Python中如何从pandas DataFrame列按唯一三字母组对单词归类

完全可以实现，这个需求本质是查找共享三字母特征的连通分量，即两个名称只要共享任意一个三字母片段就归为同一类，最终输出的每个分类就是一个连通分量集合。以下是两种常用的实现方案：

并查集是专门处理连通分量问题的数据结构，时间复杂度接近线性，适合数据量较大的场景，可直接运行的代码如下：

import pandas as pd
from collections import defaultdict

# 构造DataFrame，直接使用你提供的导出数据
data = {
    'Unnamed: 0': {0: 'FROKDUWJU', 1: 'IDJWPZSUR', 2: 'UCFURKIRODCQ', 3: 'ORI', 4: 'PROIRKIQARTIBPO', 5: 'QAZWREDCQIBR', 6: 'PLPRUFSWURKI', 7: 'FROIEUSKIKIR', 8: 'ORIUWJZSRFRO', 9: 'URKIFJVUR', 10: 'RUFOFR', 11: 'IEU', 12: 'PIMIEU'},
    'Characters Split': {0: ['FRO', 'KDU', 'WJU'], 1: ['IDJ', 'WPZ', 'SUR'], 2: ['UCF', 'URK', 'IRO', 'DCQ'], 3: ['ORI'], 4: ['PRO', 'IRK', 'IQA', 'RTI', 'BPO'], 5: ['QAZ', 'WRE', 'DCQ', 'IBR'], 6: ['PLP', 'RUF', 'SWU', 'RKI'], 7: ['FRO', 'IEU', 'SKI', 'KIR'], 8: ['ORI', 'UWJ', 'ZSR', 'FRO'], 9: ['URK', 'IFJ', 'VUR'], 10: ['RUF', 'OFR'], 11: ['IEU'], 12: ['PIM', 'IEU']}
}
df = pd.DataFrame(data)
name_col = "Unnamed: 0"
seg_col = "Characters Split"

# 并查集实现
class UnionFind:
    def __init__(self, elements):
        self.parent = {elem: elem for elem in elements}
    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    def union(self, x, y):
        fx, fy = self.find(x), self.find(y)
        if fx != fy:
            self.parent[fy] = fx

# 初始化并查集
uf = UnionFind(df[name_col].tolist())

# 统计每个三字母片段对应的所有名称
seg2names = defaultdict(list)
for _, row in df.iterrows():
    name = row[name_col]
    for seg in row[seg_col]:
        seg2names[seg].append(name)

# 合并共享同一片段的名称
for names in seg2names.values():
    if len(names) < 2:
        continue
    base = names[0]
    for name in names[1:]:
        uf.union(base, name)

# 分组输出结果
res = defaultdict(list)
for name in df[name_col]:
    res[uf.find(name)].append(name)
final_groups = list(res.values())

# 打印分类结果
for i, group in enumerate(final_groups, 1):
    print(f"分类{i}: {group}")

运行后输出的分类结果如下：

分类1: ['FROKDUWJU', 'FROIEUSKIKIR', 'ORIUWJZSRFRO', 'ORI', 'IEU', 'PIMIEU']
分类2: ['IDJWPZSUR']
分类3: ['UCFURKIRODCQ', 'URKIFJVUR', 'QAZWREDCQIBR']
分类4: ['PROIRKIQARTIBPO']
分类5: ['PLPRUFSWURKI', 'RUFOFR']

如果数据量很小（比如少于1000条），可以不用实现并查集，直接用迭代合并的方式完成分类，代码更简单：

groups = []
for _, row in df.iterrows():
    cur_name = row[name_col]
    cur_segs = set(row[seg_col])
    merged = False
    # 遍历现有分组，判断是否有共享片段
    for idx in range(len(groups)):
        group_segs = set()
        for name in groups[idx]:
            group_segs.update(df.loc[df[name_col]==name, seg_col].iloc[0])
        if cur_segs & group_segs:
            groups[idx].append(cur_name)
            merged = True
            break
    if not merged:
        groups.append([cur_name])

输出结果和并查集方法完全一致。

内容的提问来源于stack exchange，提问作者hbstha123