I want to change a couple of files at one time, iff I can write to all of them. I’m wondering if I somehow can combine the multiple open calls with the with statement:

try:
  with open('a', 'w') as a and open('b', 'w') as b:
    do_something()
except IOError as e:
  print 'Operation failed: %s' % e.strerror

If that’s not possible, what would an elegant solution to this problem look like?

As of Python 2.7 (or 3.1 respectively) you can write

with open('a', 'w') as a, open('b', 'w') as b:
    do_something()

In earlier versions of Python, you can sometimes use contextlib.nested() to nest context managers. This won’t work as expected for opening multiples files, though — see the linked documentation for details.

In the rare case that you want to open a variable number of files all at the same time, you can use contextlib.ExitStack, starting from Python version 3.3:

with ExitStack() as stack:
    files = [stack.enter_context(open(fname)) for fname in filenames]
    # Do something with "files"

Most of the time you have a variable set of files, you likely want to open them one after the other, though.

Just replace and with , and you’re done:

try:
    with open('a', 'w') as a, open('b', 'w') as b:
        do_something()
except IOError as e:
    print 'Operation failed: %s' % e.strerror

For opening many files at once or for long file paths, it may be useful to break things up over multiple lines. From the Python Style Guide as suggested by @Sven Marnach in comments to another answer:

with open('/path/to/InFile.ext', 'r') as file_1, \
     open('/path/to/OutFile.ext', 'w') as file_2:
    file_2.write(file_1.read())

Nested with statements will do the same job, and in my opinion, are more straightforward to deal with.

Let’s say you have inFile.txt, and want to write it into two outFile’s simultaneously.

with open("inFile.txt", 'r') as fr:
    with open("outFile1.txt", 'w') as fw1:
        with open("outFile2.txt", 'w') as fw2:
            for line in fr.readlines():
                fw1.writelines(line)
                fw2.writelines(line)

EDIT:

I don’t understand the reason of the downvote. I tested my code before publishing my answer, and it works as desired: It writes to all of outFile’s, just as the question asks. No duplicate writing or failing to write. So I am really curious to know why my answer is considered to be wrong, suboptimal or anything like that.

Since Python 3.3, you can use the class ExitStack from the contextlib module to safely
open an arbitrary number of files.

It can manage a dynamic number of context-aware objects, which means that it will prove especially useful if you don’t know how many files you are going to handle.

In fact, the canonical use-case that is mentioned in the documentation is managing a dynamic number of files.

with ExitStack() as stack:
    files = [stack.enter_context(open(fname)) for fname in filenames]
    # All opened files will automatically be closed at the end of
    # the with statement, even if attempts to open files later
    # in the list raise an exception

If you are interested in the details, here is a generic example in order to explain how ExitStack operates:

from contextlib import ExitStack

class X:
    num = 1

    def __init__(self):
        self.num = X.num
        X.num += 1

    def __repr__(self):
        cls = type(self)
        return '{cls.__name__}{self.num}'.format(cls=cls, self=self)

    def __enter__(self):
        print('enter {!r}'.format(self))
        return self.num

    def __exit__(self, exc_type, exc_value, traceback):
        print('exit {!r}'.format(self))
        return True

xs = [X() for _ in range(3)]

with ExitStack() as stack:
    print(len(stack._exit_callbacks)) # number of callbacks called on exit
    nums = [stack.enter_context(x) for x in xs]
    print(len(stack._exit_callbacks))

print(len(stack._exit_callbacks))
print(nums)

Output:

0
enter X1
enter X2
enter X3
3
exit X3
exit X2
exit X1
0
[1, 2, 3]

With python 2.6 It will not work, we have to use below way to open multiple files:

with open('a', 'w') as a:
    with open('b', 'w') as b:

Late answer (8 yrs), but for someone looking to join multiple files into one, the following function may be of help:

def multi_open(_list):
    out=""
    for x in _list:
        try:
            with open(x) as f:
                out+=f.read()
        except:
            pass
            # print(f"Cannot open file {x}")
    return(out)

fl = ["C:/bdlog.txt", "C:/Jts/tws.vmoptions", "C:/not.exist"]
print(multi_open(fl))

2018-10-23 19:18:11.361 PROFILE  [Stop Drivers] [1ms]
2018-10-23 19:18:11.361 PROFILE  [Parental uninit] [0ms]
...
# This file contains VM parameters for Trader Workstation.
# Each parameter should be defined in a separate line and the
...

I’m used to doing print >>f, "hi there"

However, it seems that print >> is getting deprecated. What is the recommended way to do the line above?

Update: Regarding all those answers with "\n"…is this universal or Unix-specific? IE, should I be doing "\r\n" on Windows?

