Python基于多条件规则为pandas DataFrame新增result列实现求助

方案1：使用numpy.select（推荐，大体积数据效率更高） #

该方法是向量化操作，相比逐行遍历的apply方法性能优势明显，适合绝大多数场景。

完整实现代码： #

import pandas as pd
import numpy as np

# 示例原始数据（实际使用时替换成你自己的DataFrame即可）
data = [['1', 'stp', 'salaried', '> 10 lakh'],
        ['2', 'stp', 'business', '> 10 lakh'],
        ['3', 'n_stp', 'salaried', '<= 10 lakh'],
        ['4', 'n_stp', 'other', '任意值'],
        ['5', 'stp', 'other', '任意值']]
df = pd.DataFrame(data, columns = ['s.no', 'flg', 'emp', 'inc'])

# 按规则定义条件列表和对应返回值列表，顺序要一一对应
conditions = [
    # 规则1
    df['emp'].isin(['salaried', 'business']) & (df['inc'] == '> 10 lakh') & (df['flg'] == 'stp'),
    # 规则2
    df['emp'].isin(['salaried', 'business']) & (df['inc'] == '<= 10 lakh') & (df['flg'] == 'n_stp'),
    # 规则5
    (df['emp'] == 'other') & (df['flg'] == 'n_stp'),
    # 规则3
    df['emp'].isin(['salaried', 'business']) & (df['inc'] == '> 10 lakh') & (df['flg'] == 'n_stp'),
    # 规则4
    df['emp'].isin(['salaried', 'business']) & (df['inc'] == '<= 10 lakh') & (df['flg'] == 'stp'),
    # 规则6
    (df['emp'] == 'other') & (df['flg'] == 'stp')
]

choices = [
    'no_issue',
    'no_issue',
    'no issue',
    'issue',
    'issue',
    'issue'
]

# 新增result列，default参数为不满足所有规则时的默认填充值，可按需修改
df['result'] = np.select(conditions, choices, default=np.nan)

print(df)

运行输出结果： #

s.no    flg       emp          inc    result
0    1    stp  salaried    > 10 lakh  no_issue
1    2    stp  business    > 10 lakh  no_issue
2    3  n_stp  salaried  <= 10 lakh  no_issue
3    4  n_stp     other         任意值  no issue
4    5    stp     other         任意值     issue

方案2：使用自定义函数+apply（适合逻辑后续要频繁调整的场景） #

如果后续规则还会经常迭代，自定义函数的可读性和可维护性更高，缺点是数据量超过10万行时性能较低。

实现代码： #

import pandas as pd

# 示例数据
data = [['1', 'stp', 'salaried', '> 10 lakh'],
        ['2', 'stp', 'business', '> 10 lakh'],
        ['3', 'n_stp', 'salaried', '<= 10 lakh']]
df = pd.DataFrame(data, columns = ['s.no', 'flg', 'emp', 'inc'])

# 自定义规则匹配函数
def match_rule(row):
    emp = row['emp']
    flg = row['flg']
    inc = row['inc']
    if emp in ['salaried', 'business']:
        if (inc == '> 10 lakh' and flg == 'stp') or (inc == '<= 10 lakh' and flg == 'n_stp'):
            return 'no_issue'
        else:
            return 'issue'
    elif emp == 'other':
        return 'no issue' if flg == 'n_stp' else 'issue'
    # 可补充其他emp取值的规则，默认返回None
    return None

# 逐行应用函数生成新列
df['result'] = df.apply(match_rule, axis=1)

内容的提问来源于stack exchange，提问作者KReEd