Create file but if name exists add number
Question:
Does Python have any built-in functionality to add a number to a filename if it already exists?
My idea is that it would work the way certain OS’s work – if a file is output to a directory where a file of that name already exists, it would append a number or increment it.
I.e: if “file.pdf” exists it will create “file2.pdf”, and next time “file3.pdf”.
Answers:
In a way, Python has this functionality built into the tempfile
module. Unfortunately, you have to tap into a private global variable, tempfile._name_sequence
. This means that officially, tempfile
makes no guarantee that in future versions _name_sequence
even exists — it is an implementation detail.
But if you are okay with using it anyway, this shows how you can create uniquely named files of the form file#.pdf
in a specified directory such as /tmp
:
import tempfile
import itertools as IT
import os
def uniquify(path, sep = ''):
def name_sequence():
count = IT.count()
yield ''
while True:
yield '{s}{n:d}'.format(s = sep, n = next(count))
orig = tempfile._name_sequence
with tempfile._once_lock:
tempfile._name_sequence = name_sequence()
path = os.path.normpath(path)
dirname, basename = os.path.split(path)
filename, ext = os.path.splitext(basename)
fd, filename = tempfile.mkstemp(dir = dirname, prefix = filename, suffix = ext)
tempfile._name_sequence = orig
return filename
print(uniquify('/tmp/file.pdf'))
I was trying to implement the same thing in my project but @unutbu’s answer seemed too ‘heavy’ for my needs so I came up with following code finally:
import os
index = ''
while True:
try:
os.makedirs('../hi'+index)
break
except WindowsError:
if index:
index = '('+str(int(index[1:-1])+1)+')' # Append 1 to number in brackets
else:
index = '(1)'
pass # Go and try create file again
Just in case someone stumbled upon this and requires something simpler.
Since the tempfile hack A) is a hack and B) still requires a decent amount of code anyway, I went with a manual implementation. You basically need:
- A way to Safely create a file if and only if it does not exist (this is what the tempfile hack affords us).
- A generator for filenames.
- A wrapping function to hide the mess.
I defined a safe_open that can be used just like open:
def iter_incrementing_file_names(path):
"""
Iterate incrementing file names. Start with path and add " (n)" before the
extension, where n starts at 1 and increases.
:param path: Some path
:return: An iterator.
"""
yield path
prefix, ext = os.path.splitext(path)
for i in itertools.count(start=1, step=1):
yield prefix + ' ({0})'.format(i) + ext
def safe_open(path, mode):
"""
Open path, but if it already exists, add " (n)" before the extension,
where n is the first number found such that the file does not already
exist.
Returns an open file handle. Make sure to close!
:param path: Some file name.
:return: Open file handle... be sure to close!
"""
flags = os.O_CREAT | os.O_EXCL | os.O_WRONLY
if 'b' in mode and platform.system() == 'Windows':
flags |= os.O_BINARY
for filename in iter_incrementing_file_names(path):
try:
file_handle = os.open(filename, flags)
except OSError as e:
if e.errno == errno.EEXIST:
pass
else:
raise
else:
return os.fdopen(file_handle, mode)
# Example
with safe_open("some_file.txt", "w") as fh:
print("Hello", file=fh)
I haven’t tested this yet but it should work, iterating over possible filenames until the file in question does not exist at which point it breaks.
def increment_filename(fn):
fn, extension = os.path.splitext(path)
n = 1
yield fn + extension
for n in itertools.count(start=1, step=1)
yield '%s%d.%s' % (fn, n, extension)
for filename in increment_filename(original_filename):
if not os.isfile(filename):
break
recently I encountered the same thing and here is my approach:
import os
file_name = "file_name.txt"
if os.path.isfile(file_name):
expand = 1
while True:
expand += 1
new_file_name = file_name.split(".txt")[0] + str(expand) + ".txt"
if os.path.isfile(new_file_name):
continue
else:
file_name = new_file_name
break
This works for me.
The initial file name is 0.yml
, if it exists, it will add one until meet the requirement
import os
import itertools
def increment_filename(file_name):
fid, extension = os.path.splitext(file_name)
yield fid + extension
for n in itertools.count(start=1, step=1):
new_id = int(fid) + n
yield "%s%s" % (new_id, extension)
def get_file_path():
target_file_path = None
for file_name in increment_filename("0.yml"):
file_path = os.path.join('/tmp', file_name)
if not os.path.isfile(file_path):
target_file_path = file_path
break
return target_file_path
A little bit later but there is still something like this should work properly, mb it will be useful for someone.
