关于python：嵌套函数是Pythonic吗？

Is Nesting Functions Pythonic?

我正在用R.PI学习Python，我遇到了一个小问题。在我看来，下面的代码会让"inputchecker"函数在内存中保持打开状态，同时它会返回到"getinput"函数。

这是坏代码吗？是否应该做得非常不同？

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def getinput(i):
if i == 1:
first = input("Would you like A or B?")
inputchecker(1, first)
elif i == 2:
second = input("Would you like C or D?")
inputchecker(2, second)

def inputchecker(n, userinput):
def _tryagain_(n):
usage(n)
getinput(n)
if n == 1:
if userinput in ("A","B"):
print("You chose wisely.")
getinput(2)
else:
_tryagain_(n)
elif n == 2:
if userinput in ("C","D"):
print("You chose wisely.")
else:
_tryagain_(n)

def usage(u):
if u == 1:
print("Usage: Just A or B please.")
if u == 2:
print("Usage: Just C or D please.")

getinput(1)

相关讨论

拥有两个彼此无限调用的函数并不是最好的控制流。用一个while循环会更好

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def getinput(i):
while i:
if i == 1:
first = input("Would you like A or B?")
i = inputchecker(1, first)
elif i == 2:
second = input("Would you like C or D?")
i = inputchecker(2, second)

def inputchecker(n, userinput):
if n == 1:
if userinput in ("A","B"):
print("You chose wisely.")
return 2
else:
getusage(i)
return i
elif n == 2:
if userinput in ("C","D"):
print("You chose wisely.")
else:
getusage(i)
return i

如果将其简化为单个函数，可能会更好。没有理由要分开。

我当然会避免递归调用。另外，我会让验证函数返回一个布尔值，而不是下一个问题的编号。由于您总是按顺序问问题，这似乎只会使代码的读者感到复杂。

另外，我会让验证总是返回一些东西——你永远不会知道：第一个参数也可能是错误的：

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def getinput():
valid = False
while not valid:
first = input("Would you like A or B?")
valid = inputIsValid(1, first)
valid = False
while not valid:
second = input("Would you like C or D?")
valid = inputIsValid(2, second)
return [first, second]

def inputIsValid(n, userinput):
valid = False
if n == 1:
valid = userinput in ("A","B")
elif n == 2:
valid = userinput in ("C","D")
if valid:
print("You chose wisely.")
else:
usage(n)
return valid

def usage(u):
if u == 1:
print("Usage: Just A or B please.")
elif u == 2:
print("Usage: Just C or D please.")

getinput()