散装搜索/替换使用python

Question

文件名我有：

1. Excel文件为A1：B2。
2. 一个包含200个jpeg文件的文件夹。

我试图与价值搜索文件名的文件夹中Column A如果发现不改变文件的扩展名的文件夹中，在Column B值替换它。

在这里，我坚持使用各种打滑来做到这一点，但失败了。这里是我的代码：

import os 
import xlrd 
path = r'c:\users\c_thv\desktop\x.xls' 
#collect the files in fexceler 
path1 = r'c:\users\c_thv\desktop' 
data = [] 
for name in os.listdir(path1): 
    if os.path.isfile(os.path.join(path1, name)): 
    fileName, fileExtension = os.path.splitext(name) 
    if fileExtension == '.py': 
     data.append(fileName) 
#print data 
#collect the filenames for changing 
book = xlrd.open_workbook(path) 
sheet = book.sheet_by_index(0) 
cell = sheet.cell(0,0) 
cells = sheet.row_slice(rowx=0,start_colx=0,end_colx=2) 
excel = [] 
#collect the workable data in an list 
for cell in cells: 
    excel.append(cell) 
#print excel 
#compare list for matches 
for i,j in enumerate(excel): 
    if j in data[:]: 
    os.rename(excel[i],data[i]) 

Answer 1

尝试print "Match found"后if j in data[:]:只是为了检查，如果条件是见过。我的猜测是没有匹配，因为列表data是python filemanes（if fileExtension == '.py'）上的完整列表，您正在寻找excel列表中的jpeg文件。 此外，old未定义。

编辑：

如果我理解正确的话，这将有助于：

import os, xlrd 

path = 'c:/users/c_thv/desktop' #path to jpg files 
path1 = 'c:/users/c_thv/desktop/x.xls' 

data =[] #list of jpg filenames in folder 

#lets create a filenames list without the jpg extension 
for name in os.listdir(path): 
    fileName, fileExtension = os.path.splitext(name) 
    if fileExtension =='.jpg': 
     data.append(fileName) 


#lets create a list of old filenames in the excel column a 

book = xlrd.open_workbook(path1) 
sheet = book.sheet_by_index(0) 
oldNames =[] 
for row in range(sheet.nrows): 
    oldNames.append(sheet.cell_value(row,0)) 


#lets create a list with the new names in column b 
newNames =[] 
for row in range(sheet.nrows): 
    newNames.append(sheet.cell_value(row,1)) 


#now create a dictionary with the old name in a and the corresponding new name in b 

fileNames = dict(zip(oldNames,newNames)) 

print fileNames 

#lastly rename your jpg files 

for f in data: 
    if f in fileNames.keys(): 
     os.rename(path+'/'+f+'.jpg', path+'/'+fileNames[f]+'.jpg')