Python提取DataFrame中File_Path列的第4级目录路径

原始数据 #

import pandas as pd

df = pd.DataFrame({
    'File_Path': [
        '/data/application/AANX/aanx-dataeng-slas-sysyphus/scripts/s_shell/call_iws/call_PP_NEXT_RTBA_MAU_IND_INVE_D.sh',
        '/data/application/AANX/aanx-dataeng-slas-sysyphus/scripts/s_shell/call_iws/call_PP_NEXT_RTBA_MAU_IND_EMPF_D.sh',
        '/data/app_next_best_action/call_nba_as11.sh',
        '/data/application/AAIN/aain-srv-motor-extracao-next/iws/call_run_extract_default.sh cdlc_ing',
        'sh /data/processos/current/aplicacao/AAVR/ACN10/scr/exec_fim_grupo.sh ACN10_ARQ_1'
    ]
})

需求说明 #

提取File_Path列路径树结构的第4项，特殊情况处理：

• 路径层级不足4层时，直接取整个路径
• 路径开头带有sh 前缀的，先去除该前缀再处理

实现代码 #

def extract_parent_path(file_path):
    # 移除开头的"sh "前缀
    cleaned_path = file_path.replace(r'^sh\s+', '', regex=True)
    # 分离路径与后续参数，仅保留路径部分
    path_part = cleaned_path.split(' ', 1)[0]
    # 分割路径为层级列表，过滤空元素（处理绝对路径开头的/）
    path_levels = [level for level in path_part.split('/') if level]
    
    if len(path_levels) >= 4:
        # 取前4层，拼接为带首尾/的路径
        return '/' + '/'.join(path_levels[:4]) + '/'
    else:
        # 层级不足4层，返回完整路径
        return path_part

# 应用函数生成新列
df['Parent_path'] = df['File_Path'].apply(extract_parent_path)

处理结果 #

File_Path                                        Parent_path
0  /data/application/AANX/aanx-dataeng-slas-sysyphus/scripts/s_shel...  /data/application/AANX/aanx-dataeng-slas-sysyphus/
1  /data/application/AANX/aanx-dataeng-slas-sysyphus/scripts/s_shel...  /data/application/AANX/aanx-dataeng-slas-sysyphus/
2                          /data/app_next_best_action/call_nba_as11.sh                          /data/app_next_best_action/call_nba_as11.sh
3  /data/application/AAIN/aain-srv-motor-extracao-next/iws/call_run...       /data/application/AAIN/aain-srv-motor-extracao-next/
4  sh /data/processos/current/aplicacao/AAVR/ACN10/scr/exec_fim_gr...               /data/processos/current/aplicacao/

内容的提问来源于stack exchange，提问作者gfernandes