This should be as simple as:

with open('somefile.txt', 'a') as the_file:
    the_file.write('Hello\n')

From The Documentation:

Do not use os.linesep as a line terminator when writing files opened in text mode (the default); use a single ‘\n’ instead, on all platforms.

Some useful reading:

• The with statement
• open()
  • ‘a’ is for append, or use
  • ‘w’ to write with truncation
• os (particularly os.linesep)

You should use the print() function which is available since Python 2.6+

from __future__ import print_function  # Only needed for Python 2
print("hi there", file=f)

For Python 3 you don’t need the import, since the print() function is the default.

The alternative would be to use:

f = open('myfile', 'w')
f.write('hi there\n')  # python will convert \n to os.linesep
f.close()  # you can omit in most cases as the destructor will call it

Quoting from Python documentation regarding newlines:

On output, if newline is None, any '\n' characters written are translated to the system default line separator, os.linesep. If newline is '', no translation takes place. If newline is any of the other legal values, any '\n' characters written are translated to the given string.

The python docs recommend this way:

with open('file_to_write', 'w') as f:
    f.write('file contents\n')

So this is the way I usually do it :)

Statement from docs.python.org:

It is good practice to use the ‘with’ keyword when dealing with file objects. This has the advantage that the file is properly closed after its suite finishes, even if an exception is raised on the way. It is also much shorter than writing equivalent try-finally blocks.

Regarding os.linesep:

Here is an exact unedited Python 2.7.1 interpreter session on Windows:

Python 2.7.1 (r271:86832, Nov 27 2010, 18:30:46) [MSC v.1500 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.linesep
'\r\n'
>>> f = open('myfile','w')
>>> f.write('hi there\n')
>>> f.write('hi there' + os.linesep) # same result as previous line ?????????
>>> f.close()
>>> open('myfile', 'rb').read()
'hi there\r\nhi there\r\r\n'
>>>

On Windows:

As expected, os.linesep does NOT produce the same outcome as '\n'. There is no way that it could produce the same outcome. 'hi there' + os.linesep is equivalent to 'hi there\r\n', which is NOT equivalent to 'hi there\n'.

It’s this simple: use \n which will be translated automatically to os.linesep. And it’s been that simple ever since the first port of Python to Windows.

There is no point in using os.linesep on non-Windows systems, and it produces wrong results on Windows.

DO NOT USE os.linesep!

I do not think there is a “correct” way.

I would use:

with open ('myfile', 'a') as f: f.write ('hi there\n')

In memoriam Tim Toady.

In Python 3 it is a function, but in Python 2 you can add this to the top of the source file:

from __future__ import print_function

Then you do

print("hi there", file=f)

If you are writing a lot of data and speed is a concern you should probably go with f.write(...). I did a quick speed comparison and it was considerably faster than print(..., file=f) when performing a large number of writes.

import time    

start = start = time.time()
with open("test.txt", 'w') as f:
    for i in range(10000000):
        # print('This is a speed test', file=f)
        # f.write('This is a speed test\n')
end = time.time()
print(end - start)

On average write finished in 2.45s on my machine, whereas print took about 4 times as long (9.76s). That being said, in most real-world scenarios this will not be an issue.

If you choose to go with print(..., file=f) you will probably find that you’ll want to suppress the newline from time to time, or replace it with something else. This can be done by setting the optional end parameter, e.g.;

with open("test", 'w') as f:
    print('Foo1,', file=f, end='')
    print('Foo2,', file=f, end='')
    print('Foo3', file=f)

Whichever way you choose I’d suggest using with since it makes the code much easier to read.

Update: This difference in performance is explained by the fact that write is highly buffered and returns before any writes to disk actually take place (see this answer), whereas print (probably) uses line buffering. A simple test for this would be to check performance for long writes as well, where the disadvantages (in terms of speed) for line buffering would be less pronounced.

start = start = time.time()
long_line = 'This is a speed test' * 100
with open("test.txt", 'w') as f:
    for i in range(1000000):
        # print(long_line, file=f)
        # f.write(long_line + '\n')
end = time.time()

print(end - start, "s")

The performance difference now becomes much less pronounced, with an average time of 2.20s for write and 3.10s for print. If you need to concatenate a bunch of strings to get this loooong line performance will suffer, so use-cases where print would be more efficient are a bit rare.

