递归：返回列表中具有偶数个字符的项目数

Question

我很难理解递归的概念。有人可以帮助批评我的代码吗？我正在尝试以递归方式返回具有偶数位数的列表中的项目数。

alist = ["hello", "is", "there", "anybody", "out", "there?"]


def evenItems(alist):
  
    if len(alist[0]) == 0:
        return 0
    if len(alist[0]) % 2 == 0:
        
        return evenItems(alist[1:len(alist)-1] + 1 ) 
        
    else:
        return evenItems(alist[1:len(alist)-1])

Answer 1

批评：在 cmets 中：

alist = ["hello", "is", "there", "anybody", "out", "there?"]


def evenItems(alist):
  
    if len(alist[0]) == 0: # The base case is when alist is empty. So use 'alist' instead of 'alist[0]' (alist[0] is the first element)
        return 0
    if len(alist[0]) % 2 == 0:
        # A few things here:
        # - the '+1' should be on the outside of the parenthesis because it should not be part of the recursive evenItems function call
        # - To get the rest of the list you could use alist[1:] (which is from index one to the end of the list)
        return evenItems(alist[1:len(alist)-1] + 1 ) 
       
    else:
        return evenItems(alist[1:len(alist)-1])

---

我的建议

我喜欢制作不同的变量，以便我可以清楚地看到我正在处理的内容。你看我使用（currentWord 和 restOfList）。对于递归函数，我还喜欢将程序分成基本案例和递归调用（再次，只是为了更清楚）。

alist = ["hello", "is", "there", "anybody", "out", "there?"]


def evenItems(alist):

    # Base Case: If alist is empty  
    if len(alist) == 0:
        return 0
        
    # Recursive call
    currentWord = alist[0]
    restOfList = alist[1:]
            
    if len(currentWord) % 2 == 0:
        
        return evenItems(restOfList)  + 1
        
    else:
        return evenItems(restOfList)

print(evenItems(alist))

---

调试：

递归很困难。我通常喜欢打印出每次通话发生的情况，以便更好地理解。

alist = ["hello", "is", "there", "anybody", "out", "there?"]


def evenItems(alist, level):

    print("  " * level + "Enter evenItems. Level: " + str(level))
    # Base Case: If alist is empty  
    if len(alist) == 0:
        print("  " * level + "Exit evenItems list empty. Level: " + str(level))
        return 0
        
    # Recursive call
    currentWord = alist[0]
    print("  " * level + "Current Word is: " + currentWord)
    restOfList = alist[1:]
    print("  " * level + "Rest of List is: " + str(restOfList))

    currentEvenItemsCount = 0        
    if len(currentWord) % 2 == 0:
        
        currentEvenItemsCount = evenItems(restOfList, level + 1)  + 1
        
    else:
        currentEvenItemsCount = evenItems(restOfList, level + 1)

    print("  " * level + "Exit evenItems even word. Level: " + str(level))
    return currentEvenItemsCount

print(evenItems(alist, 0))

输出

Enter evenItems. Level: 0
Current Word is: hello
Rest of List is: ['is', 'there', 'anybody', 'out', 'there?']
  Enter evenItems. Level: 1
  Current Word is: is
  Rest of List is: ['there', 'anybody', 'out', 'there?']
    Enter evenItems. Level: 2
    Current Word is: there
    Rest of List is: ['anybody', 'out', 'there?']
      Enter evenItems. Level: 3
      Current Word is: anybody
      Rest of List is: ['out', 'there?']
        Enter evenItems. Level: 4
        Current Word is: out
        Rest of List is: ['there?']
          Enter evenItems. Level: 5
          Current Word is: there?
          Rest of List is: []
            Enter evenItems. Level: 6
            Exit evenItems list empty. Level: 6
          Exit evenItems even word. Level: 5
        Exit evenItems even word. Level: 4
      Exit evenItems even word. Level: 3
    Exit evenItems even word. Level: 2
  Exit evenItems even word. Level: 1
Exit evenItems even word. Level: 0
2

Answer 2

您可以考虑改为使用 negative indexing 在列表的后面执行检查，以获取最后一个元素和不包括最后一个元素的子列表，以便下一次递归调用：

def numEvenItems(alist):
    if len(alist) == 0:
        return 0
    return (1 if len(alist[-1]) % 2 == 0 else 0) + numEvenItems(alist[:-1])

lst = ["hello", "is", "there", "anybody", "out", "there?"]
print(numEvenItems(lst))

输出：

2

试试看here。