Python：按列表元素首段ID分组统计对应元素个数

给定如下列表：

['21,*,1', '21,stuff,1', '21,*,10', '21,stuff,4', '21,*,1', '21,*,114', '21,stuff,1', 
 '28,*,37', '28,stuff,1', '28,*,4', '28,stuff,4', '28,*,4', '28,XXX,3', '28,stuff,1', '28,XXX,23', '28,XXX,87',
 '38,*,72', '38,stuff,9', '38,*,6', '38,stuff,32', '38,stuff,3', '38,stuff,3', '38,XXX,1', '38,*,23',
 '42,*,37', '42,XXX,1', '42,stuff,4']

需要按每个字符串逗号分割后的首段ID（ID始终升序排列）分组，统计每个ID对应的元素数量，示例输出为：

[['21', 7], ['28', 9], ['38', 8], ['42', 3]]

作为Python新手，寻求简洁优雅的实现方法。

方法一：使用collections.defaultdict统计 #

这是最直观的计数方式，适合新手理解：

from collections import defaultdict

data = ['21,*,1', '21,stuff,1', '21,*,10', '21,stuff,4', '21,*,1', '21,*,114', '21,stuff,1', 
        '28,*,37', '28,stuff,1', '28,*,4', '28,stuff,4', '28,*,4', '28,XXX,3', '28,stuff,1', '28,XXX,23', '28,XXX,87',
        '38,*,72', '38,stuff,9', '38,*,6', '38,stuff,32', '38,stuff,3', '38,stuff,3', '38,XXX,1', '38,*,23',
        '42,*,37', '42,XXX,1', '42,stuff,4']

# 初始化默认值为0的字典，用于计数
count_dict = defaultdict(int)
for s in data:
    # 拆分字符串，取首段作为ID
    id_ = s.split(',')[0]
    count_dict[id_] += 1

# 按ID升序排序，转成要求的二维列表格式
result = [[k, v] for k, v in sorted(count_dict.items())]
print(result)

方法二：使用itertools.groupby（最简洁） #

因为题目说明ID始终是升序排列的，groupby可以直接对连续相同ID的元素分组，代码更简洁：

from itertools import groupby

data = ['21,*,1', '21,stuff,1', '21,*,10', '21,stuff,4', '21,*,1', '21,*,114', '21,stuff,1', 
        '28,*,37', '28,stuff,1', '28,*,4', '28,stuff,4', '28,*,4', '28,XXX,3', '28,stuff,1', '28,XXX,23', '28,XXX,87',
        '38,*,72', '38,stuff,9', '38,*,6', '38,stuff,32', '38,stuff,3', '38,stuff,3', '38,XXX,1', '38,*,23',
        '42,*,37', '42,XXX,1', '42,stuff,4']

# 定义分组键：提取每个字符串的首段ID
get_id = lambda x: x.split(',')[0]
# 分组后统计每组长度，直接生成结果
result = [[key, len(list(group))] for key, group in groupby(data, get_id)]
print(result)

方法三：使用collections.Counter #

先提取所有ID，再用Counter统计次数，逻辑清晰：

from collections import Counter

data = ['21,*,1', '21,stuff,1', '21,*,10', '21,stuff,4', '21,*,1', '21,*,114', '21,stuff,1', 
        '28,*,37', '28,stuff,1', '28,*,4', '28,stuff,4', '28,*,4', '28,XXX,3', '28,stuff,1', '28,XXX,23', '28,XXX,87',
        '38,*,72', '38,stuff,9', '38,*,6', '38,stuff,32', '38,stuff,3', '38,stuff,3', '38,XXX,1', '38,*,23',
        '42,*,37', '42,XXX,1', '42,stuff,4']

# 提取所有ID组成列表
id_list = [s.split(',')[0] for s in data]
# 统计每个ID的出现次数
counts = Counter(id_list)
# 排序后转成目标格式
result = [[k, v] for k, v in sorted(counts.items())]
print(result)

内容的提问来源于stack exchange，提问作者Seb