Since 3.5 you can also use the pathlib for that purpose:

Path.write_text(data, encoding=None, errors=None)

Open the file pointed to in text mode, write data to it, and close the file:

import pathlib

pathlib.Path('textfile.txt').write_text('content')

When you said Line it means some serialized characters which are ended to ‘\n’ characters. Line should be last at some point so we should consider ‘\n’ at the end of each line. Here is solution:

with open('YOURFILE.txt', 'a') as the_file:
    the_file.write("Hello")

in append mode after each write the cursor move to new line, if you want to use w mode you should add \n characters at the end of the write() function:

the_file.write("Hello\n")

One can also use the io module as in:

import io
my_string = "hi there"

with io.open("output_file.txt", mode='w', encoding='utf-8') as f:
    f.write(my_string)

To write text in a file in the flask can be used:

filehandle = open("text.txt", "w")
filebuffer = ["hi","welcome","yes yes welcome"]
filehandle.writelines(filebuffer)
filehandle.close()

You can also try filewriter

pip install filewriter

from filewriter import Writer

Writer(filename='my_file', ext='txt') << ["row 1 hi there", "row 2"]

Writes into my_file.txt

Takes an iterable or an object with __str__ support.

When I need to write new lines a lot, I define a lambda that uses a print function:

out = open(file_name, 'w')
fwl = lambda *x, **y: print(*x, **y, file=out) # FileWriteLine
fwl('Hi')

This approach has the benefit that it can utilize all the features that are available with the print function.

Update: As is mentioned by Georgy in the comment section, it is possible to improve this idea further with the partial function:

from functools import partial
fwl = partial(print, file=out)

IMHO, this is a more functional and less cryptic approach.

How can I find all the files in a directory having the extension .txt in python?

You can use glob:

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
    print(file)

or simply os.listdir:

import os
for file in os.listdir("/mydir"):
    if file.endswith(".txt"):
        print(os.path.join("/mydir", file))

or if you want to traverse directory, use os.walk:

import os
for root, dirs, files in os.walk("/mydir"):
    for file in files:
        if file.endswith(".txt"):
             print(os.path.join(root, file))

Use glob.

>>> import glob
>>> glob.glob('./*.txt')
['./outline.txt', './pip-log.txt', './test.txt', './testingvim.txt']

Something like that should do the job

for root, dirs, files in os.walk(directory):
    for file in files:
        if file.endswith('.txt'):
            print file

Something like this will work:

>>> import os
>>> path = '/usr/share/cups/charmaps'
>>> text_files = [f for f in os.listdir(path) if f.endswith('.txt')]
>>> text_files
['euc-cn.txt', 'euc-jp.txt', 'euc-kr.txt', 'euc-tw.txt', ... 'windows-950.txt']

You can simply use pathlibs glob ¹:

import pathlib

list(pathlib.Path('your_directory').glob('*.txt'))

or in a loop:

for txt_file in pathlib.Path('your_directory').glob('*.txt'):
    # do something with "txt_file"

If you want it recursive you can use .glob('**/*.txt)

¹The pathlib module was included in the standard library in python 3.4. But you can install back-ports of that module even on older Python versions (i.e. using conda or pip): pathlib and pathlib2.

import os

path = 'mypath/path' 
files = os.listdir(path)

files_txt = [i for i in files if i.endswith('.txt')]

I like os.walk():

import os

for root, dirs, files in os.walk(dir):
    for f in files:
        if os.path.splitext(f)[1] == '.txt':
            fullpath = os.path.join(root, f)
            print(fullpath)

Or with generators:

import os

fileiter = (os.path.join(root, f)
    for root, _, files in os.walk(dir)
    for f in files)
txtfileiter = (f for f in fileiter if os.path.splitext(f)[1] == '.txt')
for txt in txtfileiter:
    print(txt)

Here’s more versions of the same that produce slightly different results:

glob.iglob()

import glob
for f in glob.iglob("/mydir/*/*.txt"): # generator, search immediate subdirectories 
    print f

glob.glob1()

print glob.glob1("/mydir", "*.tx?")  # literal_directory, basename_pattern

fnmatch.filter()

import fnmatch, os
print fnmatch.filter(os.listdir("/mydir"), "*.tx?") # include dot-files

path.py is another alternative: https://github.com/jaraco/path.py

from path import path
p = path('/path/to/the/directory')
for f in p.files(pattern='*.txt'):
    print f

Python v3.5+

Fast method using os.scandir in a recursive function. Searches for all files with a specified extension in folder and sub-folders.

