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I hope this is not trivial but I am wondering the following:
If I have a specific folder with n csv files, how could I iteratively read all of them, one at a time, and perform some calculations on their values?
For a single file, for example, I do something like this and perform some calculations on the x array:
import csv
import os
directoryPath=raw_input('Directory path for native csv file: ')
csvfile = numpy.genfromtxt(directoryPath, delimiter=",")
x=csvfile[:,2] #Creates the array that will undergo a set of calculations
I know that I can check how many csv files there are in a given folder (check here):
import glob
for files in glob.glob("*.csv"):
print files
But I failed to figure out how to possibly nest the numpy.genfromtxt() function in a for loop, so that I read in all the csv files of a directory that it is up to me to specify.
EDIT
The folder I have only has jpg and csv files. The latter are named eventX.csv, where X ranges from 1 to 50. The for loop I am referring to should therefore consider the file names the way they are.
That's how I'd do it:
import os
directory = os.path.join("c:\\","path")
for root,dirs,files in os.walk(directory):
for file in files:
if file.endswith(".csv"):
f=open(file, 'r')
# perform calculation
f.close()
Using pandas and glob as the base packages
import glob
import pandas as pd
glued_data = pd.DataFrame()
for file_name in glob.glob(directoryPath+'*.csv'):
x = pd.read_csv(file_name, low_memory=False)
glued_data = pd.concat([glued_data,x],axis=0)
I think you look for something like this
import glob
for file_name in glob.glob(directoryPath+'*.csv'):
x = np.genfromtxt(file_name,delimiter=',')[:,2]
# do your calculations
Edit
If you want to get all csv files from a folder (including subfolder) you could use subprocess instead of glob (note that this code only works on linux systems)
import subprocess
file_list = subprocess.check_output(['find',directoryPath,'-name','*.csv']).split('\n')[:-1]
for i,file_name in enumerate(file_list):
x = np.genfromtxt(file_name,delimiter=',')[:,2]
# do your calculations
# now you can use i as an index
It first searches the folder and sub-folders for all file_names using the find command from the shell and applies your calculations afterwards.
According to the documentation of numpy.genfromtxt(), the first argument can be a
File, filename, or generator to read.
That would mean that you could write a generator that yields the lines of all the files like this:
def csv_merge_generator(pattern):
for file in glob.glob(pattern):
for line in file:
yield line
# then using it like this
numpy.genfromtxt(csv_merge_generator('*.csv'))
should work. (I do not have numpy installed, so cannot test easily)
Here's a more succinct way to do this, given some path = "/path/to/dir/".
import glob
import pandas as pd
pd.concat([pd.read_csv(f) for f in glob.glob(path+'*.csv')])
Then you can apply your calculation to the whole dataset, or, if you want to apply it one by one:
pd.concat([process(pd.read_csv(f)) for f in glob.glob(path+'*.csv')])
The function below will return a dictionary containing a dataframe for each .csv file in the folder within your defined path.
import pandas as pd
import glob
import os
import ntpath
def panda_read_csv(path):
pd_csv_dict = {}
csv_files = glob.glob(os.path.join(path, "*.csv"))
for csv_file in csv_files:
file_name = ntpath.basename(csv_file)
pd_csv_dict['pd_' + file_name] = pd.read_csv(csv_file, sep=";", encoding='mac_roman')
locals().update(pd_csv_dict)
return pd_csv_dict
You can use pathlib glob functionality to list all .csv in a path, and pandas to read them.
Then it's only a matter of applying whatever function you want (which, if systematic, can also be done within the list comprehension)
import pands as pd
from pathlib import Path
path2csv = Path("/your/path/")
csvlist = path2csv.glob("*.csv")
csvs = [pd.read_csv(g) for g in csvlist ]
Another answer using list comprehension:
from os import listdir
files= [f for f in listdir("./") if f.endswith(".csv")]
You need to import the glob library and then use it like following:
import glob
path='C:\\Users\\Admin\\PycharmProjects\\db_conection_screenshot\\seclectors_absent_images'
filenames = glob.glob(path + "\*.png")
print(len(filenames))
There is an mkv file in a folder named "export". What I want to do is to make a python script which fetches the file name from that export folder.
Let's say the folder is at "C:\Users\UserName\Desktop\New_folder\export".
How do I fetch the name?
I tried using this os.path.basename and os.path.splitext .. well.. didn't work out like I expected.
os.path implements some useful functions on pathnames. But it doesn't have access to the contents of the path. For that purpose, you can use os.listdir.
The following command will give you a list of the contents of the given path:
os.listdir("C:\Users\UserName\Desktop\New_folder\export")
Now, if you just want .mkv files you can use fnmatch(This module provides support for Unix shell-style wildcards) module to get your expected file names:
import fnmatch
import os
print([f for f in os.listdir("C:\Users\UserName\Desktop\New_folder\export") if fnmatch.fnmatch(f, '*.mkv')])
Also as #Padraic Cunningham mentioned as a more pythonic way for dealing with file names you can use glob module :
map(path.basename,glob.iglob(pth+"*.mkv"))
You can use glob:
from glob import glob
pth ="C:/Users/UserName/Desktop/New_folder/export/"
print(glob(pth+"*.mkv"))
path+"*.mkv" will match all the files ending with .mkv.
