当我尝试做数学运算时，Python重新启动shell

我对以下代码有问题。我尝试了很多网页来寻找一些解决方案，但我没能找到。 问题是，当我在打印了4行之后运行这个脚本时，Python会重新启动shell。在

jupyter

中，它发送消息：

内核似乎已死亡。它将自动重新启动。

我尝试在脚本的第一行读取一个文件，其中包含（

300000

）行数据。然后在计算完

ODE

和其他函数后，我希望打印结果以进行比较

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

hr, dr, cr, br = np.genfromtxt('outputs/new.txt',unpack=True)

def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1]                   
    return od   

def dec(H0, z, od0, c, b):
    od = ant(H0, z, od0, c, b)
    q = -1 - (-2 + od/(6 * c))
    return q

for i in range(len(hr)):
    for z in range (0,1):
        print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i])

这是一个简单的代码，但我不知道最终的问题是什么。 我真的很感谢你的帮助

输入文件（

new.txt

）可以是

---

好的，问题在于

odeint

。建议改为切换到

现在，我完全不知道这意味着什么，数学和事物的意义是我无法理解的。然而，我试图用solve_ivp最好地模仿odeint行为，它开箱即用，不接受args。现在，贫民区lambda来营救

od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1] #before
od = solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1] #after

请注意，这并不是完全的替换，solve ivp的结果以这种方式呈现一个浮点，您希望将其包装为

[od]

，以完全匹配旧结果。 至于我用来缩小odeint的最小代码，你可以看到下面的战场

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)




def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    #od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1] #this solver crashed
    od = solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1] #this worked out. perhaps wrap in square brackets [od] if needed.
    #od = OD_H(od0,z,c,b)  #this alone without the solvers worked fine              
    return od   

#def dec(H0, z, od0, c, b): #remove the middleman
#    od = ant(H0, z, od0, c, b)
#    q = -1 - (-2 + od/(6 * c))
#    return od

for i in range(5): #simple check instead
    z = 0 #you do not need a loop here
    res = ant(hr[i], z, dr[i], cr[i], br[i])
    print(res)
        #print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i]) print was not the culprit

不过，归根结底，我还不太清楚为什么odeint会以这种方式崩溃

编辑：

对于op，下面是代码更新的方式

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)

def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    #od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1]  
    od = [solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1]]
    return od   

def dec(H0, z, od0, c, b):
    od = ant(H0, z, od0, c, b)
    q = -1 - (-2 + od/(6 * c))
    return q

for i in range(len(hr)):
    for z in range (0,1):
        print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i])

上述数据的输出如下所示：

[0.16771346] 71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
[0.18113212] 71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
[0.11404428] 71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
[-0.07627332] 70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486
[0.03981973] 70.181278149341 0.44888833570037 0.077917371645449 0.04777288009128
[-0.13031641] 70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
[-0.13031641] 70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
[0.06395836] 70.011812869146 0.44210637315163 0.07871914246357 0.048393990901086
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.06809821] 70.123419349447 0.43961350279409 0.07862300319627 0.048607832896286
[0.14293214] 70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
[0.14293214] 70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
[0.14029196] 70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
[0.14029196] 70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
[-0.27587903] 71.551080656041 0.51047305096688 0.066682530098241 0.0446474321235

我可以用来比较的代码如下。请注意旧的odeint解算器，但仅打印了4个值

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)

def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1]  
    #od = [solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1]]
    return od   

def dec(H0, z, od0, c, b):
    od = ant(H0, z, od0, c, b)
    q = -1 - (-2 + od/(6 * c))
    return q

for i in range(4): #simplified to avoid crash
    for z in range (0,1):
        print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i])

以及输出：

[0.16771346] 71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
[0.18113212] 71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
[0.11404428] 71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
[-0.07627332] 70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486

---

您提供的代码并不总是生成负值。我不想理解您为什么期望或希望他们这样做，但结果应该与原始代码相匹配。

好的，所以问题在于

odeint

。建议改为切换到

现在，我完全不知道这意味着什么，数学和事物的意义是我无法理解的。然而，我试图用solve_ivp最好地模仿odeint行为，它开箱即用，不接受args。现在，贫民区lambda来营救

od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1] #before
od = solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1] #after

请注意，这并不是完全的替换，solve ivp的结果以这种方式呈现一个浮点，您希望将其包装为

[od]

，以完全匹配旧结果。 至于我用来缩小odeint的最小代码，你可以看到下面的战场

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)




def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    #od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1] #this solver crashed
    od = solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1] #this worked out. perhaps wrap in square brackets [od] if needed.
    #od = OD_H(od0,z,c,b)  #this alone without the solvers worked fine              
    return od   

#def dec(H0, z, od0, c, b): #remove the middleman
#    od = ant(H0, z, od0, c, b)
#    q = -1 - (-2 + od/(6 * c))
#    return od

for i in range(5): #simple check instead
    z = 0 #you do not need a loop here
    res = ant(hr[i], z, dr[i], cr[i], br[i])
    print(res)
        #print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i]) print was not the culprit

不过，归根结底，我还不太清楚为什么odeint会以这种方式崩溃

编辑：

对于op，下面是代码更新的方式

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)

def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    #od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1]  
    od = [solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1]]
    return od   

def dec(H0, z, od0, c, b):
    od = ant(H0, z, od0, c, b)
    q = -1 - (-2 + od/(6 * c))
    return q

for i in range(len(hr)):
    for z in range (0,1):
        print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i])