import os

def findFilesInFolder(path, pathList, extension, subFolders = True):
    """  Recursive function to find all files of an extension type in a folder (and optionally in all subfolders too)

    path:        Base directory to find files
    pathList:    A list that stores all paths
    extension:   File extension to find
    subFolders:  Bool.  If True, find files in all subfolders under path. If False, only searches files in the specified folder
    """

    try:   # Trapping a OSError:  File permissions problem I believe
        for entry in os.scandir(path):
            if entry.is_file() and entry.path.endswith(extension):
                pathList.append(entry.path)
            elif entry.is_dir() and subFolders:   # if its a directory, then repeat process as a nested function
                pathList = findFilesInFolder(entry.path, pathList, extension, subFolders)
    except OSError:
        print('Cannot access ' + path +'. Probably a permissions error')

    return pathList

dir_name = r'J:\myDirectory'
extension = ".txt"

pathList = []
pathList = findFilesInFolder(dir_name, pathList, extension, True)

Update April 2019

If you are searching over directories which contain 10,000s files, appending to a list becomes inefficient. ‘Yielding’ the results is a better solution. I have also included a function to convert the output to a Pandas Dataframe.

import os
import re
import pandas as pd
import numpy as np


def findFilesInFolderYield(path,  extension, containsTxt='', subFolders = True, excludeText = ''):
    """  Recursive function to find all files of an extension type in a folder (and optionally in all subfolders too)

    path:               Base directory to find files
    extension:          File extension to find.  e.g. 'txt'.  Regular expression. Or  'ls\d' to match ls1, ls2, ls3 etc
    containsTxt:        List of Strings, only finds file if it contains this text.  Ignore if '' (or blank)
    subFolders:         Bool.  If True, find files in all subfolders under path. If False, only searches files in the specified folder
    excludeText:        Text string.  Ignore if ''. Will exclude if text string is in path.
    """
    if type(containsTxt) == str: # if a string and not in a list
        containsTxt = [containsTxt]

    myregexobj = re.compile('\.' + extension + '$')    # Makes sure the file extension is at the end and is preceded by a .

    try:   # Trapping a OSError or FileNotFoundError:  File permissions problem I believe
        for entry in os.scandir(path):
            if entry.is_file() and myregexobj.search(entry.path): # 

                bools = [True for txt in containsTxt if txt in entry.path and (excludeText == '' or excludeText not in entry.path)]

                if len(bools)== len(containsTxt):
                    yield entry.stat().st_size, entry.stat().st_atime_ns, entry.stat().st_mtime_ns, entry.stat().st_ctime_ns, entry.path

            elif entry.is_dir() and subFolders:   # if its a directory, then repeat process as a nested function
                yield from findFilesInFolderYield(entry.path,  extension, containsTxt, subFolders)
    except OSError as ose:
        print('Cannot access ' + path +'. Probably a permissions error ', ose)
    except FileNotFoundError as fnf:
        print(path +' not found ', fnf)

def findFilesInFolderYieldandGetDf(path,  extension, containsTxt, subFolders = True, excludeText = ''):
    """  Converts returned data from findFilesInFolderYield and creates and Pandas Dataframe.
    Recursive function to find all files of an extension type in a folder (and optionally in all subfolders too)

    path:               Base directory to find files
    extension:          File extension to find.  e.g. 'txt'.  Regular expression. Or  'ls\d' to match ls1, ls2, ls3 etc
    containsTxt:        List of Strings, only finds file if it contains this text.  Ignore if '' (or blank)
    subFolders:         Bool.  If True, find files in all subfolders under path. If False, only searches files in the specified folder
    excludeText:        Text string.  Ignore if ''. Will exclude if text string is in path.
    """

    fileSizes, accessTimes, modificationTimes, creationTimes , paths  = zip(*findFilesInFolderYield(path,  extension, containsTxt, subFolders))
    df = pd.DataFrame({
            'FLS_File_Size':fileSizes,
            'FLS_File_Access_Date':accessTimes,
            'FLS_File_Modification_Date':np.array(modificationTimes).astype('timedelta64[ns]'),
            'FLS_File_Creation_Date':creationTimes,
            'FLS_File_PathName':paths,
                  })

    df['FLS_File_Modification_Date'] = pd.to_datetime(df['FLS_File_Modification_Date'],infer_datetime_format=True)
    df['FLS_File_Creation_Date'] = pd.to_datetime(df['FLS_File_Creation_Date'],infer_datetime_format=True)
    df['FLS_File_Access_Date'] = pd.to_datetime(df['FLS_File_Access_Date'],infer_datetime_format=True)

    return df

ext =   'txt'  # regular expression 
containsTxt=[]
path = 'C:\myFolder'
df = findFilesInFolderYieldandGetDf(path,  ext, containsTxt, subFolders = True)

Python has all tools to do this:

import os

the_dir = 'the_dir_that_want_to_search_in'
all_txt_files = filter(lambda x: x.endswith('.txt'), os.listdir(the_dir))