Python 积分计算器不打印值
Python integral calculator doesn't print a value
我制作了一个计算器,可以将任何给定函数近似作为输入。他们后来我想让它计算一个积分,但是写完之后:
function = str(input("The function that must be expanded and integrated: "))
它不打印数字,而是打印一个值。这是我的代码:
from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import *
from sympy import series
from math import *
function = str(input("The function that must be expanded and integrated: "))
x0 = int(input("Point of development: "))
n = int(input("Amount of expressions: "))
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(N, a, b):
def f(x):
return series(function, x, x0, n)
value=0
value=2
for n in range(1, N+1):
value += f(a+((n-(1/2))*((b-a)/N)))
value2 = ((b-a)/N)*value
return value2
print("...................")
print("Here is your answer: ")
print(integrate(N, a, b))
我想,这是因为我的输入是一个字符串。但是我不能选择我的输入是一个整数,因为 exp(-x**2)
不是一个整数。如果是这样,我怎样才能在我的计算器中输入任何函数并仍然得到一个值?
您的代码中存在一些重大问题:
- 在
integrate
里面,你使用的是局部变量n
,但是在f(x)
里面你认为它是全局变量n
(但是使用的是局部变量,这是你想要的,只需在 f(x)
内打印 n
)。 x
作为f(x)
中的全局变量和参数也是如此。如果要在同一范围内使用全局变量和局部变量,请不要使用相同的名称。
f(x)
的 return 值是一个 sympy
表达式,而不是单个值,这就是为什么你得到你得到的输出。
from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import series
function = str(input("The function to be expanded and integrated: "))
x0 = int(input("Point of development: "))
n = 1 + int(input("Degree: "))
# input 0 -> n=1 -> constant (1 term, constant)
# input 1 -> n=2 -> linear (2 terms, constant + linear)
# input 2 -> n=3 -> quadratic (3 terms, constant + linear + quadratic)
# ...
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(function, x0, n, N, a, b): # using the approach with all variables as parameters
taylor = series(function, x, x0, n) # the same expression for the function, create it once
taylor = taylor.removeO() # do not use O term (may corrups subs below)
dx = (b-a)/N # also computed just once
def f(v):
return taylor.subs(x,v) # taylor is expression, return value is float evaluated with substituted x by v
return dx * sum(f(a+(i+1/2)*dx) for i in range(N)) # simple sum function, can be rewriten using a for loop
print("...................")
print("Here is your answer: ")
print(integrate(function, x0, n, N, a, b))
x**2
的一些输出从 x=0
集成到 x=2
在 x=1
处扩展。解析结果为8/3=2.6666666...
.
x**2, 1, 0, 5, 0, 2 => 2.0 # constant approximation
x**2, 1, 1, 5, 0, 2 => 2.0 # linear approximation
x**2, 1, 2, 5, 0, 2 => 2.64 # quadratic approximation - exact function
x**2, 1, 2, 10, 0, 2 => 2.66
x**2, 1, 2, 100, 0, 2 => 2.6666
x**2, 1, 2, 1000, 0, 2 => 2.666666
您可以使用 lambdify
to "convert a SymPy expression into a function that allows for fast numeric evaluation"。对于 N=1000
的情况,加速是显着的。
from sympy.utilities.lambdify import lambdify
def integrate(function, x0, n, N, a, b):
taylor = series(function, x, x0, n)
taylor = lambdify(x,taylor.removeO()) # here
dx = (b-a)/N
def f(v):
return taylor(v) # here
return dx * sum(f(a+(i+1/2)*dx) for i in range(N))
我制作了一个计算器,可以将任何给定函数近似作为输入。他们后来我想让它计算一个积分,但是写完之后:
function = str(input("The function that must be expanded and integrated: "))
它不打印数字,而是打印一个值。这是我的代码:
from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import *
from sympy import series
from math import *
function = str(input("The function that must be expanded and integrated: "))
x0 = int(input("Point of development: "))
n = int(input("Amount of expressions: "))
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(N, a, b):
def f(x):
return series(function, x, x0, n)
value=0
value=2
for n in range(1, N+1):
value += f(a+((n-(1/2))*((b-a)/N)))
value2 = ((b-a)/N)*value
return value2
print("...................")
print("Here is your answer: ")
print(integrate(N, a, b))
我想,这是因为我的输入是一个字符串。但是我不能选择我的输入是一个整数,因为 exp(-x**2)
不是一个整数。如果是这样,我怎样才能在我的计算器中输入任何函数并仍然得到一个值?
您的代码中存在一些重大问题:
- 在
integrate
里面,你使用的是局部变量n
,但是在f(x)
里面你认为它是全局变量n
(但是使用的是局部变量,这是你想要的,只需在f(x)
内打印n
)。x
作为f(x)
中的全局变量和参数也是如此。如果要在同一范围内使用全局变量和局部变量,请不要使用相同的名称。 f(x)
的 return 值是一个sympy
表达式,而不是单个值,这就是为什么你得到你得到的输出。
from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import series
function = str(input("The function to be expanded and integrated: "))
x0 = int(input("Point of development: "))
n = 1 + int(input("Degree: "))
# input 0 -> n=1 -> constant (1 term, constant)
# input 1 -> n=2 -> linear (2 terms, constant + linear)
# input 2 -> n=3 -> quadratic (3 terms, constant + linear + quadratic)
# ...
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(function, x0, n, N, a, b): # using the approach with all variables as parameters
taylor = series(function, x, x0, n) # the same expression for the function, create it once
taylor = taylor.removeO() # do not use O term (may corrups subs below)
dx = (b-a)/N # also computed just once
def f(v):
return taylor.subs(x,v) # taylor is expression, return value is float evaluated with substituted x by v
return dx * sum(f(a+(i+1/2)*dx) for i in range(N)) # simple sum function, can be rewriten using a for loop
print("...................")
print("Here is your answer: ")
print(integrate(function, x0, n, N, a, b))
x**2
的一些输出从 x=0
集成到 x=2
在 x=1
处扩展。解析结果为8/3=2.6666666...
.
x**2, 1, 0, 5, 0, 2 => 2.0 # constant approximation
x**2, 1, 1, 5, 0, 2 => 2.0 # linear approximation
x**2, 1, 2, 5, 0, 2 => 2.64 # quadratic approximation - exact function
x**2, 1, 2, 10, 0, 2 => 2.66
x**2, 1, 2, 100, 0, 2 => 2.6666
x**2, 1, 2, 1000, 0, 2 => 2.666666
您可以使用 lambdify
to "convert a SymPy expression into a function that allows for fast numeric evaluation"。对于 N=1000
的情况,加速是显着的。
from sympy.utilities.lambdify import lambdify
def integrate(function, x0, n, N, a, b):
taylor = series(function, x, x0, n)
taylor = lambdify(x,taylor.removeO()) # here
dx = (b-a)/N
def f(v):
return taylor(v) # here
return dx * sum(f(a+(i+1/2)*dx) for i in range(N))