Python实现《One man went to mow》童谣输出格式问题求助

问题说明 #

需要用Python实现童谣《One man went to mow》的输出，但现有代码无法生成符合要求的格式，尤其无法正确拼接出类似“Four men, three men, two men, one man”的倒序列表内容。

现有代码 #

num = ['One man','Two men','Three men','Four men','Five men',
       'Six men','Seven men','Eight men','Nine men','Ten men']

for i in range(0,len(num)):
    print(num[i],'went to mow, went to mow a meadow,')
    # Check if i is greater than 0, to avoid index out of range errors
    if i > 0:
        # Loop through the previous elements in the num list
        for n in range(i-1,-1,-1):
            print(num[i],num[n],'and his dog, went to mow a meadow.')

期望输出示例 #

One man went to mow, went to mow a meadow
One man and his dog
Went to mow a meadow

Two men went to mow
Went to mow a meadow
Two men, one man and his dog
Went to mow a meadow

Three men went to mow
Went to mow a meadow
Three men, two men, one man
And his dog
Went to mow a meadow

Four men went to mow
Went to mow a meadow
Four men, three men, two men, one man
And his dog
Went to mow a meadow

...（以此类推至第十段）

解决方案 #

核心问题是现有代码未正确拼接倒序人员列表，且重复输出当前项。以下是修正后的代码，针对不同段落的格式差异做了针对性处理：

num = ['One man', 'Two men', 'Three men', 'Four men', 'Five men',
       'Six men', 'Seven men', 'Eight men', 'Nine men', 'Ten men']

for i in range(len(num)):
    current = num[i]
    # 处理每段开头的换行逻辑
    if i == 0:
        print(f"{current} went to mow, went to mow a meadow")
    else:
        print(f"{current} went to mow")
        print("Went to mow a meadow")
    
    # 处理人员列表和"his dog"部分的格式
    if i == 0:
        print(f"{current} and his dog")
    else:
        # 生成从当前项到第一项的倒序列表，并拼接成字符串
        people_list = num[i::-1]
        people_str = ", ".join(people_list)
        if i == 1:
            print(f"{people_str} and his dog")
        else:
            print(people_str)
            print("And his dog")
    
    # 每段结尾固定行
    print("Went to mow a meadow")
    # 段落间添加空行（最后一段除外）
    if i != len(num) - 1:
        print()

代码关键说明 #

• num[i::-1]：生成从当前索引i到索引0的倒序列表，比如i=3时，得到['Four men', 'Three men', 'Two men', 'One man']
• ", ".join(people_list)：将列表元素用逗号连接成符合要求的连续字符串
• 分三段逻辑处理不同段落的格式差异，完全匹配期望输出的换行和拼接规则
• 通过判断是否为最后一段，控制段落间的空行分隔

内容的提问来源于stack exchange，提问作者Tommy