字符串列表去重合并与分组统计：指定输出格式实现求助

我有如下格式的字符串列表：

"a 05/13/22 apple"        
"a 05/13/22 apple"        
"b 05/13/22 apple"        
"b 05/13/22 apple"        
"c 05/13/22 apple"        
"c 05/13/22 apple"        
"a 05/27/22 strawberry"   
"a 05/27/22 strawberry"   
"b 05/27/22 strawberry"   
"b 05/27/22 strawberry"   
"c 05/27/22 strawberry"   
"c 05/27/22 strawberry"   
"a 07/29/22 banana"       
"a 07/29/22 banana"       
"b 07/29/22 banana"       
"b 07/29/22 banana"       
"c 07/29/22 banana"       
"c 07/29/22 banana"       

我已将每个字符串拆分为letter、date、fruit三个独立值，希望输出以下格式的结果：

6 occurrences found for apple 
05/13/22 apple [a,b,c]

6 occurrences found for strawberry
05/27/22 strawberry [a,b,c]

6 occurrences found for banana
07/29/22 banana [a,b,c]  

尝试用以下循环代码处理，但逻辑有误，无法得到正确结果：

fruit_exists = []
pexists = []
pfruit, pdate, pletters = '','',[]
for instance in fruit_info:
        letter, date, fruit = instance.split()
        if fruit not in fruit_exists:
            fruit_exists.append(fruit)    
        if pdate == '':
            pfruit = fruit
            pdate = date
        if pdate+pfruit not in pexists:
            if letter not in pletters:
                pletters.append(letter)
            if pdate != date:
                print(f'{pdate} - {pfruit} for {", ".join(pletters)}')
                pexists.append(pdate+pfruit)
                pfruit = fruit
                pdate = date
                pletters = []
    print(f'{pdate} - {pfruit} for {", ".join(pletters)}')

解决思路 #

原代码逻辑过于复杂，状态判断容易出错。可以用字典分组存储的方式简化处理：

1. 按水果名称作为key，统一管理对应水果的总出现次数、日期、关联字母集合
2. 遍历数据时，自动统计次数、用集合去重字母，无需手动判断重复
3. 最后遍历字典，按要求格式输出

实现代码 #

# 假设fruit_info是已拆分前的原始字符串列表
fruit_info = [
    "a 05/13/22 apple",
    "a 05/13/22 apple",
    "b 05/13/22 apple",
    "b 05/13/22 apple",
    "c 05/13/22 apple",
    "c 05/13/22 apple",
    "a 05/27/22 strawberry",
    "a 05/27/22 strawberry",
    "b 05/27/22 strawberry",
    "b 05/27/22 strawberry",
    "c 05/27/22 strawberry",
    "c 05/27/22 strawberry",
    "a 07/29/22 banana",
    "a 07/29/22 banana",
    "b 07/29/22 banana",
    "b 07/29/22 banana",
    "c 07/29/22 banana",
    "c 07/29/22 banana"
]

# 初始化存储结构：key为水果名，value存储统计数据
fruit_stats = {}

for item in fruit_info:
    letter, date, fruit = item.split()
    if fruit not in fruit_stats:
        # 首次遇到该水果，初始化统计项
        fruit_stats[fruit] = {
            "total": 1,
            "date": date,
            "letters": {letter}
        }
    else:
        # 更新已有水果的统计数据
        fruit_stats[fruit]["total"] += 1
        fruit_stats[fruit]["letters"].add(letter)

# 按要求格式输出结果
for fruit, stats in fruit_stats.items():
    print(f"{stats['total']} occurrences found for {fruit}")
    # 将字母集合转为逗号分隔的字符串，包裹在[]中
    letters_str = ",".join(sorted(stats["letters"]))
    print(f"{stats['date']} {fruit} [{letters_str}]\n")

代码说明 #

• 用字典fruit_stats集中管理每个水果的所有统计数据，逻辑清晰，避免复杂的状态切换
• 使用集合letters自动去重字母，省去手动判断重复的步骤
• 最终遍历字典输出，完全匹配期望的格式

内容的提问来源于stack exchange，提问作者paul m