2017-07-15 124 views
2

我有以下代码:如何从文件路径中提取文件名?

os.listdir("staging") 

# Seperate filename from extension 
sep = os.sep 

# Change the casing 
for n in os.listdir("staging"): 
    print(n) 
    if os.path.isfile("staging" + sep + n): 
     filename_one, extension = os.path.splitext(n) 
     os.rename("staging" + sep + n, "staging" + sep + filename_one.lower() + extension) 

# Show the new file names 
print ('\n--------------------------------\n') 
for n in os.listdir("staging"): 
    print (n) 

# Remove the blanks, -, %, and/
for n in os.listdir("staging"): 
    print (n) 
    if os.path.isfile("staging" + sep + n): 
     filename_zero, extension = os.path.splitext(n) 
     os.rename("staging" + sep + n , "staging" + sep + filename_zero.replace(' ','_').replace('-','_').replace('%','pct').replace('/','_') + extension) 

# Show the new file names 
print ('\n--------------------------------\n') 
for n in os.listdir("staging"): 
    print (n) 

""" 
In order to fix all of the column headers and to solve the encoding issues and remove nulls, 
first read in all of the CSV's to python as dataframes, then make changes and rewrite the old files 
""" 
import os 
import glob 
import pandas as pd 

files = glob.glob(os.path.join("staging" + "/*.csv")) 

print(files) 

# Create an empty dictionary to hold the dataframes from csvs 
dict_ = {} 

# Write the files into the dictionary 
for file in files: 
    dict_[file] = pd.read_csv(file, header = 0, dtype = str, encoding = 'cp1252').fillna('') 

在词典中,dataframes被命名为“文件夹/名称(CSV)”我希望做的是从键删除前缀“分期/”词典。

我该怎么做?

回答

4

如果你想要做的是截断的文件路径,只是文件名,你可以使用os.path.basename

for file in files: 
    fname = os.path.basename(file) 
    dict_[fname] = pd.read_csv(file, header = 0, dtype = str, encoding = 'cp1252').fillna('') 

例子:

In [465]: os.path.basename('Desktop/test.txt') 
Out[465]: 'test.txt' 
+0

@COLDSPEED,我不知道如何执行你的suggestion-我有61键在字典中,他们被标记为“staging /文件名”我想将键改为“文件名” - 我试着用files = glob.glob(os.path.join(“staging”+“/ * .csv“)) print(files) 对于文件中的文件: file = os.path.basename(file) print(files) – zsad512

+0

请看如何提供[mcve]。 –

0

正如ColdSpeed说,你可以使用“操作系统.path.basename“截断文件到它的名字,但我认为你所指的是pycache数据的能力?

例如,下面是我的目录: My Directory for the Game I'm Making Atm In the Assets folder..

你看pycache文件夹?将其初始化为一个模块。 然后,您可以从该模块导入文件(例如staging.txt文件并对其进行操作)。 For Example 我使用资产文件夹级别(或应该是)的IpConfig.txt文件,并占用一行信息。

import pygame as pyg 
import sys 
import os 
import math 
import ssl 
import socket as sock 
import ipaddress as ipad 
import threading 
import random 
print("Modules Installed!") 

class two: 
    # Find out how to refer to class super construct 
    def main(Display, SecSock, ipadd, clock): 
     # I have code here that has nothing to do with the question... 


    def __init__(): 
     print("Initializing[2]...") 
     # Initialization of Pygame and SSL Socket goes here 

     searchQuery = open("IpConfig.txt", 'r') #Opening the File IpConfig(Which now should open on the top level of the game files) 

     step2 = searchQuery.readlines()# read the file 
     ipadd = step2[6] # This is what you should have or something similar where you reference the line you want to copy or manipulate. 

     main(gameDisplay, SSLSock, ipadd, clock)# Im having issues here myself - (main() is not defined it says) 
     print(ipadd) 
     print("Server Certificate Configuration Enabled...") 








    __init__() # Start up the procedure 
+0

感谢您的详细回应,但@COLDSPEED得到它的权利,我需要做的只是截断文件名插入到数据库 – zsad512

+0

我只是澄清你将如何去阅读该数据库。 – JerryPlayz101

1

article在这里工作了就好了,我

import os 
inputFilepath = 'path/to/file/foobar.txt' 
filename_w_ext = os.path.basename(inputFilepath) 
filename, file_extension = os.path.splitext(filename_w_ext) 
#filename = foobar 
#file_extension = .txt 

path, filename = os.path.split(path/to/file/foobar.txt) 
# path = path/to/file 
# filename = foobar.txt 

希望它可以帮助别人寻找这个答案