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 

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:

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5:

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6:

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7:

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 

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.

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