问题:如何获取按python创建日期排序的目录列表?
获取目录中所有文件的列表的最佳方法是什么,按日期[创建| 修改],在Windows机器上使用python?
What is the best way to get a list of all files in a directory, sorted by date [created | modified], using python, on a windows machine?
回答 0
更新:dirpath
在Python 3中按修改日期对条目进行排序:
import os
from pathlib import Path
paths = sorted(Path(dirpath).iterdir(), key=os.path.getmtime)
(在这里输入@Pygirl的答案以提高知名度)
如果您已经有了一个文件名列表files
,则可以在Windows上按创建时间对其进行排序:
files.sort(key=os.path.getctime)
例如,您可以使用@Jay的答案中glob
所示的文件列表。
老答案
这里有一个更详细的版本@Greg Hewgill
的答案。这是最符合问题要求的。它区分了创建日期和修改日期(至少在Windows上如此)。
#!/usr/bin/env python
from stat import S_ISREG, ST_CTIME, ST_MODE
import os, sys, time
# path to the directory (relative or absolute)
dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'
# get all entries in the directory w/ stats
entries = (os.path.join(dirpath, fn) for fn in os.listdir(dirpath))
entries = ((os.stat(path), path) for path in entries)
# leave only regular files, insert creation date
entries = ((stat[ST_CTIME], path)
for stat, path in entries if S_ISREG(stat[ST_MODE]))
#NOTE: on Windows `ST_CTIME` is a creation date
# but on Unix it could be something else
#NOTE: use `ST_MTIME` to sort by a modification date
for cdate, path in sorted(entries):
print time.ctime(cdate), os.path.basename(path)
例:
$ python stat_creation_date.py
Thu Feb 11 13:31:07 2009 stat_creation_date.py
Update: to sort dirpath
‘s entries by modification date in Python 3:
import os
from pathlib import Path
paths = sorted(Path(dirpath).iterdir(), key=os.path.getmtime)
(put @Pygirl’s answer here for greater visibility)
If you already have a list of filenames files
, then to sort it inplace by creation time on Windows:
files.sort(key=os.path.getctime)
The list of files you could get, for example, using glob
as shown in @Jay’s answer.
old answer
Here’s a more verbose version of @Greg Hewgill
‘s answer. It is the most conforming to the question requirements. It makes a distinction between creation and modification dates (at least on Windows).
#!/usr/bin/env python
from stat import S_ISREG, ST_CTIME, ST_MODE
import os, sys, time
# path to the directory (relative or absolute)
dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'
# get all entries in the directory w/ stats
entries = (os.path.join(dirpath, fn) for fn in os.listdir(dirpath))
entries = ((os.stat(path), path) for path in entries)
# leave only regular files, insert creation date
entries = ((stat[ST_CTIME], path)
for stat, path in entries if S_ISREG(stat[ST_MODE]))
#NOTE: on Windows `ST_CTIME` is a creation date
# but on Unix it could be something else
#NOTE: use `ST_MTIME` to sort by a modification date
for cdate, path in sorted(entries):
print time.ctime(cdate), os.path.basename(path)
Example:
$ python stat_creation_date.py
Thu Feb 11 13:31:07 2009 stat_creation_date.py
回答 1
过去,我是使用Python脚本来确定目录中最近更新的文件的方式:
import glob
import os
search_dir = "/mydir/"
# remove anything from the list that is not a file (directories, symlinks)
# thanks to J.F. Sebastion for pointing out that the requirement was a list
# of files (presumably not including directories)
files = list(filter(os.path.isfile, glob.glob(search_dir + "*")))
files.sort(key=lambda x: os.path.getmtime(x))
这应该可以根据文件mtime执行您想要的操作。
编辑:请注意,如果需要,也可以使用os.listdir()代替glob.glob()-我在原始代码中使用glob的原因是我想使用glob仅搜索具有特定集合的文件文件扩展名,glob()更适合。要使用listdir,结果如下所示:
import os
search_dir = "/mydir/"
os.chdir(search_dir)
files = filter(os.path.isfile, os.listdir(search_dir))
files = [os.path.join(search_dir, f) for f in files] # add path to each file
files.sort(key=lambda x: os.path.getmtime(x))
I’ve done this in the past for a Python script to determine the last updated files in a directory:
import glob
import os
search_dir = "/mydir/"
# remove anything from the list that is not a file (directories, symlinks)
# thanks to J.F. Sebastion for pointing out that the requirement was a list
# of files (presumably not including directories)
files = list(filter(os.path.isfile, glob.glob(search_dir + "*")))
files.sort(key=lambda x: os.path.getmtime(x))
That should do what you’re looking for based on file mtime.
