# How can I open multiple files using “with open” in Python?

## The Question :

714 people think this question is useful

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?

1112 people think this answer is useful

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.

106 people think this answer is useful

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



65 people think this answer is useful

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:



19 people think this answer is useful

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:
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.

16 people think this answer is useful

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]



3 people think this answer is useful

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:



1 people think this answer is useful

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:
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
...