You can use built-in iterator to do this ( image downloader as example for you ):
def image_downloader():
image_url = 'some_image_url'
for count in range(10):
image_data = requests.get(image_url).content
with open(f'image_{count}.jpg', 'wb') as handler:
handler.write(image_data)
Files will increment properly. Result is:
image.jpg
image_0.jpg
image_1.jpg
image_2.jpg
image_3.jpg
image_4.jpg
image_5.jpg
image_6.jpg
image_7.jpg
image_8.jpg
image_9.jpg
If all files being numbered isn’t a problem, and you know beforehand the name of the file to be written, you could simply do:
import os
counter = 0
filename = "file{}.pdf"
while os.path.isfile(filename.format(counter)):
counter += 1
filename = filename.format(counter)
import os
class Renamer():
def __init__(self, name):
self.extension = name.split('.')[-1]
self.name = name[:-len(self.extension)-1]
self.filename = self.name
def rename(self):
i = 1
if os.path.exists(self.filename+'.'+self.extension):
while os.path.exists(self.filename+'.'+self.extension):
self.filename = '{} ({})'.format(self.name,i)
i += 1
return self.filename+'.'+self.extension
I’ve implemented a similar solution with pathlib
:
Create file-names that match the pattern path/<file-name>-dd.ext
. Perhaps this solution can help…
import pathlib
from toolz import itertoolz as itz
def file_exists_add_number(path_file_name, digits=2):
pfn = pathlib.Path(path_file_name)
parent = pfn.parent # parent-dir of file
stem = pfn.stem # file-name w/o extension
suffix = pfn.suffix # NOTE: extension starts with '.' (dot)!
try:
# search for files ending with '-dd.ext'
last_file = itz.last(parent.glob(f"{stem}-{digits * '?'}{suffix}"))
except:
curr_no = 1
else:
curr_no = int(last_file.stem[-digits:]) + 1
# int to string and add leading zeros
curr_no = str(last_no).zfill(digits)
path_file_name = parent / f"{stem}-{curr_no}{suffix}"
return str(path_file_name)
Pls note: That solution starts at 01
and will only find file-pattern containing -dd
!
I ended up writing my own simple function for this. Primitive, but gets the job done:
def uniquify(path):
filename, extension = os.path.splitext(path)
counter = 1
while os.path.exists(path):
path = filename + " (" + str(counter) + ")" + extension
counter += 1
return path
I found that the os.path.exists() conditional function did what I needed. I’m using a dictionary-to-csv saving as an example, but the same logic could work for any file type:
import os
def smart_save(filename, dict):
od = filename + '_' # added underscore before number for clarity
for i in np.arange(0,500,1): # I set an arbitrary upper limit of 500
d = od + str(i)
if os.path.exists(d + '.csv'):
pass
else:
with open(d + '.csv', 'w') as f: #or any saving operation you need
for key in dict.keys():
f.write("%s,%sn"%(key, dictionary[key]))
break
Note: this appends a number (starting at 0) to the file name by default, but it’s easy to shift that around.
Let’s say you already have those files:
This function generates the next available non-already-existing filename, by adding a _1, _2, _3, … suffix before the extension if necessary:
import os
def nextnonexistent(f):
fnew = f
root, ext = os.path.splitext(f)
i = 0
while os.path.exists(fnew):
i += 1
fnew = '%s_%i%s' % (root, i, ext)
return fnew
print(nextnonexistent('foo.txt')) # foo_3.txt
print(nextnonexistent('bar.txt')) # bar_1.txt
print(nextnonexistent('baz.txt')) # baz.txt
Easy way for create new file if this name in your folder
if 'sample.xlsx' in os.listdir('testdir/'):
i = 2
while os.path.exists(f'testdir/sample ({i}).xlsx'):
i += 1
wb.save(filename=f"testdir/sample ({i}).xlsx")
else:
wb.save(filename=f"testdir/sample.xlsx")
This function validates if the file name exists using regex expresion and recursion
def validate_outfile_name(input_path):
filename, extension = os.path.splitext(input_path)
if os.path.exists(input_path):
output_path = ""
pattern = '([0-9])'
match = re.search(pattern, filename)
if match:
version = filename[match.start() + 1]
try: new_version = int(version) + 1
except: new_version = 1
output_path = f"{filename[:match.start()]}({new_version}){extension}"
output_path = validate_outfile_name(output_path)
else:
version = 1
output_path = f"{filename}({version}){extension}"
return output_path
else:
return input_path
def create_file():
counter = 0
filename = "file"
while os.path.isfile(f"dir/{filename}{counter}.txt"):
counter += 1
print(f"{filename}{counter}.txt")
Does Python have any built-in functionality to add a number to a filename if it already exists?