To just get the basenames you can use map or a list comp with iglob:
from glob import iglob
print(list(map(path.basename,iglob(pth+"*.mkv"))))
print([path.basename(f) for f in iglob(pth+"*.mkv")])
iglob returns an iterator so you don't build a list for no reason.
I assume you're basically asking how to list files in a given directory. What you want is:
import os
print os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
If there's multiple files and you want the one(s) that have a .mkv end you could do:
import os
files = os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
mkv_files = [_ for _ in files if _[-4:] == ".mkv"]
print mkv_files
If you are searching for recursive folder search, this method will help you to get filename using os.walk, also you can get those file's path and directory using this below code.
import os, fnmatch
for path, dirs, files in os.walk(os.path.abspath(r"C:/Users/UserName/Desktop/New_folder/export/")):
for filename in fnmatch.filter(files, "*.mkv"):
print(filename)
You can use glob
import glob
for file in glob.glob('C:\Users\UserName\Desktop\New_folder\export\*.mkv'):
print(str(file).split('\')[-1])
This will list out all the files having extention .mkv as
file.mkv, file2.mkv and so on.
From os.walk you can read file paths as a list
files = [ file_path for _, _, file_path in os.walk(DIRECTORY_PATH)]
for file_name in files[0]: #note that it has list of lists
print(file_name)
I have files that I want only 'foo' and 'bar' left from split.
dn = "C:\\X\\Data\\"
files
f= C:\\X\\Data\\foo.txt
f= C:\\X\\Dats\\bar.txt
I have tried f.split(".",1)[0]
I thought since dn and .txt are pre-defined I could subtract, nope.
Split does not work for me.
How about using the proper path handling methods from os.path?
>>> f = 'C:\\X\\Data\\foo.txt'
>>> import os
>>> os.path.basename(f)
'foo.txt'
>>> os.path.dirname(f)
'C:\\X\\Data'
>>> os.path.splitext(f)
('C:\\X\\Data\\foo', '.txt')
>>> os.path.splitext(os.path.basename(f))
('foo', '.txt')
To deal with path and file names, it is best to use the built-in module os.path in Python. Please look at function dirname, basename and split in that module.
simple Example for your Help.
import os
from os import path
path_to_directory = "C:\\X\\Data"
for f in os.listdir(path_to_directory):
name , extension = path.splitext(f)
print(name)
Output
foo
bar
These two lines return a list of file names without extensions:
import os
[fname.rsplit('.', 1)[0] for fname in os.listdir("C:\\X\\Data\\")]
It seems you've left out some code. From what I can tell you're trying to split the contents of the file.
To fix your problem, you need to operate on a list of the files in the directory. That is what os.listdir does for you. I've also added a more sophisticated split. rsplit operates from the right, and will only split the first . it finds. Notice the 1 as the second argument.
another example:
f.split('\\')[-1].split('.')[0]
Using python3 and pathlib:
import pathlib
f = 'C:\\X\\Data\\foo.txt'
print(pathlib.PureWindowsPath(f).stem)
will print: 'foo'
I have a folder in Windows 7 which contains multiple .txt files. How would one get every file in said directory as a list?
You can use os.listdir(".") to list the contents of the current directory ("."):
for name in os.listdir("."):
if name.endswith(".txt"):
print(name)
If you want the whole list as a Python list, use a list comprehension:
a = [name for name in os.listdir(".") if name.endswith(".txt")]
import os
import glob
os.chdir('c:/mydir')
files = glob.glob('*.txt')
All of the answers here don't address the fact that if you pass glob.glob() a Windows path (for example, C:\okay\what\i_guess\), it does not run as expected. Instead, you need to use pathlib:
from pathlib import Path
glob_path = Path(r"C:\okay\what\i_guess")
file_list = [str(pp) for pp in glob_path.glob("**/*.txt")]
import fnmatch
import os
return [file for file in os.listdir('.') if fnmatch.fnmatch(file, '*.txt')]
If you just need the current directory, use os.listdir.
>>> os.listdir('.') # get the files/directories
>>> [os.path.abspath(x) for x in os.listdir('.')] # gets the absolute paths
>>> [x for x in os.listdir('.') if os.path.isfile(x)] # only files
>>> [x for x in os.listdir('.') if x.endswith('.txt')] # files ending in .txt only
You can also use os.walk if you need to recursively get the contents of a directory. Refer to the python documentation for os.walk.
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?
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))
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 (make sure that list contains absolute path):
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
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)
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.
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.
In python 3.5+
from pathlib import Path
sorted(Path('.').iterdir(), key=lambda f: f.stat().st_mtime)
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
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.
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)
# *** 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))
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.
For completeness with os.scandir (2x faster over pathlib):
import os
sorted(os.scandir('/tmp/test'), key=lambda d: d.stat().st_mtime)
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
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.
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
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
Add the file directory/folder in path, if you want to have specific file type add the file extension, and then get file name in chronological order.
This works for me.
import glob, os
from pathlib import Path
path = os.path.expanduser(file_location+"/"+date_file)
os.chdir(path)
saved_file=glob.glob('*.xlsx')
saved_file.sort(key=os.path.getmtime)
print(saved_file)
Turns out os.listdir sorts by last modified but in reverse so you can do:
import os
last_modified=os.listdir()[::-1]
Maybe you should use shell commands. In Unix/Linux, find piped with sort will probably be able to do what you want.