我正在尝试将.Xlsx中的一组文件重命名为.xls。这是我到目前为止所做的:
allFiles = glob.glob("/*.Xlsx") # Folder that has all the files
renamed = []
for filename in allFiles:
if filename.endswith(".Xlsx"):
os.rename(filename, filename[:-5])
renamed.append(filename[:-5]) # removed the .Xlsx extension
os.path.join(renamed, '.xls') # this fails
我试图看看如何将.xls添加到上面的列表renamed
如果我逐行阅读,我想
这是删除磁盘上文件的所有.xlsx扩展名
os.rename(filename, filename[:-5]) # Problem 1
然后将没有扩展名的名称添加到列表中
renamed.append(filename[:-5])
然后尝试在整个数组上加入a)和b)文件及其扩展而不是两个路径
os.path.join(renamed, '.xls') # Problem 2 and 3
你宁愿
newname = filename[:-5] # remove extension
newname = newname + ".xls" # add new extension
os.rename(filename, newname) # rename correctly
renamed.append( ... ) # Whatever name you want in the list
另请注意,对于以小写if filename.endswith(".Xlsx"):
结尾的所有文件,False
可能是.xlsx
。
您可以使用操作系统的帮助,而不是[:-5]
:
import glob
import os
allFiles = glob.glob("c:/test/*.xlsx")
renamed = []
for filename in allFiles:
path, filename = os.path.split(filename)
basename, extension = os.path.splitext(filename)
if extension == ".xlsx":
destination = os.path.join(path, basename+".xls")
os.rename(filename, destination)
仅供参考:如果重命名是程序的唯一目的,请在Windows命令提示符下尝试ren *.xlsx *.xls
。
if filename.endswith(".Xlsx"):
应该永远是真的。os.rename(filename, filename[:-5]) # this renames foo.Xlsx to foo, everything after it is too late.
renamed.append(filename[:-5]) # This adds the file w/o the extension, but also w/o the new extension.
os.path.join(renamed, '.xls') # This is a statement which would produce a result if called correctly (i. e. with a string instead of a list), but the result is discarded.
相反,做
basename = filename[:-5]
newname = os.path.join(basename, '.xls')
os.rename(filename, newname)
renamed.append(basename) # or newname? Choose what you need.
如果我理解正确,您目前正在按以下步骤划分流程:
os.path.join
不会将列表作为输入)为了保持简单,我只会重命名为新的扩展名,如果你需要renamed
列表,请填充它。像这样:
allFiles = glob.glob("/*.Xlsx") <- Folder that has all the files
renamed = []
for filename in allFiles:
if filename.endswith(".Xlsx"):
new_name = filename[:-5]+'.xls'
os.rename(filename, new_name)
renamed.append(new_name)
os.path.join
不重命名该文件。您应该使用os.rename
方法直接重命名:
os.rename(filename, filename[:-5]+'.xls')