使用递归来反转列表，并使用python计算列表中元素的出现次数答案

【问题标题】：Using recursion to reverse a list, and count the number of occurances of an element within the list, using python使用递归来反转列表，并使用python计算列表中元素的出现次数
【发布时间】：2013-04-22 00:23:14
【问题描述】：

我正在尝试使用递归和列表来练习，方法是编写一个函数，该函数接受一个字符串，将其拆分为一个列表，然后使用递归来生成字符串的反转版本，并计算次数搜索词出现在列表中。换句话说，如果用户输入“The quick brown fox”，并且想知道“duck”在该句子中出现的频率，程序会输出“fox brown quick The”，然后是“duck is找到（在本例中为 0）。看起来我的代码目前正在运行，但是，它不会打印出 REVERSE 和 COUNT 函数的结果。有人可以向我解释一下原因吗？

此函数反转字符串中元素的顺序

def REVERSE(strng):
    new_list=''
#base case
    if len(strng)==0:
        return new_list+""
#recursive call
#returns the first element in the string, and adds it to the end of the rest of the string called recursively from the second element
    else:
        new_list+=REVERSE(strng[1:]) + strng[0]
        return new_list
    print (new_list)

计算字符串中子字符串出现次数的函数

def COUNT(strng,srch):
        count=0
    #base case
        if len(strng)==0:
            return count
    #recursive call in event search term found in the first element of the list
        elif strng[0]==srch:
            count+=1
            return COUNT(strng[1:],srch)
    #recursive call in event search term not found in first element of list
        else:
            count+=0
            return COUNT(strng[1:],srch)   
        print ("The term" + srch + "occurs" + count + "times")

这是调用这两个函数的程序。我将它们保存在单独的文件中以练习导入等

from functions import *
def main():
        terms = input ("Enter the list of terms:\n").split(" ")
    query = input("Enter a query term:\n")
    print("List in reverse order:")
    REVERSE(terms)
    print()
    COUNT(terms, query)
main()

【问题讨论】：

标签： python list recursion count reverse

【解决方案1】：

REVERSE 和 COUNT 都永远不会执行这些打印语句 - 在执行到达 prints 之前，它们将始终 return。

您可能希望从REVERSE 和COUNT 中删除无用的prints，并在main 中打印它们的返回值：

def main():
terms = input ("Enter the list of terms:\n").split(" ")
query = input("Enter a query term:\n")
print("List in reverse order:")
print(REVERSE(terms))
print()
print(COUNT(terms, query))

【讨论】：

【解决方案2】：

到目前为止，在我看来，以下内容可能会对您有所帮助。

#!/usr/bin/python

def REVERSE(strng):
    new_list=''
    #base case
    if len(strng)==0:
        new_list+=""
    #recursive call
    #returns the first element in the string, and adds it to the end of the rest of the string called recursively from the second element
    else:
        new_list+=REVERSE(strng[1:]) + strng[0] + " "
    return new_list


def COUNT(strng,srch):
        count=0
        #base case
        if len(strng)==0:
            count = 0
        #recursive call in event search term found in the first element of the list
        elif strng[0]==srch:
            count+=1
            COUNT(strng[1:],srch)
        #recursive call in event search term not found in first element of list
        else:
            count+=0
            COUNT(strng[1:],srch)
        return count

if __name__ == '__main__':
    terms = input ("Enter the list of terms:\n").split(" ")
    query = input("Enter a query term:\n")
    print("List in reverse order: %s" % REVERSE(terms))
    print ("The term '%s' occurs %d times." % (query, COUNT(terms, query)))

【讨论】：