My idea is that it would work the way certain OS’s work – if a file is output to a directory where a file of that name already exists, it would append a number or increment it.
I.e: if “file.pdf” exists it will create “file2.pdf”, and next time “file3.pdf”.
In a way, Python has this functionality built into the tempfile
module. Unfortunately, you have to tap into a private global variable, tempfile._name_sequence
. This means that officially, tempfile
makes no guarantee that in future versions _name_sequence
even exists — it is an implementation detail.
But if you are okay with using it anyway, this shows how you can create uniquely named files of the form file#.pdf
in a specified directory such as /tmp
:
import tempfile
import itertools as IT
import os
def uniquify(path, sep = ''):
def name_sequence():
count = IT.count()
yield ''
while True:
yield '{s}{n:d}'.format(s = sep, n = next(count))
orig = tempfile._name_sequence
with tempfile._once_lock:
tempfile._name_sequence = name_sequence()
path = os.path.normpath(path)
dirname, basename = os.path.split(path)
filename, ext = os.path.splitext(basename)
fd, filename = tempfile.mkstemp(dir = dirname, prefix = filename, suffix = ext)
tempfile._name_sequence = orig
return filename
print(uniquify('/tmp/file.pdf'))
I was trying to implement the same thing in my project but @unutbu’s answer seemed too ‘heavy’ for my needs so I came up with following code finally:
import os
index = ''
while True:
try:
os.makedirs('../hi'+index)
break
except WindowsError:
if index:
index = '('+str(int(index[1:-1])+1)+')' # Append 1 to number in brackets
else:
index = '(1)'
pass # Go and try create file again
Just in case someone stumbled upon this and requires something simpler.
Since the tempfile hack A) is a hack and B) still requires a decent amount of code anyway, I went with a manual implementation. You basically need:
- A way to Safely create a file if and only if it does not exist (this is what the tempfile hack affords us).
- A generator for filenames.
- A wrapping function to hide the mess.
I defined a safe_open that can be used just like open:
def iter_incrementing_file_names(path):
"""
Iterate incrementing file names. Start with path and add " (n)" before the
extension, where n starts at 1 and increases.
:param path: Some path
:return: An iterator.
"""
yield path
prefix, ext = os.path.splitext(path)
for i in itertools.count(start=1, step=1):
yield prefix + ' ({0})'.format(i) + ext
def safe_open(path, mode):
"""
Open path, but if it already exists, add " (n)" before the extension,
where n is the first number found such that the file does not already
exist.
Returns an open file handle. Make sure to close!
:param path: Some file name.
:return: Open file handle... be sure to close!
"""
flags = os.O_CREAT | os.O_EXCL | os.O_WRONLY
if 'b' in mode and platform.system() == 'Windows':
flags |= os.O_BINARY
for filename in iter_incrementing_file_names(path):
try:
file_handle = os.open(filename, flags)
except OSError as e:
if e.errno == errno.EEXIST:
pass
else:
raise
else:
return os.fdopen(file_handle, mode)
# Example
with safe_open("some_file.txt", "w") as fh:
print("Hello", file=fh)
I haven’t tested this yet but it should work, iterating over possible filenames until the file in question does not exist at which point it breaks.
def increment_filename(fn):
fn, extension = os.path.splitext(path)
n = 1
yield fn + extension
for n in itertools.count(start=1, step=1)
yield '%s%d.%s' % (fn, n, extension)
for filename in increment_filename(original_filename):
if not os.isfile(filename):
break
recently I encountered the same thing and here is my approach:
import os
file_name = "file_name.txt"
if os.path.isfile(file_name):
expand = 1
while True:
expand += 1
new_file_name = file_name.split(".txt")[0] + str(expand) + ".txt"
if os.path.isfile(new_file_name):
continue
else:
file_name = new_file_name
break
This works for me.
The initial file name is 0.yml
, if it exists, it will add one until meet the requirement
import os
import itertools
def increment_filename(file_name):
fid, extension = os.path.splitext(file_name)
yield fid + extension
for n in itertools.count(start=1, step=1):
new_id = int(fid) + n
yield "%s%s" % (new_id, extension)
def get_file_path():
target_file_path = None
for file_name in increment_filename("0.yml"):
file_path = os.path.join('/tmp', file_name)
if not os.path.isfile(file_path):
target_file_path = file_path
break
return target_file_path
A little bit later but there is still something like this should work properly, mb it will be useful for someone.
