How do I create a incrementing filename in Python?

问题:

I'm creating a program that will create a file and save it to the directory with the filename sample.xml. Once the file is saved when i try to run the program again it overwrites the old file into the new one because they do have the same file name. How do I increment the file names so that whenever I try to run the code again it will going to increment the file name. and will not overwrite the existing one. I am thinking of checking the filename first on the directory and if they are the same the code will generate a new filename:

fh = open("sample.xml", "w")
rs = [blockresult]
fh.writelines(rs)
fh.close()

回答1:

I would iterate through sample[int].xml for example and grab the next available name that is not used by a file or directory.

import os

i = 0
while os.path.exists("sample%s.xml" % i):
    i += 1

fh = open("sample%s.xml" % i, "w")
....

That should give you sample0.xml initially, then sample1.xml, etc.

Note that the relative file notation by default relates to the file directory/folder you run the code from. Use absolute paths if necessary. Use os.getcwd() to read your current dir and os.chdir(path_to_dir) to set a new current dir.

回答2:

def get_nonexistant_path(fname_path):
    """
    Get the path to a filename which does not exist by incrementing path.

    Examples
    --------
    >>> get_nonexistant_path('/etc/issue')
    '/etc/issue-1'
    >>> get_nonexistant_path('whatever/1337bla.py')
    'whatever/1337bla.py'
    """
    if not os.path.exists(fname_path):
        return fname_path
    filename, file_extension = os.path.splitext(fname_path)
    i = 1
    new_fname = "{}-{}{}".format(filename, i, file_extension)
    while os.path.exists(new_fname):
        i += 1
        new_fname = "{}-{}{}".format(filename, i, file_extension)
    return new_fname

Before you open the file, call

fname = get_nonexistant_path("sample.xml")

This will either give you 'sample.xml' or - if this alreay exists - 'sample-i.xml' where i is the lowest positive integer such that the file does not already exist.

I recommend using os.path.abspath("sample.xml"). If you have ~ as home directory, you might need to expand it first.

Please note that race conditions might occur with this simple code if you have multiple instances running at the same time. If this might be a problem, please check this question.

回答3:

Try setting a count variable, and then incrementing that variable nested inside the same loop you write your file in. Include the count loop inside the name of the file with an escape character, so every loop ticks +1 and so does the number in the file.

Some code from a project I just finished:

numberLoops = #some limit determined by the user
currentLoop = 1
while currentLoop < numberLoops:
    currentLoop = currentLoop + 1

    fileName = ("log%d_%d.txt" % (currentLoop, str(now())))

For reference:

from time import mktime, gmtime

def now(): 
   return mktime(gmtime()) 

which is probably irrelevant in your case but i was running multiple instances of this program and making tons of files. Hope this helps!

回答4:

Sequentially checking each file name to find the next available one works fine with small numbers of files, but quickly becomes slower as the number of files increases.

Here is a version that finds the next available file name in log(n) time:

import os

def next_path(path_pattern):
    """
    Finds the next free path in an sequentially named list of files

    e.g. path_pattern = 'file-%s.txt':

    file-1.txt
    file-2.txt
    file-3.txt

    Runs in log(n) time where n is the number of existing files in sequence
    """
    i = 1

    # First do an exponential search
    while os.path.exists(path_pattern % i):
        i = i * 2

    # Result lies somewhere in the interval (i/2..i]
    # We call this interval (a..b] and narrow it down until a + 1 = b
    a, b = (i / 2, i)
    while a + 1 < b:
        c = (a + b) / 2 # interval midpoint
        a, b = (c, b) if os.path.exists(path_pattern % c) else (a, c)

    return path_pattern % b

To measure the speed improvement I wrote a small test function that creates 10,000 files:

for i in range(1,10000):
    with open(next_path('file-%s.foo'), 'w'):
        pass

And implemented the naive approach:

def next_path_naive(path_pattern):
    """
    Naive (slow) version of next_path
    """
    i = 1
    while os.path.exists(path_pattern % i):
        i += 1
    return path_pattern % i

And here are the results:

Fast version:

real    0m2.132s
user    0m0.773s
sys 0m1.312s

Naive version:

real    2m36.480s
user    1m12.671s
sys 1m22.425s

Finally, note that either approach is susceptible to race conditions if multiple actors are trying to create files in the sequence at the same time.