EDIT: Note that you can also use os.listdir() in place of glob.glob() if desired – the reason I used glob in my original code was that I was wanting to use glob to only search for files with a particular set of file extensions, which glob() was better suited to. To use listdir here’s what it would look like:
import os
search_dir = "/mydir/"
os.chdir(search_dir)
files = filter(os.path.isfile, os.listdir(search_dir))
files = [os.path.join(search_dir, f) for f in files] # add path to each file
files.sort(key=lambda x: os.path.getmtime(x))
回答 2
有一个os.path.getmtime
函数可以指定自epoch以来的秒数,并且应快于os.stat
。
import os
os.chdir(directory)
sorted(filter(os.path.isfile, os.listdir('.')), key=os.path.getmtime)
There is an os.path.getmtime
function that gives the number of seconds since the epoch
and should be faster than os.stat
.
import os
os.chdir(directory)
sorted(filter(os.path.isfile, os.listdir('.')), key=os.path.getmtime)
回答 3
这是我的版本:
def getfiles(dirpath):
a = [s for s in os.listdir(dirpath)
if os.path.isfile(os.path.join(dirpath, s))]
a.sort(key=lambda s: os.path.getmtime(os.path.join(dirpath, s)))
return a
首先,我们建立文件名列表。isfile()用于跳过目录;如果应包括目录,则可以省略。然后,我们使用修改日期作为关键字对列表进行排序。
Here’s my version:
def getfiles(dirpath):
a = [s for s in os.listdir(dirpath)
if os.path.isfile(os.path.join(dirpath, s))]
a.sort(key=lambda s: os.path.getmtime(os.path.join(dirpath, s)))
return a
First, we build a list of the file names. isfile() is used to skip directories; it can be omitted if directories should be included. Then, we sort the list in-place, using the modify date as the key.
回答 4
这里是单线:
import os
import time
from pprint import pprint
pprint([(x[0], time.ctime(x[1].st_ctime)) for x in sorted([(fn, os.stat(fn)) for fn in os.listdir(".")], key = lambda x: x[1].st_ctime)])
这将调用os.listdir()以获得文件名列表,然后为每个文件名调用os.stat()以获得创建时间,然后根据创建时间进行排序。
请注意,此方法仅对每个文件调用os.stat()一次,这比对某种比较中的每个比较调用它更有效。
Here’s a one-liner:
import os
import time
from pprint import pprint
pprint([(x[0], time.ctime(x[1].st_ctime)) for x in sorted([(fn, os.stat(fn)) for fn in os.listdir(".")], key = lambda x: x[1].st_ctime)])
This calls os.listdir() to get a list of the filenames, then calls os.stat() for each one to get the creation time, then sorts against the creation time.
Note that this method only calls os.stat() once for each file, which will be more efficient than calling it for each comparison in a sort.
回答 5
不更改目录:
import os
path = '/path/to/files/'
name_list = os.listdir(path)
full_list = [os.path.join(path,i) for i in name_list]
time_sorted_list = sorted(full_list, key=os.path.getmtime)
print time_sorted_list
# if you want just the filenames sorted, simply remove the dir from each
sorted_filename_list = [ os.path.basename(i) for i in time_sorted_list]
print sorted_filename_list
Without changing directory:
import os
path = '/path/to/files/'
name_list = os.listdir(path)
full_list = [os.path.join(path,i) for i in name_list]
time_sorted_list = sorted(full_list, key=os.path.getmtime)
print time_sorted_list
# if you want just the filenames sorted, simply remove the dir from each
sorted_filename_list = [ os.path.basename(i) for i in time_sorted_list]
print sorted_filename_list
回答 6
在python 3.5+
from pathlib import Path
sorted(Path('.').iterdir(), key=lambda f: f.stat().st_mtime)
In python 3.5+
from pathlib import Path
sorted(Path('.').iterdir(), key=lambda f: f.stat().st_mtime)
回答 7
如果您想按日期顺序读取具有某些扩展名的文件,这是我使用不带过滤器的glob的答案(Python 3)。
dataset_path='/mydir/'
files = glob.glob(dataset_path+"/morepath/*.extension")
files.sort(key=os.path.getmtime)
Here’s my answer using glob without filter if you want to read files with a certain extension in date order (Python 3).
dataset_path='/mydir/'
files = glob.glob(dataset_path+"/morepath/*.extension")
files.sort(key=os.path.getmtime)
回答 8
# *** the shortest and best way ***
# getmtime --> sort by modified time
# getctime --> sort by created time
import glob,os
lst_files = glob.glob("*.txt")
lst_files.sort(key=os.path.getmtime)
print("\n".join(lst_files))
# *** the shortest and best way ***
# getmtime --> sort by modified time
# getctime --> sort by created time
import glob,os
lst_files = glob.glob("*.txt")
lst_files.sort(key=os.path.getmtime)
print("\n".join(lst_files))
回答 9
sorted(filter(os.path.isfile, os.listdir('.')),
key=lambda p: os.stat(p).st_mtime)
您可以使用os.walk('.').next()[-1]
而不是进行过滤os.path.isfile
,但这会在列表中留下os.stat
无效的符号链接,从而使它们失败。
sorted(filter(os.path.isfile, os.listdir('.')),
key=lambda p: os.stat(p).st_mtime)
You could use os.walk('.').next()[-1]
instead of filtering with os.path.isfile
, but that leaves dead symlinks in the list, and os.stat
will fail on them.