You can use built-in iterator to do this ( image downloader as example for you ):
def image_downloader():
image_url = 'some_image_url'
for count in range(10):
image_data = requests.get(image_url).content
with open(f'image_{count}.jpg', 'wb') as handler:
handler.write(image_data)
Files will increment properly. Result is:
image.jpg
image_0.jpg
image_1.jpg
image_2.jpg
image_3.jpg
image_4.jpg
image_5.jpg
image_6.jpg
image_7.jpg
image_8.jpg
image_9.jpg
If all files being numbered isn’t a problem, and you know beforehand the name of the file to be written, you could simply do:
import os
counter = 0
filename = "file{}.pdf"
while os.path.isfile(filename.format(counter)):
counter += 1
filename = filename.format(counter)
import os
class Renamer():
def __init__(self, name):
self.extension = name.split('.')[-1]
self.name = name[:-len(self.extension)-1]
self.filename = self.name
def rename(self):
i = 1
if os.path.exists(self.filename+'.'+self.extension):
while os.path.exists(self.filename+'.'+self.extension):
self.filename = '{} ({})'.format(self.name,i)
i += 1
return self.filename+'.'+self.extension
I’ve implemented a similar solution with pathlib
:
Create file-names that match the pattern path/<file-name>-dd.ext
. Perhaps this solution can help…
import pathlib
from toolz import itertoolz as itz
def file_exists_add_number(path_file_name, digits=2):
pfn = pathlib.Path(path_file_name)
parent = pfn.parent # parent-dir of file
stem = pfn.stem # file-name w/o extension
suffix = pfn.suffix # NOTE: extension starts with '.' (dot)!
try:
# search for files ending with '-dd.ext'
last_file = itz.last(parent.glob(f"{stem}-{digits * '?'}{suffix}"))
except:
curr_no = 1
else:
curr_no = int(last_file.stem[-digits:]) + 1
# int to string and add leading zeros
curr_no = str(last_no).zfill(digits)
path_file_name = parent / f"{stem}-{curr_no}{suffix}"
return str(path_file_name)
Pls note: That solution starts at 01
and will only find file-pattern containing -dd
!
I ended up writing my own simple function for this. Primitive, but gets the job done:
def uniquify(path):
filename, extension = os.path.splitext(path)
counter = 1
while os.path.exists(path):
path = filename + " (" + str(counter) + ")" + extension
counter += 1
return path
I found that the os.path.exists() conditional function did what I needed. I’m using a dictionary-to-csv saving as an example, but the same logic could work for any file type:
import os
def smart_save(filename, dict):
od = filename + '_' # added underscore before number for clarity
for i in np.arange(0,500,1): # I set an arbitrary upper limit of 500
d = od + str(i)
if os.path.exists(d + '.csv'):
pass
else:
with open(d + '.csv', 'w') as f: #or any saving operation you need
for key in dict.keys():
f.write("%s,%sn"%(key, dictionary[key]))
break
Note: this appends a number (starting at 0) to the file name by default, but it’s easy to shift that around.
Let’s say you already have those files:
This function generates the next available non-already-existing filename, by adding a _1, _2, _3, … suffix before the extension if necessary:
import os
def nextnonexistent(f):
fnew = f
root, ext = os.path.splitext(f)
i = 0
while os.path.exists(fnew):
i += 1
fnew = '%s_%i%s' % (root, i, ext)
return fnew
print(nextnonexistent('foo.txt')) # foo_3.txt
print(nextnonexistent('bar.txt')) # bar_1.txt
print(nextnonexistent('baz.txt')) # baz.txt
Easy way for create new file if this name in your folder
if 'sample.xlsx' in os.listdir('testdir/'):
i = 2
while os.path.exists(f'testdir/sample ({i}).xlsx'):
i += 1
wb.save(filename=f"testdir/sample ({i}).xlsx")
else:
wb.save(filename=f"testdir/sample.xlsx")
This function validates if the file name exists using regex expresion and recursion
def validate_outfile_name(input_path):
filename, extension = os.path.splitext(input_path)
if os.path.exists(input_path):
output_path = ""
pattern = '([0-9])'
match = re.search(pattern, filename)
if match:
version = filename[match.start() + 1]
try: new_version = int(version) + 1
except: new_version = 1
output_path = f"{filename[:match.start()]}({new_version}){extension}"
output_path = validate_outfile_name(output_path)
else:
version = 1
output_path = f"{filename}({version}){extension}"
return output_path
else:
return input_path
def create_file():
counter = 0
filename = "file"
while os.path.isfile(f"dir/{filename}{counter}.txt"):
counter += 1
print(f"{filename}{counter}.txt")