解析某些文件类型的目录 [英] parsing directory for certain filetypes

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问题描述



i写了一个函数来解析一个给定的目录并制作一个排序列表

的.txt文件,.doc扩展名.it工作,但我想要知道是不是这个b $ b太膨胀了......这可以用更有效的方式重写吗?


这里是...

来自字符串导入拆分

来自os.path导入isdir,join,normpath

来自os import listdir


def parsefolder(dirname):

filenms = []

folder = dirname

isadr = isdir(文件夹)

if(isadr):

dirlist = listdir(文件夹)

filenm =""

for d in dirlist:

filenm = x

if(filenm.endswith((" txt"," doc"))):

nmparts = []

nmparts = split(filenm,''。'')

if((nmparts [1] ==''txt'')或(nmparts [1] == ''doc'')):

filenms.append(filenm)

filenms.sort()

filenameslist = []

filenameslist = [normpa th(在文件中为y加入(文件夹,y))

numifiles = len(文件名列表)

打印文件名列表

返回文件名列表

folder =''F:/ mysys / code / tstfolder''

parsefolder(文件夹)

谢谢,

RG

hi
i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?

here it is...

from string import split
from os.path import isdir,join,normpath
from os import listdir

def parsefolder(dirname):
filenms=[]
folder=dirname
isadr=isdir(folder)
if (isadr):
dirlist=listdir(folder)
filenm=""
for x in dirlist:
filenm=x
if(filenm.endswith(("txt","doc"))):
nmparts=[]
nmparts=split(filenm,''.'' )
if((nmparts[1]==''txt'') or (nmparts[1]==''doc'')):
filenms.append(filenm)
filenms.sort()
filenameslist=[]
filenameslist=[normpath(join(folder,y)) for y in filenms]
numifiles=len(filenameslist)
print filenameslist
return filenameslist
folder=''F:/mysys/code/tstfolder''
parsefolder(folder)
thanks,
RG

推荐答案

royG napisa ??(a):
royG napisa??(a):

i写了一个函数解析一个给定的目录并制作一个已排序的列表

的.txt文件扩展名.doc有效。但是我想知道它是否b / b
太过膨胀..这可以用更有效的方式重写吗?
i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?



可能这应该重写并且应该非常紧凑。也许你应该

抢字符串:


find

Probably this should be rewriten and should be very compact. Maybe you should
grab string:

find


dirname -type f -a \(-name' '* .txt'' - o-name''* .doc''\)


并按\ n分开?

-

UFO职业
www.totalizm.org
dirname -type f -a \( -name ''*.txt'' -o -name ''*.doc'' \)

and split by "\n"?
--
UFO Occupation
www.totalizm.org


3月10日上午8:57,royG< roygeor ... @ gmail.comwrote:
On Mar 10, 8:57 am, royG <roygeor...@gmail.comwrote:

i写了一个函数来解析一个给定的目录并制作一个排序列表

的.txt文件,.doc扩展名.it工作,但我想知道它是否

太臃肿了......这可以用更有效的方式重写吗?
i wrote a function to parse a given directory and make a sorted list
of files with .txt,.doc extensions .it works,but i want to know if it
is too bloated..can this be rewritten in more efficient manner?



尝试''glob''模块。


....

Jay

Try the ''glob'' module.

....
Jay


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