[5],fl[6],fl[7],fl[8],fl[9]]]
    l2=[[fl[0]],[fl[1]],[fl[2]],[fl[3]],[fl[4]],[fl[5]],[fl[6]],[fl[7]],[fl[8]],[fl[9]]]

    mm=np.dot(l2,Cov)
    zz=np.dot(mm,l2)
    hh=np.linalg.det(zz)   #calculation of determinant

print (hh)

感谢您的帮助和关注

Answer 1

对于矩阵的组件，您可以使用以下结构：很清楚，不需要任何解释。

第二段分别选择每个组件并在矩阵中正确使用它。

  def new_calculation(n):    
    yn = odeint(HDE,y0,[z0,xx[n]])  
    yyn=yn[-1,0]
    O_Dn=1-O_m-(1/(2*rc*yyn))
    q=quad(ant,0,xx[n],args=(O_m,O_Dn))[0]     
    h=log10((1+xx[n])*(299000/70)*q)+25     
    fn=(ss[n]-h)
    return fn


f_list = []
    for i in range(2):  # the value '2' reflects matrix size
        f_list.append(new_calculation(i))

    rdag=[f_list]
    rmat=[[f] for f in f_list]

    mm=np.dot(rdag,Cov)
    zz=np.dot(mm,rmat)
    hh=np.linalg.det(zz)

ValueError：使用ODEINT

1 个答案: