所以说我有一个较低的值和一个来自用户输入的积分的上限值。我首先要求下限,然后检查其有效性。然后为了比较我的上限值和我的下限我做了一个嵌套函数,这样我也可以要求用户输入上限,检查其有效性并确保我的上限大于我的下限(因为你知道整合),用下面的代码显示。
def LowLimCheck():
while True:
try:
a = float(input("Please enter the lower limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
print("You have chosen the lower limit: ", a)
def UppLimCheck():
b = -1
while b <= a:
while True:
try:
b = float(input("Please enter the upper limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
if b <= a:
print("The upper limit must be bigger than the lower limit!")
print("You have chosen the upper limit: ", b)
return b
UppLimCheck()
return a
现在这一切都很好,直到我真的需要使用值a和b,因为我需要将这些值放入我设置的积分中。它基本上是辛普森规则制定的一般整体,现在很容易进入。所以我将函数定义为:
def simpsonsRule(func, a, b, N):
<insert code here>
<insert code here>
return d
好的,所以我的功能基本上允许用户插入任意函数,上限(a),下限(b)和N(Simpson规则中的条带数)并且它吐出d这是积分的评估Simpson规则的任意函数。我现在的问题是,当我尝试打印答案时,我可以将变量a取出并放入积分但我不能将变量b取出,因为它在函数中!例如,如果我现在打印积分值(例如在sin(x)和N = 20的情况下)
print(simpsonsRule(lambda x:np.sin(x), a, b, 20)
所以我知道a和b值在它们自己的函数中是局部的。现在,对于a的值,我可以轻松地执行此操作以获得值
k = 0 #initialising the variable
k = LowLimCheck()
print(simpsonsRule(lambda x:np.sin(x), k, b, 20)
因为k调用LowLimCheck(),它返回我可以放入函数的值。但是如何获得嵌套在第一个函数中的b值?我基本上想用b。有没有办法解决这个问题?
为这个冗长的问题道歉并提前致谢!
答案 0 :(得分:1)
你可以从LowLimCheck()返回一个元组:
def LowLimCheck():
...
b = UppLimCheck()
return (a,b)
然后在调用LowLimCheck()
时解压缩它们a, b = LowLimCheck()
<强>更新强>
在对您的问题的最直接回答中,LowLimCheck()
变为:
def LowLimCheck():
while True:
try:
a = float(input("Please enter the lower limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
print("You have chosen the lower limit: ", a)
def UppLimCheck():
b = -1
while b <= a:
while True:
try:
b = float(input("Please enter the upper limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
if b <= a:
print("The upper limit must be bigger than the lower limit!")
print("You have chosen the upper limit: ", b)
return b
b = UppLimCheck() # Storing the b
return (a,b) # Passing b out with a in a tuple
然后致电
a, b = LowLimCheck()
最后,
print(simpsonsRule(lambda x:np.sin(x), a, b, 20)
替代解决方案(更多实质性更改,但更好的代码结构 - 如原始评论中所述;更平坦,更易读,更少考虑范围):
def LowLimCheck():
while True:
try:
a = float(input("Please enter the lower limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
print("You have chosen the lower limit: ", a)
return a
def UppLimCheck(a):
b = -1
while b <= a:
while True:
try:
b = float(input("Please enter the upper limit of the integral: "))
break
except ValueError:
print("Invalid input. Please enter a number.")
if b <= a:
print("The upper limit must be bigger than the lower limit!")
print("You have chosen the upper limit: ", b)
return b
然后:
lowLim = LowLimCheck()
upLim = UppLimCheck(lowLim) # pass in lowLim as an argument
print(simpsonsRule(lambda x:np.sin(x), lowLim, upLim, 20)