回答5:

Without storing state data in an extra file, a quicker solution to the ones presented here would be to do the following:

from glob import glob
import os

files = glob("somedir/sample*.xml")
files = files.sorted()
cur_num = int(os.path.basename(files[-1])[6:-4])
cur_num += 1
fh = open("somedir/sample%s.xml" % cur_num, 'w')
rs = [blockresult]
fh.writelines(rs)
fh.close()

This will also keep incrementing, even if some of the lower numbered files disappear.

The other solution here that I like (pointed out by Eiyrioü) is the idea of keeping a temporary file that contains your most recent number:

temp_fh = open('somedir/curr_num.txt', 'r')
curr_num = int(temp_fh.readline().strip())
curr_num += 1
fh = open("somedir/sample%s.xml" % cur_num, 'w')
rs = [blockresult]
fh.writelines(rs)
fh.close()

回答6:

Another example using recursion

import os
def checkFilePath(testString, extension, currentCount):
    if os.path.exists(testString + str(currentCount) +extension):
        return checkFilePath(testString, extension, currentCount+1)
    else:
        return testString + str(currentCount) +extension

Use:

checkFilePath("myfile", ".txt" , 0)

回答7:

The two ways to do it are:

1. Check for the existence of the old file and if it exists try the next file name +1
2. save state data somewhere

---

an easy way to do it off the bat would be:

import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(pth.abspath(filename+str(filenum)+".py")):
    filenum+=1
my_next_file = open(filename+str(filenum)+".py",'w')

as a design thing, while True slows things down and isn't a great thing for code readability

---

edited: @EOL contributions/ thoughts

so I think not having .format is more readable at first glance - but using .format is better for generality and convention so.

import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(pth.abspath(filename+str(filenum)+".py")):
    filenum+=1
my_next_file = open("{}{}.py".format(filename, filenum),'w')
# or 
my_next_file = open(filename + "{}.py".format(filenum),'w')

and you don't have to use abspath - you can use relative paths if you prefer, I prefer abs path sometimes because it helps to normalize the paths passed :).

import os.path as pth
filename = "myfile"
filenum = 1
while (pth.exists(filename+str(filenum)+".py"):
    filenum+=1
##removed for conciseness

回答8:

Another solution that avoids the use of while loop is to use os.listdir() function which returns a list of all the files and directories contained in a directory whose path is taken as an argument.

To answer the example in the question, supposing that the directory you are working in only contains "sample_i.xlm" files indexed starting at 0, you can easily obtain the next index for the new file with the following code.

import os

new_index = len(os.listdir('path_to_file_containing_only_sample_i_files'))
new_file = open('path_to_file_containing_only_sample_i_files/sample_%s.xml' % new_index, 'w')

回答9:

You can use a while loop with a counter which checks if a file with a name and the counter's value exists if it does then move on else break and make a file.

I have done it in this way for one of my projects:`

from os import path
import os

i = 0
flnm = "Directory\\Filename" + str(i) + ".txt"
while path.exists(flnm) :
    flnm = "Directory\\Filename" + str(i) + ".txt"
    i += 1
f = open(flnm, "w") #do what you want to with that file...
f.write(str(var))
f.close() # make sure to close it.

`

Here the counter i starts from 0 and a while loop checks everytime if the file exists, if it does it moves on else it breaks out and creates a file from then you can customize. Also make sure to close it else it will result in the file being open which can cause problems while deleting it. I used path.exists() to check if a file exists. Don't do from os import * it can cause problem when we use open() method as there is another os.open() method too and it can give the error. TypeError: Integer expected. (got str) Else wish u a Happy New Year and to all.