上述数据的输出如下所示：

[0.16771346] 71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
[0.18113212] 71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
[0.11404428] 71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
[-0.07627332] 70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486
[0.03981973] 70.181278149341 0.44888833570037 0.077917371645449 0.04777288009128
[-0.13031641] 70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
[-0.13031641] 70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
[0.06395836] 70.011812869146 0.44210637315163 0.07871914246357 0.048393990901086
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.15976794] 69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
[0.06809821] 70.123419349447 0.43961350279409 0.07862300319627 0.048607832896286
[0.14293214] 70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
[0.14293214] 70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
[0.14029196] 70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
[0.14029196] 70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
[-0.27587903] 71.551080656041 0.51047305096688 0.066682530098241 0.0446474321235

我可以用来比较的代码如下。请注意旧的odeint解算器，但仅打印了4个值

import numpy as np
from scipy.integrate import odeint, solve_ivp
#from math import *
from scipy.integrate import quad  
#import math
#import pandas as pd

hr, dr, cr, br = np.genfromtxt('new.txt',unpack=True)

def OD_H(od, z, c, b):
   Omegai = (1-od)
   div1 = np.divide(1, (1 + z), where = (1 + z)!= 0)     
   dMdt = -(div1) * (2 *(1-od)* (-2 + (od/(6 * c))) + 3 - 3 * b**2 * Omegai - 3 * od)
   return dMdt

def ant(H0, z, od0, c, b):
    z1 = 0
    od = odeint(OD_H, od0, [z1, z], args=(c, b))[-1]  
    #od = [solve_ivp(lambda od, z: OD_H(od, z, c, b), t_span = [z1, z], y0 = [od0])['y'][-1][-1]]
    return od   

def dec(H0, z, od0, c, b):
    od = ant(H0, z, od0, c, b)
    q = -1 - (-2 + od/(6 * c))
    return q

for i in range(4): #simplified to avoid crash
    for z in range (0,1):
        print(dec(hr[i], z, dr[i], cr[i], br[i]),hr[i], dr[i], cr[i], br[i])

以及输出：

[0.16771346] 71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
[0.18113212] 71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
[0.11404428] 71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
[-0.07627332] 70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486

---

您提供的代码并不总是生成负值。我不理解您为什么期望或希望他们这样做，但结果应该与原始代码匹配。

您的代码将按预期工作，只需一个小改动：避免零长度的积分间隔，如果遇到这种情况，直接返回初始值

def ant(H0, z, od0, c, b):
    z1 = 0
    if type(od0) is np.float64: od0 = np.array([od0]); # for uniform output
    od = od0 if abs(z-z1) < 1e-15 else odeint(OD_H, od0, [z1, z], args=(c, b))[-1]                   
    return od   

定义剂（H0、z、od0、c、b）： z1=0 如果类型（od0）是np.float64:od0=np.array（[od0]）；#用于均匀输出 如果abs（z-z1）<1e-15，则od=od0，否则odint（od_H，od0，[z1，z]，args=（c，b））[-1] 返回od

---

您的代码将按预期工作，只需做一个小改动：避免零长度的积分间隔，如果遇到这种情况，直接返回初始值

def ant(H0, z, od0, c, b):
    z1 = 0
    if type(od0) is np.float64: od0 = np.array([od0]); # for uniform output
    od = od0 if abs(z-z1) < 1e-15 else odeint(OD_H, od0, [z1, z], args=(c, b))[-1]                   
    return od   

定义剂（H0、z、od0、c、b）： z1=0 如果类型（od0）是np.float64:od0=np.array（[od0]）；#用于均匀输出 如果abs（z-z1）<1e-15，则od=od0，否则odint（od_H，od0，[z1，z]，args=（c，b））[-1] 返回od

---

创建一个不同的new.txt副本，其中只包含30行数据或其他内容。如果它加载，您可以确认我的猜测：这很可能是因为您没有足够的ram来保存所有数据。这只是一个典型的内存问题。@Paritossingh是的，我以前曾尝试过一些有这个问题的计算，在添加了4 GB额外的ddr3 RAM之后，我的问题就解决了。但现在我有8GB内存。但现在让我来做你的建议。等等…@Paritossingh不，一个文本文件又有17行了，我也有同样的问题。你能提供最小的输入，仍然允许复制这个问题吗？@Paritossingh你是什么意思，我抓不到它创建一个新的.txt副本，只有30行数据或其他东西。如果它加载，您可以确认我的猜测：这很可能是因为您没有足够的ram来保存所有数据。这只是一个典型的内存问题。@Paritossingh是的，我以前曾尝试过一些有这个问题的计算，在添加了4 GB额外的ddr3 RAM之后，我的问题就解决了。但现在我有8GB内存。但现在让我来做你的建议。等等…@Paritossingh不，一个文本文件又有17行，我也有同样的问题。你能提供最小的输入，仍然允许复制这个问题吗？@Paritossingh你是什么意思，在我问一个问题之前，我听不懂。如果使用

def dec（）

函数，则打印的结果为

正

，而全部结果必须为负。当然，这样我就不知道最后的答案了。我的意思是它们是错误的结果如果结果是错误的，那么它们就是你给出的代码的结果。你必须检查方程的逻辑等等。检查更新，看看这是否给你任何线索@恐怕我无法帮助你理解所有数学运算背后的逻辑。不，不，代码是正确的，即使在前面的代码中，我们可以看到5行的负值。但是这个代码的答案是完全肯定的。使用此代码