Spyder/Jupyter内核崩溃重启、Atom代码无图像输出问题求助

我原本运行正常的代码，再次运行时在Atom中代码执行完成但图像无法弹出。尝试在Spyder和Jupyter中运行时，两者均出现"kernel died, restarting"错误提示。我试过其他问题提到的解决方法但都无效，还更新了Anaconda Navigator。我电脑操作经验不多，不确定下一步该怎么做，当前使用Python 3.6。非常感谢您的帮助。

编辑：补充代码
编辑2：问题已解决，是我的代码存在错误（当然不是Python的问题 :D）

原代码 #

#CIE4365-16 #Assignment 1 - Heat flow 
import numpy as np 
import matplotlib.pyplot as plt 
import scipy.integrate as spin 
from collections import namedtuple 

#grid/discretization 
def topTbound(t, bPar): 
    '''Temperature for top boundary, where t is time''' 
    boundT = bPar.avgT - bPar.rangeT * np.cos(2 * np.pi * (t - bPar.tmin) / 365.25) 
    return boundT 

def heatflux(t, T, sPar, mDim, bPar): 
    boundT = topTbound(t, bPar) #temperature for top boundary funtion 
    locT = T.copy() #copies T, doesn't change T in the original, can now change it here 
    locT[mDim.nin-2, 0] = boundT #setting the boundary Temperature 
    q = np.zeros((mDim.nin, 1)) # ,1 to make it a column 
    for i in range(1, mDim.nin-1): 
        q[i, 0] = - sPar.lambdaIN[i, 0] * (locT[i, 0] - locT[i - 1, 0]) / mDim.dzn[i - 1, 0] 
    # Robin boundary at the top 
    q[mDim.nin - 1, 0] = - bPar.kRobinTop * (boundT - T[mDim.nin - 2, 0]) 
    #q[nin - 1, 0] = q[nin-2, 0] #setting the top 
    q[0, 0] = 0 #to show that the bottom layer has no flow, so q must be zero 
    return q 

#divergence of heatflux through nodes 
def divheatflux(t, T, sPar, mDim, bPar): 
    locT = T.copy().reshape(nn, 1) 
    qH = heatflux(locT, t, sPar, mDim, bPar) #heatflux equation defined earlier 
    divqH = np.zeros((mDim.nn, 1)) 
    for i in range(0, mDim.nn): 
        divqH[i, 0] = -(qH[i + 1, 0] - qH[i, 0]) / (mDim.dzin[i, 0] * sPar.zetaBN[i, 0]) 
    divqHRet = divqH.reshape(mDim.nn) 
    return divqHRet 

#reshape to columns so it is a vertical profile 
depth = -15 
dz = 0.1 #distance between internode layers 
nin = 151 #number of internodes 
nn = nin - 1 #number of nodes is always one less that number of internodes 

#zin is the internodes boundaries depth location 
zin = np.arange(depth, 0 + dz, dz).reshape(nin, 1) 
#nr, nc = np.shape(zin) #nr (number of rows) is the number of internodes 

#number of nodes, will be filled by the for loop 
zn = np.zeros([nin-1, 1]) 
for i in range(0, len(zn)): 
    zn[i] = (zin[i, 0] + zin[i+1, 0]) / 2 
zn[0, 0] = zin[0, 0] #this is to change the first value in the array from 0 to -10 
zn[len(zn)-1] = zin[len(zin)-1] #this is to change the last value in the array from -0.25 to 0 

#dzn will be the distance between the node 
dzn = np.zeros([len(zn)-1, 1]) 
for i in range(0, len(zn)-1): 
    dzn[i, 0]=zn[i, 0]-zn[i+1, 0] 
#dzn = np.reshape(dzn, ((nin - 2), 1)) 

#dzin will be the distance between the internodes 
dzin = np.zeros(len(zin)-1) 
for i in range(0, len(dzin)): 
    dzin[i]=zin[i]-zin[i+1] 
dzin = np.reshape(dzin, ((nin-1), 1)) 

#use namedtuple to put dimensions in same place 
modDim = namedtuple('ModDim',['zn', 'zin', 'dzn', 'dzin', 'nn' , 'nin']) 
mDim = modDim(zn=zn,zin=zin,dzn=dzn,dzin=dzin,nn=nn,nin=nin) 

#Properties of the soil 
zetasolid = 2.235e6 # [J m^-3 K^-1] volumetric heat capacity of solid 
zetawater = 4.154e6 # [J m^-3 K^-1] volumetric heat capacity of water 
qu = 0.75 # quartz content 
sw = 0 #sw is the water saturation 
rhowater = 1000 # [kg/m^3] density of water 
rhosolid = 2650 # [kg/m^3] density of solid 
rhobulk = 1700 # [kg/m^3] dry bulk density of soil 
p = (1 - rhobulk) / rhosolid # [-] porosity of soil = saturated water content. 
lambdaQuartz = 6 # [W/(mK)] thermal conductivity of quartz 

#combine soil parameters in one place 
soilPar = namedtuple('soilPar',['zetaBN','lambdaIN']) 
sPar = soilPar(zetaBN = np.ones(np.shape(zn)) * ((1 - p) * zetasolid + p * zetawater * sw), lambdaIN = np.ones(np.shape(zin)) * lambdaQuartz * (24 * 3600)) 

# boundary parameters 
boundPar = namedtuple('boundPar',['avgT', 'rangeT', 'tmin', 'kRobinTop', 'kRobinBot']) 
bPar = boundPar(avgT = 283, rangeT = 15, tmin = 46, kRobinTop = 1, kRobinBot = 0) 

# Initial Conditions 
Tini = np.ones(np.shape(zn)) * (283) # degrees K 
tOut = np.linspace(0, 365.25 * 10, 365) # time, *10 = 10 years, steps per day 
nOut = np.shape(tOut)[0] 
T0 = Tini.copy().squeeze() 
t_span = [tOut[0], tOut[-1]] 

#TODE = spin.solve_ivp(divheatflux, t_span, T0, sPar, mDim, bPar, 
# method='RK45', t_eval=tOut, vectorized=True, rtol=1e-5) 
TODE = spin.odeint(divheatflux, tOut, T0, args = (sPar,mDim,bPar)) 

for i in range(0, nOut-1): 
    TODE[i, nn - 1] = topTbound(tOut[i], bPar) 

#plotting 
plt.close('all') #closing the previous plots 
fig1, ax1 = plt.subplots(figsize=(9,4)) 
ax1.plot(tOut,TODE[:, 0:nn-1:20], '.-') 
ax1.set_title('Temperature') 

fig2, ax2 = plt.subplots(figsize=(5,8)) 
for i in np.arange(0, nOut-1, 10): 
    ax2.plot(TODE[i, :], zn.squeeze(),'-') 
ax2.set_title('Temperature vs. depth') 

plt.show() 

问题根源说明 #

排查代码后发现，divheatflux函数调用heatflux时参数顺序搞反了：

# 错误调用：把温度数组locT放在了时间参数t的位置
qH = heatflux(locT, t, sPar, mDim, bPar)

而heatflux的定义明确要求第一个参数是时间t，第二个是温度数组T：

def heatflux(t, T, sPar, mDim, bPar):

这种参数不匹配会导致后续数值计算出现异常，直接引发Spyder/Jupyter内核崩溃，Atom中虽然代码执行完成，但后续绘图逻辑也受影响无法正常弹出图像。修正参数顺序后，代码就能正常运行并生成预期的温度曲线了。

内容的提问来源于stack exchange，提问作者J.S.