Pandas按条件批量替换多列值的高效实现方法

目前我使用如下代码在满足特定条件时替换列取值：

df.loc[df['event'] == 1, 'pre'] = 0
df.loc[df['event'] == 1, 'post'] = 1

df.loc[df['event'] == 2, 'pre'] = 1
df.loc[df['event'] == 2, 'post'] = 0

df.loc[df['event'] == 4, 'pre'] = 1
df.loc[df['event'] == 4, 'post'] = 0

但该写法可扩展性差，无法适配更多条件场景，请问有没有更高效的实现方式？

完整测试代码如下：

import numpy as np
import pandas as pd
nfreq=500
arr=np.array([[11850,0,1],
[12310,0,3],
[13924,0,4],
[16690,0,1],
[17082,0,3],
[18746,0,4],
[21956,0,2],
[22324,0,3],
[23694,0,4],
[25382,0,1],
[25776,0,3],
[28592,0,4],
[31676,0,2],
[32028,0,3],
[33498,0,4]])
trange = np.where(arr == 3)[0]
val=np.array([arr[trange, 0],arr[trange, 2],
          (arr[trange, 0]-arr[trange-1, 0])/nfreq,
          (arr[trange+1, 0]-arr[trange, 0])/nfreq]).T


trange= np.where(arr[:,2] != 3)[0]
val_oth=np.array([arr[trange, 0],arr[trange, 2],arr[trange, 2],arr[trange, 2]]).T
val_oth[:,2]=1
val_oth[:,-1]=1
df = pd.DataFrame(np.vstack((val,val_oth)),columns=['timepoint','event','pre','post'])
df.loc[df['event'] == 1, 'pre'] = 0
df.loc[df['event'] == 1, 'post'] = 1

df.loc[df['event'] == 2, 'pre'] = 1
df.loc[df['event'] == 2, 'post'] = 0

df.loc[df['event'] == 4, 'pre'] = 1
df.loc[df['event'] == 4, 'post'] = 0
df.sort_values(by='timepoint', ascending=True,inplace=True)
df.reset_index(drop=True,inplace=True)

你可以将事件匹配规则集中管理，用矢量化操作替代重复的.loc语句，大幅提升扩展性和执行效率，以下是几种常用实现方式：

方案1：映射字典循环赋值（最易维护，适合规则不多的场景） #

把所有事件对应的pre、post取值统一存在字典里，新增规则仅需修改字典即可：

# 集中管理所有事件匹配规则，新增事件直接加键值对就行
event_rule = {
    1: (0, 1),
    2: (1, 0),
    4: (1, 0)
}

for event, (pre_val, post_val) in event_rule.items():
    df.loc[df['event'] == event, ['pre', 'post']] = pre_val, post_val

方案2：merge映射表（性能最优，适合事件规则非常多的场景） #

将规则转成DataFrame后和原表合并，全程矢量化操作，数据量越大优势越明显：

rule_df = pd.DataFrame.from_dict(
    {1: [0,1], 2:[1,0], 4:[1,0]},
    orient='index',
    columns=['pre_rule', 'post_rule']
)

# 合并规则直接覆盖原列
df = df.merge(rule_df, left_on='event', right_index=True, how='left')
df['pre'] = df['pre_rule']
df['post'] = df['post_rule']
df.drop(columns=['pre_rule', 'post_rule'], inplace=True)

方案3：np.select多条件匹配（适合规则逻辑复杂，不是简单等值匹配的场景） #

如果后续规则会加入大于、小于等复杂判断，用np.select更灵活：

conditions = [
    df['event'] == 1,
    df['event'].isin([2,4])
]
pre_values = [0, 1]
post_values = [1, 0]

df['pre'] = np.select(conditions, pre_values, default=df['pre'])
df['post'] = np.select(conditions, post_values, default=df['post'])

三种方案都能完全替代你原本的重复.loc写法，最终输出结果和原代码完全一致。

内容的提问来源于stack exchange，提问作者rpb