回答 10
from pathlib import Path
import os
sorted(Path('./').iterdir(), key=lambda t: t.stat().st_mtime)
要么
sorted(Path('./').iterdir(), key=os.path.getmtime)
要么
sorted(os.scandir('./'), key=lambda t: t.stat().st_mtime)
其中,m时间为修改时间。
from pathlib import Path
import os
sorted(Path('./').iterdir(), key=lambda t: t.stat().st_mtime)
or
sorted(Path('./').iterdir(), key=os.path.getmtime)
or
sorted(os.scandir('./'), key=lambda t: t.stat().st_mtime)
where m time is modified time.
回答 11
这是学习的基本步骤:
import os, stat, sys
import time
dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'
listdir = os.listdir(dirpath)
for i in listdir:
os.chdir(dirpath)
data_001 = os.path.realpath(i)
listdir_stat1 = os.stat(data_001)
listdir_stat2 = ((os.stat(data_001), data_001))
print time.ctime(listdir_stat1.st_ctime), data_001
this is a basic step for learn:
import os, stat, sys
import time
dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'
listdir = os.listdir(dirpath)
for i in listdir:
os.chdir(dirpath)
data_001 = os.path.realpath(i)
listdir_stat1 = os.stat(data_001)
listdir_stat2 = ((os.stat(data_001), data_001))
print time.ctime(listdir_stat1.st_ctime), data_001
回答 12
如果文件是到不存在文件的符号链接,则Alex Coventry的答案将产生异常,以下代码更正了该答案:
import time
import datetime
sorted(filter(os.path.isfile, os.listdir('.')),
key=lambda p: os.path.exists(p) and os.stat(p).st_mtime or time.mktime(datetime.now().timetuple())
如果文件不存在,则使用now(),符号链接将位于列表的最后。
Alex Coventry’s answer will produce an exception if the file is a symlink to an unexistent file, the following code corrects that answer:
import time
import datetime
sorted(filter(os.path.isfile, os.listdir('.')),
key=lambda p: os.path.exists(p) and os.stat(p).st_mtime or time.mktime(datetime.now().timetuple())
When the file doesn’t exist, now() is used, and the symlink will go at the very end of the list.
回答 13
这是一条简单的几行,用于查找扩展并提供排序选项
def get_sorted_files(src_dir, regex_ext='*', sort_reverse=False):
files_to_evaluate = [os.path.join(src_dir, f) for f in os.listdir(src_dir) if re.search(r'.*\.({})$'.format(regex_ext), f)]
files_to_evaluate.sort(key=os.path.getmtime, reverse=sort_reverse)
return files_to_evaluate
Here is a simple couple lines that looks for extention as well as provides a sort option
def get_sorted_files(src_dir, regex_ext='*', sort_reverse=False):
files_to_evaluate = [os.path.join(src_dir, f) for f in os.listdir(src_dir) if re.search(r'.*\.({})$'.format(regex_ext), f)]
files_to_evaluate.sort(key=os.path.getmtime, reverse=sort_reverse)
return files_to_evaluate
回答 14
为了os.scandir
确保完整性(比快2倍pathlib
):
import os
sorted(os.scandir('/tmp/test'), key=lambda d: d.stat().st_mtime)
For completeness with os.scandir
(2x faster over pathlib
):
import os
sorted(os.scandir('/tmp/test'), key=lambda d: d.stat().st_mtime)
回答 15
这是我的版本:
import os
folder_path = r'D:\Movies\extra\new\dramas' # your path
os.chdir(folder_path) # make the path active
x = sorted(os.listdir(), key=os.path.getctime) # sorted using creation time
folder = 0
for folder in range(len(x)):
print(x[folder]) # print all the foldername inside the folder_path
folder = +1
This was my version:
import os
folder_path = r'D:\Movies\extra\new\dramas' # your path
os.chdir(folder_path) # make the path active
x = sorted(os.listdir(), key=os.path.getctime) # sorted using creation time
folder = 0
for folder in range(len(x)):
print(x[folder]) # print all the foldername inside the folder_path
folder = +1
回答 16
也许您应该使用shell命令。在Unix / Linux中,使用sort传递的find可能可以执行您想要的操作。
Maybe you should use shell commands. In Unix/Linux, find piped with sort will probably be able to do what you want.