[英]Rename the name of files in a folder with the help of Python Script, using name map from Excel sheet
[英]Rename the file names in the folder with the help of Python Script, using name map from Excel sheet
我希望將其重命名的文件夾中有許多CSV文件。 有一個Excel工作表,其中包含要重命名為文件夾的文件的名稱。
文件夾中的文件名為
TestData_30April.csv
TestData_20April.csv
TestData_18April.csv etc
而excel工作表包含的名稱為
0.25-TestData_30April
0.98-TestData_20April
0.33-TestData_20April etc
此外,excel工作表中的第一行包含標題名稱,而病房的第二行包含要重命名的文件名。
我的目標是將所有其他文件的TestData_30April.csv重命名為0.25-TestData_30April.csv。
這是代碼:
#Excel Sheet containing name of files to be renamed in that folder
path="C:\\Users\\Desktop\\Test_Data\\Test_Summary.csv"
#Folder Containg all orginal file names
dir = "C:\\Users\\Desktop\\Wear_Data"
wb = xlrd.open_workbook(path)
sheet = wb.sheet_by_index(0)
sheet.cell_value(0, 0)
#In excel sheet column X or col_values(23) contains the file name to be renamed
print(sheet.col_values(23))
list_of_filename_in_folder = [] # name of the files in the folder
list_of_filename_in_excel = [] #name of the files in excel
path_to_folder = '' # base path of folder
for name in list_of_filename_in_excel:
excel_file_name = os.path.join(path_to_folder, name,'.csv')
newname = name
if '-' in name:
newname = name.split('-')[1]
dir_file_name = os.path.join(path_to_folder,newname,'.csv' )
if os.path.exists(dir_file_name):
print('changing file name {} to {}'.format(newname,name))
os.rename(dir_file_name, excel_file_name)
else:
print('no file {} with name found in location'.format(newname+'.csv')
這是錯誤:
XLRDError: Unsupported format, or corrupt file: Expected BOF record;
請幫助解決此錯誤。
盡管您可以使用Excel打開csv文件,但.csv
文件與通常的excel文件不同(以.xlsx
結尾)。 Python附帶了一種非常方便的處理csv文件的方式: csv
模塊 。
假設您的數據看起來像您的示例,則可以執行以下操作:
import csv
import os
path= 'C:\\Users\\Desktop\\Test_Data\\Test_Summary.csv'
dir = 'C:\\Users\\Desktop\\Wear_Data'
# open the .csv file with the csv module
with open(path, 'r') as f:
csv_file = csv.reader(f)
# read the new file name from every row
for row in csv_file:
# assuming the new file path is stored in the first column (= row[0])
new_file_name = row[0] + '.csv'
# your old file should always have the same pattern according to your example
old_file_name = new_file_name.split('-')[1] + '.csv'
old_file = os.path.join(dir, old_file_name)
new_file = os.path.join(dir, new_file_name)
# rename the file
os.rename(old_file, new_file)
我沒有測試此代碼段,但我認為它顯示了使用csv
模塊如何工作的基本原理。
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