Python 带有迭代的inf和nan值
之前的代码中有Rs值Python 带有迭代的inf和nan值,python,Python,之前的代码中有Rs值 from numpy import exp as e Ri = 9 Rr = 19/2 Rs = 10 i = 0 Er = 0 Rra = 0 x = 1 def F(n): return 745*(1-e(-x/10))-49*x #函数准计算误差相对论 def错误(Rra、Rrn): Erel=abs((Rrn-Rra)/Rrn)*100 回程鱼 print ('{:^15}{:^15}{:^15}{:^15}{:^15}{:^15}'.format('#
from numpy import exp as e
Ri = 9
Rr = 19/2
Rs = 10
i = 0
Er = 0
Rra = 0
x = 1
def F(n):
return 745*(1-e(-x/10))-49*x
#函数准计算误差相对论
def错误(Rra、Rrn):
Erel=abs((Rrn-Rra)/Rrn)*100
回程鱼
print ('{:^15}{:^15}{:^15}{:^15}{:^15}{:^15}'.format('# iter','Ri','Rs','Rr','F(Ri)','Erel(%)'))
while (i < 10):
Er = error(Rra,Rr)
Rra = Rr `#Rra sera el Rr anterior para determinar el error relativo`
if F(Ri).all() * F(Rr).all() < 0:
Rs = Rr
elif F(Ri).all() * F(Rr).all() > 0:
Ri = Rr
elif F(Rr) == 0:
print('La raiz es',Rr)
Rr = (Ri + Rs) / 2
i = i + 1
#Condicional para el primer error relativo
if i == 1:
print ('{:^15}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.6f}'.format(i-1,Ri,Rs,Rr,F(Rr)))
elif i > 1:
print ('{:^15}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.10f}'.format(i-1,Ri,Rs,Rr,F(Rr),Er))
x=1
def F(n):
返回745*(1-e(-x/10))-49*x
在您的F(n)
方法中,x
是一个赋值为1且永不更改的值。因此F(n)
将始终返回相同的值21.896123563210153
如果F(Ri)*F(Rr)<0:
Rs=Rr
elif(Ri)*F(Rr)>0:
Ri=Rr
elif(Rr)==0:
打印('La raiz es',右后)
因此,F(Ri)*F(Rr)
始终为正值。因此,Ri
具有相同的Rr
值
Rr=(F(Rs)*Ri-F(Ri)*Rs)/(F(Rs)-F(Ri))
通过将F(Rs)-F(Ri)
除以0来计算Rr
。这在数学上是违法的。所以在你的例子中,Rr
是nan
在第一个循环之后,Ri
被分配Rr
的值
def错误(Rra、Rrn):
Erel=abs((Rrn-Rra)/Rrn)*100
回程鱼
Er=错误(Rra,Rr)
Rrn
的值为Rr
。在(Rrn-Rra)/Rrn
中,将nan
值除以。这将导致nan
总之,根本原因是方法
F(n)
中的x
是一个常量。在您编写的Rr=(F(Rs)*Ri-F(Ri)*Rs)/(F(Rs)-F(Ri))
中,您希望首次运行时Rs
和Ri
的值是什么?因此,F(Rs)-F(Ri)
的结果应该是什么?你知道这是怎么引起问题的吗?另外:Rs
的初始值应该来自哪里?我加上Rs值的来源
from numpy import exp as e
Ri = 9
Rr = 10
i = 0
Er = 0
x = 1
def F(n):
return 745*(1-e(-x/10))-49*x
#Funcion para calcular el error relativo
def error (Rra,Rrn):
Erel = abs((Rrn - Rra) / Rrn) * 100
return Erel
print ('{:^15}{:^15}{:^15}{:^15}{:^15}{:^25}'.format('# iter','Ri','Rs','Rr','F(Rr)','Erel(%)'))
while (i < 10):
Er = error(Rra,Rr)
Rra = Rr #Rra sera el Rr anterior para determinar el error relativo
if F(Ri) * F(Rr) < 0:
Rs = Rr
elif F(Ri) * F(Rr) > 0:
Ri = Rr
elif F(Rr) == 0:
print('La raiz es',Rr)
Rr = (F(Rs)*Ri -F(Ri)*Rs)/(F(Rs) - F(Ri))
i = i + 1
#Condicional para el primer error relativo
if i == 1:
print ('{:^15}{:^15.10f}{:^15.10f}{:^15.10f}{:^15.10f}'.format(i-1,Ri,Rs,Rr,F(Rr)))
elif i > 1:
print ('{:^15}{:^15.10f}{:^15.10f}{:^15.10f}{:^15.10f}{:^25.10f}'.format(i-1,Ri,Rs,Rr,F(Rr),Er))
# iter Ri Rs Rr F(Rr) Erel(%)
0 10.0000000000 10.0000000000 nan 21.8961235632
1 nan 10.0000000000 nan 21.8961235632 nan
2 nan 10.0000000000 nan 21.8961235632 nan
3 nan 10.0000000000 nan 21.8961235632 nan
4 nan 10.0000000000 nan 21.8961235632 nan
5 nan 10.0000000000 nan 21.8961235632 nan
6 nan 10.0000000000 nan 21.8961235632 nan
7 nan 10.0000000000 nan 21.8961235632 nan
8 nan 10.0000000000 nan 21.8961235632 nan
9 nan 10.0000000000 nan 21.8961235632 nan
/usr/local/lib/python3.7/dist-packages/ipykernel_launcher.py:30: RuntimeWarning: invalid value encountered in double_scalars