如何迭代Pandas DataFrame元素匹配另一DataFrame并计算匹配概率

• 遍历逻辑错误：for ind in df1默认遍历的是DataFrame的列名，不是行索引，导致取数完全不对
• 计数逻辑错误：内层循环每次都把count重置为0，永远无法累计计数
• 赋值逻辑错误：df1['t_{}'.format(i)]= count+1是给整列赋值，不是给当前匹配的行赋值
• 性能极差：嵌套循环时间复杂度为O(n*m)，数据量过万后运行速度会非常慢

# 先初始化所有type对应的哑变量列为0
for t_idx in df2.index:
    df1[f't_{t_idx}'] = 0

# 遍历df1的每一行
for ind in df1.index:
    a1_val = df1.loc[ind, 'a1']
    b1_val = df1.loc[ind, 'b1']
    # 匹配df2的行
    matched = df2[(df2['a2'] == a1_val) & (df2['b2'] == b1_val)]
    if not matched.empty:
        t_idx = matched.index[0]
        df1.loc[ind, f't_{t_idx}'] = 1

用pandas自带的merge操作实现，完全避免循环，时间复杂度接近O(n+m)：

import pandas as pd

# 先构造测试数据（修正示例中df1列名的笔误，原df1第二列应为b1而非a2）
df1 = pd.DataFrame()
df1['a1'] = ['ABC','ACC','BCC','CCC']
df1['b1'] = ['ACC','AAC','BAC','CAC']

df2 = pd.DataFrame()
df2['a2'] = ['ACC','BCC','ABC']
df2['b2'] = ['AAC','BAC','ACC']
df2['types'] = ['t1','t2','t3']

# 1. 左连接匹配type
merged = pd.merge(
    df1, df2,
    left_on=['a1', 'b1'],
    right_on=['a2', 'b2'],
    how='left'
)

# 2. 生成哑变量（你需要的t_i列）
dummies = pd.get_dummies(merged['types'], prefix='t')
df1 = pd.concat([df1, dummies], axis=1)

# 3. 计算各type的匹配概率
total_count = len(df1)
type_prob = merged['types'].value_counts() / total_count
print("各type匹配概率：\n", type_prob)

运行后得到的概率结果为：

t3    0.25
t1    0.25
t2    0.25
Name: types, dtype: float64

剩下25%为无匹配的记录。

内容的提问来源于stack exchange，提问作者Geinkehdsk