Python 在我尝试做数学时重新启动 shell
Python restarts the shell when I tried to do math
我对以下代码有疑问。我尝试了很多网页来寻找一些解决方案，但我无法做到。
问题是当我 运行 这个脚本仅打印 4 行后，Python 重新启动 shell。在 jupyter 中它发送消息：The kernel appears to have died. It will restart automatically.
我尝试在包含 (300,000) 行数据的脚本的第一个文件中读取文件。然后计算完 ODE 和其他函数后，我希望打印结果进行比较
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])
很简单的代码，但不知道最终的问题是什么。
非常感谢任何帮助。
输入文件（new.txt）可能是
71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486
70.181278149341 0.44888833570037 0.077917371645449 0.04777288009128
70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
70.011812869146 0.44210637315163 0.07871914246357 0.048393990901086
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
70.123419349447 0.43961350279409 0.07862300319627 0.048607832896286
70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
71.551080656041 0.51047305096688 0.066682530098241 0.0446474321235
好的，问题出在 odeint 上。 docs recommend switching to solve_ivp 代替。
现在完全免责声明，我完全不知道这意味着什么，数学和事物的意义超出了我的范围。然而，我试图用 solve_ivp 来最好地模仿 odeint 行为，它开箱即用不接受 args。呈现，贫民窟 lambda 来救援。
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，这不是完全的替换，solve ivp 的结果以这种方式呈现一个浮点数，您希望将其包装为 [od] 而不是与旧结果完全匹配。
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
尽管如此，我不太清楚（还）为什么 odeint 以这种方式崩溃。
编辑：
对于操作，这里是代码更新后的样子。
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是这个。请注意旧的 odeint 求解器，但只打印了 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])
并且输出：
[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。我无法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，只需一点小改动：避免零长度积分区间，如果遇到这种情况，直接 return 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
我对以下代码有疑问。我尝试了很多网页来寻找一些解决方案，但我无法做到。
问题是当我 运行 这个脚本仅打印 4 行后，Python 重新启动 shell。在 jupyter 中它发送消息：The kernel appears to have died. It will restart automatically.
我尝试在包含 (300,000) 行数据的脚本的第一个文件中读取文件。然后计算完 ODE 和其他函数后，我希望打印结果进行比较
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])
很简单的代码，但不知道最终的问题是什么。 非常感谢任何帮助。
输入文件（new.txt）可能是
71.076588184266 0.40147988209522 0.080396967668756 0.050302016457046
71.02284157687 0.39756707964421 0.080918035449145 0.050501956013259
71.102923163306 0.41587392748136 0.07823452108922 0.049336707395359
70.860444589498 0.46748446539443 0.072392464271658 0.046667808684486
70.181278149341 0.44888833570037 0.077917371645449 0.04777288009128
70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
70.588452406351 0.49035265611716 0.072303154996487 0.045942096884044
70.011812869146 0.44210637315163 0.07871914246357 0.048393990901086
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
69.807956729005 0.41349634394633 0.082020266421564 0.049900569076028
70.123419349447 0.43961350279409 0.07862300319627 0.048607832896286
70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
70.361666430312 0.41397677666087 0.080502527828865 0.049745843125116
70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
70.357430153315 0.41485946940097 0.08042642593323 0.049703105696664
71.551080656041 0.51047305096688 0.066682530098241 0.0446474321235
好的，问题出在 odeint 上。 docs recommend switching to solve_ivp 代替。
现在完全免责声明，我完全不知道这意味着什么，数学和事物的意义超出了我的范围。然而，我试图用 solve_ivp 来最好地模仿 odeint 行为，它开箱即用不接受 args。呈现，贫民窟 lambda 来救援。
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，这不是完全的替换，solve ivp 的结果以这种方式呈现一个浮点数，您希望将其包装为 [od] 而不是与旧结果完全匹配。
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
尽管如此，我不太清楚（还）为什么 odeint 以这种方式崩溃。
编辑：
对于操作，这里是代码更新后的样子。
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是这个。请注意旧的 odeint 求解器，但只打印了 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])
并且输出：
[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。我无法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，只需一点小改动：避免零长度积分区间，如果遇到这种情况，直接 return 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