如何在python中编写递归函数

Question

我已经被这个问题的解决方案困住了很长一段时间，我已经编写了一段按要求工作的代码，但我似乎在编译结束时遇到了一个错误

您需要设计一个迭代和递归函数，称为 replicate_iter 和 replicate_recur 分别接收两个 参数： times 是重复的次数和 data which 是要重复的数字或字符串。

函数应该返回一个包含重复数据的数组 争论。例如，replicate_recur(3, 5) 或 replicate_iter(3,5) 应该返回 [5,5,5]。如果时间参数是负数或零， 返回一个空数组。如果参数无效，则引发 ValueError。

我的代码如下：

def replicate_iter(times,data):

    emptyArray = []
    if not isinstance(data, int) and not isinstance(data, str):
        raise ValueError
    if times <= 0:
        print emptyArray
    if not isinstance(times,int):
        raise ValueError
    else:
        while times > 0:
            emptyArray.append(data)
            times -= 1
        return emptyArray

array = []

def replicate_recur(times,data):

    if not isinstance(data,int) and not isinstance(data,str):
        raise ValueError
    if not isinstance(times,int):
        raise ValueError
    if times <= 0 and len(array) != 0:
        return array
    elif times <= 0 and len(array) <=0:
        return []
    else:
        array.append(data)
        replicate_recur(times - 1,data)

请多多指教

错误信息：

Answer 1

首先，考虑一下：

def f(times,data):
  return [] if times == 0 else f(times - 1,data) + [data]

print(f(3,5)) # [5,5,5]

现在，关于您的递归解决方案，(1) 为了访问 array，replicate_recur 需要在其开头声明“global array”，因为变量 array 已声明超出函数范围； (2) 将递归调用“replicate_recur(times - 1,data)”修改为“return replicate_recur(times - 1,data)”，以便函数在times 大于零时实际返回一个值。 （也就是说，据我了解，为递归函数使用全局累加器通常被认为是不受欢迎的形式。）

Answer 2

根据您提供的示例，也许这就是您想要的：

array = []
def replicate_recur(times, val):
    if(isinstance(times, int) and times > 0):
        array.append(val)
        return replicate_recur(times-1, val)
    return array
print(replicate_recur(3, 5)) # [5,5,5]



array = []
def replicate_iter(times, val):
    if(isinstance(times, int) and times > 0):
        for i in range(times):
            array.append(val)
        return array
    raise ValueError
print(replicate_iter(3, 5)) #[5,5,5]

Answer 3

您不能使用全局变量。第二次调用你的函数会导致错误的结果。请改用内部函数或可选参数。

def replicate_recur(times,data, array=None):
    if array is None:
        array = []
    if times <= 0:
        return array
    array.append(data)
    return replicate_recur(times - 1, data, array)

Answer 4

也许你应该拆分逻辑，总是变得更具可读性，所以它变得更容易实现。你应该避免使用全局变量，尤其是全局列表，它们会给你带来很大的痛苦。

def rep_rec(times, data, pack=[]):
    pack += [data]
    return pack if times < 2 else rep_rec(times-1, data, pack)


def replicate_recur(times, data):
    if not times or not data:
        return []
    if isinstance(times, int) and (isinstance(data, int) or isinstance(data, str)):
        return rep_rec(times, data) if times > 0 else []
    raise ValueError

Answer 5

"""
A.  In recursions, in general, some input is minimized until some minimum condition.
    "eg: in f(x-2), x is the input" 



B.  In the following recursion, multiples of x "eg: x TIMES f(x-2)" are
    computed as x gets smaller and smaller by 2 "eg: f(x MINUS 2)", 
    until x finally reaches 1 "eg: if x==ONE return 1", the minimum CONDITION value.



C.  The recursion multiplies by many values generated by calls of function "f",
    until mininum is reached.
    "eg: x TIMES value-AT-f-CALL_a(x-2) TIMES value-AT-f-CALL_b(x-2)....



D.  When mininum CONDITION is reached, or in other words, when x reaches 1, the recursion 
    multiplies by 1, instead of another value generated by function "f" call.
    "eg: x TIMES value-AT-f-CALL_a(x-2) TIMES value-AT-f-CALL_b(x-2) TIMES 1"

    NOTE: This is why the recursion ends, as it no longer multiplies by value at f-call.



E.  In "TIMES 1" in D above, because 1 is the minimum RETURN value
    "eg: if x==1: return ONE", this means the minimum CONDITION is reached.

    NOTE: if we had x PLUS f(x-2), then the mininum value would be 
    ADDED TO (instead of multiplied by) the SUM (instead of the product).
"""

def f(x):
    """This is a recursive function f to generate product of x, as x decreases by 2"""
    if x == 1:
        return 1
    else:
        return (x * f(x-2))