问题描述
我正在尝试设计一个函数，通过传入一些 Wizard 来生成与他们的父母在同一所房子里的巫师的名字，这些Wizard 聚集在一起形成了一个家谱，如中所述文档字符串。
## Data DeFinitions:
## Wizard is (name,house,children)
## - name(str) is name of the wizard.
## - house(str) is name of the house the wizard in.
## - children(listof str) is a list of the wizard's children.
wa = ('A','S',[])
wb = ('B','G',[])
wc = ('C',[])
wd = ('D','H',[])
we = ('E','R',[wb])
wf = ('F',[we,wa])
wg = ('G',[wc,wd])
wh = ('H',[wg,wf])
这些是要传递给函数的数据的一些示例。
def get_same_house(wiz):
"""Produce the names of wizards who be in the same house as their parent.
Args:
wiz(tuple[str,str,list]): name,house and list of children of the wizard.
Returns:
list: names of wizards
"""
result = []
def fn_for_wiz(wiz,house):
if wiz[1] == house:
nonlocal result
result.append(wiz[0])
else:
house = wiz[1]
fn_for_low(wiz[2],house)
def fn_for_low(low,house):
if not low:
return None
fn_for_wiz(low[0],house),fn_for_low(low[1:],house)
fn_for_wiz(wiz,'')
return result
在上面的函数中，我在非局部范围内使用了一个变量 result 来存储想要的值并使递归助手不返回任何内容。
def get_same_house_as_parent(wiz):
"""Produce names of wizard which be in the same house as their parent.
Args:
wiz(tuple[str,house and children of a wizard.
Returns:
list: names of the wizards
todo: list of wizard to traverse next contains (wizard,parent name)
"""
result = []
def fn_for_wiz(wiz,ph,todo):
if wiz[1] == ph:
nonlocal result
result.append(wiz[0])
todo.extend([(w,wiz[1]) for w in wiz[2]])
return fn_for_low(todo)
def fn_for_low(todo):
if not todo:
return result
return fn_for_wiz(todo[0][0],todo[0][1],todo[1:])
return fn_for_wiz(wiz,'',[])
这个我使用了辅助函数的返回和非局部变量result的组合。
def get_same_house_nostoring(wiz):
"""Produce the names of wizards who be in the same house as their parent.
Args:
wiz(tuple[str,house and list of children of the wizard.
Returns:
list: names of wizards
"""
def fn_for_wiz(wiz,house):
if wiz[1] == house:
return [wiz[0]] + fn_for_low(wiz[2],house)
else:
house = wiz[1]
return fn_for_low(wiz[2],house):
if not low:
return []
return fn_for_wiz(low[0],house) + fn_for_low(low[1:],house)
return fn_for_wiz(wiz,'')
对于这个函数，我使用 return 语句直接返回想要的值。
所以我想知道我应该使用哪一个。 我可以使用累加器结束第三个 fn 以形成尾递归以获得更好的性能。但是，对我来说，第一个和第二个函数的方法要简单得多，也更容易理解或调试，这使得它们更好用，尤其是当设计函数变得更大更复杂时。 但是有什么约定吗？它们之间有什么区别？
补充问题：使用与第一个 fn 相同方法的函数是否可以视为尾递归？
解决方法
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