如何从文件路径中提取文件名？

如何解决如何从文件路径中提取文件名？

如果您只想将文件路径截断为文件名，则可以使用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：

os.path.basename('Desktop/test.txt')
# 'test.txt'

解决方法

我有以下代码：

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,"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('')

在字典中，数据帧被命名为“文件夹/名称（csv）”，我要从字典中的键中删除前缀“ staging /”。

我怎样才能做到这一点？