我有一个pandas数据框，希望删除所有\之前和之后的所有内容，以便剩下的都是可执行文件。这就是我想要实现的目标：C:\Windows\System32\services.exe到services.exe
Parent Child PID PID System_Or_User
0 C:\Windows\System32\services.exe C:\Windows\System32\svchost.exe 10396 752 System
1 C:\Windows\System32\services.exe C:\Windows\System32\svchost.exe 11688 752 System
2 C:\Windows\System32\services.exe C:\Windows\System32\svchost.exe 11624 752 System
我已经尝试了一些类似的操作，但是似乎无法解决问题，可能是因为Windows和python中使用的\不喜欢它：
PID['Parent'] = PID['Parent'].apply(lambda x: x[0].split('\ ')[-1])
PID['Parent'] = PID['Parent'].apply(lambda x: x[0].split(' \ ')[+1])
答案 0 :(得分：3)
使用struct PlayerData: Decodable {
let playerID: Int?
let platIDRef: Int?
let EnrolledDateTime: String?
let PlayerType: String?
let LastCheckIn: Double?
let SerialNo: Int?
let Offered: Double?
let PlayerReferralCode: Int?
let PlayerStats: Double?
let PlayerUpstream: Int?
}
和[[1,1,1532890155915,0,7618.37,1,8047.76,1,0.01197066,0],[2,1,1532890155915,0,7618.37,1,8046.87,1,0.27672821,0],[3,1,1532890155915,0,7618.37,1,8045.26,1,0.27672821,0],[4,1,1532890155915,0,7618.37,1,8038.04,1,0.11222798,0],[5,1,1532890155915,0,7618.37,1,8037.94,1,0.09222818,0],[6,1,1532890155915,0,7618.37,1,8036.3,1,0.1994227,0],[7,1,1532890155915,0,7618.37,1,8035.4,1,0.44272289,0],[8,1,1532890155915,0,7618.37,1,8034.52,1,0.04333567,0],[9,1,1532890155915,0,7618.37,1,8034.51,1,0.40587824,0],[10,1,1532890155915,0,7618.37,1,8032.3,1,0.10256018,0],[11,1,1532890155915,0,7618.37,1,8031.13,1,0.00637662,0],[12,1,1532890155915,0,7618.37,1,8031.12,1,0.46122825,0],[13,1,1532890155915,0,7618.37,1,8030.34,1,0.12184043,0],[14,1,1532890155915,0,7618.37,1,8030.08,1,0.01381566,0],[15,1,1532890155915,0,7618.37,1,8028.18,1,0.01001035,0],[16,1,1532890155915,0,7618.37,1,8028.16,1,0.37072738,0],[17,1,1532890155915,0,7618.37,1,8028.14,1,0.13835319,0],[18,1,1532890155915,0,7618.37,1,8027.71,1,0.4302254,0],[19,1,1532890155915,0,7618.37,1,8026.13,1,0.002,0],[20,1,1532890155915,0,7618.37,1,8025.71,1,0.04610317,0],[21,1,1532890155915,0,7618.37,1,8379.29,1,0.00301798,1],[22,1,1532890155915,0,7618.37,1,8379.3,1,0.0698277,1],[23,1,1532890155915,0,7618.37,1,8380.3,1,0.32028728,1],[24,1,1532890155915,0,7618.37,1,8380.63,1,0.04697547,1],[25,1,1532890155915,0,7618.37,1,8382.57,1,0.01067624,1],[26,1,1532890155915,0,7618.37,1,8385.38,1,0.01161041,1],[27,1,1532890155915,0,7618.37,1,8385.4,1,0.16014364,1],[28,1,1532890155915,0,7618.37,1,8385.52,1,0.1241647,1],[29,1,1532890155915,0,7618.37,1,8386.13,1,0.12978041,1],[30,1,1532890155915,0,7618.37,1,8386.82,1,0.36803144,1],[31,1,1532890155915,0,7618.37,1,8387.9,1,0.06358541,1],[32,1,1532890155915,0,7618.37,1,8387.94,1,0.56588357,1],[33,1,1532890155915,0,7618.37,1,8389.71,1,0.10676243,1],[34,1,1532890155915,0,7618.37,1,8390.52,1,0.05338121,1],[35,1,1532890155915,0,7618.37,1,8390.61,1,0.64999101,1],[36,1,1532890155915,0,7618.37,1,8393.96,1,0.00368495,1],[37,1,1532890155915,0,7618.37,1,8393.97,1,0.32669303,1],[38,1,1532890155915,0,7618.37,1,8394.6,1,0.01363635,1],[39,1,1532890155915,0,7618.37,1,8395.11,1,0.53381214,1],[40,1,1532890155915,0,7618.37,1,8395.79,1,0.07345255,1]]
，并通过键入str.split来转义反斜杠
str.get答案 1 :(得分：3)
使用str.split并通过索引转义forEach来使用-选择void validateValue() throws ValidationException {
for(String k : aList) {
validateValue(k);
}
}
s的最后一个值：
\另一个类似的想法-将str.rsplit与list一起使用，以最后一个PID['Parent'] = PID['Parent'].str.split('\\').str[-1]
PID['Child'] = PID['Child'].str.split('\\').str[-1]
进行拆分，以提高性能：
n=1详细信息：
\PID['Parent'] = PID['Parent'].str.rsplit('\\', n=1).str[-1]
PID['Child'] = PID['Child'].str.rsplit('\\', n=1).str[-1]
答案 2 :(得分：3)
您正在处理路径。如果您需要跨平台解决方案，建议将拆分内容留给os.path。
这应该与str.方法一样快（或更快）。
import os
df['Parent'] = [os.path.basename(v) for v in df['Parent']]
df['Child'] = [os.path.basename(v) for v in df['Child']]
或者，您可以使用os.path.split。
df['Parent'] = [os.path.split(v)[-1] for v in df['Parent']]
df['Child'] = [os.path.split(v)[-1] for v in df['Child']]
答案 3 :(得分：2)
或者str.replace
df.Parent.str.replace(r"C:\\Windows\\System32\\","")
Out[25]:
0 services.exe
1 services.exe
2 services.exe
Name: Parent, dtype: object