How can I read files with similar names on python and then work with them?

Question

So I'm working in a project where I need to read a bunch of files, and they have the same name except for a number, something like:

thing1.txt
thing2.txt
thing3.txt

Then I have to work with them giving them another name:

example1='.\path\thing1.txt'

Is there any way that the code can read me these files differentiating between the numbers and then assign them to that new name in a numbered form?

import fnmatch

for filename in os.listdir('.'):
    if fnmatch.fnmatch(filename, 'thing*.txt'):
        print(filename)

With the code I'm using right now I can read the files with the same name and different number, but I can't rename them in a loop to work with them after.

I want something like a loop that does this:

example*=thing*

Where * should be the number.

Edit: I should have said so, but the files I work with (thing1/2/3) have numerical values that I need to use in some operations later in the code, so that's why I need to 'rename' them.

Answer 1

You can use formatted text in Python 3.

for i in range(10):
    txt = f'text{i}.txt'
    print(txt)

Answer 2

try using os.walk(). os.walk

for (root,dirs,files) in os.walk(main_folder): 
#iterate over all the files in the folders and sub folders
    for file in files:
         filepath = os.path.join(root,file)
         # read the file from thing and write the file in example
         with open(filepath ) as f:
              with open(filepath.replace('thing',example),'w') as g:
                      g.write(f)

Answer 3

You can dynamically create and execute Python by using string formatting to actually write a new python file, and then execute it as a new process from your existing script. That would let you actually assign variable names dynamically as you're asking.

Really though, you should probably just put them in a dictionary:

import os
import fnmatch

examples = {}

for filename in os.listdir('.'):
    if fnmatch.fnmatch(filename, 'thing*.txt'):
        examples[filename[:6]] = filename

output:

examples
{'thing1': 'thing1.txt', 'thing2': 'thing2.txt', 'thing3': 'thing3.txt'}

Answer 4

You could use a string variable and a for loop to iterate:

for n in range(1, 5):
    filename="file" + str(n) + ".csv"
    print (filename)

To rename a file you can use os module

import os

os.rename('file.txt', 'newfile.txt')

Answer 5

To find the number inside a filename you can do:

fname = "some_filename1.txt"
number = "".join(x for x in fname if x.isdigit())
# This would, however, use any digits found in the filename as the number, so you can perhaps do something like:
import os
# The following code will find the number of file that is at the end of its name only
fnameonly, ext = os.path.splitext(fname)
if not fnameonly:
    fnameonly = ext # Fiilenames beginning with extension separator
    ext = ""
fnameonly = fnameonly.rstrip()
number = []
for x in range(len(fnameonly)-1, -1, -1):
    if not fnameonly[x].isdigit(): break
    number.append(fnameonly[x])
number.reverse()
number = "".join(number)
# In both cases, if number is empty, then the filename does not contain any numbers in it and you can skip this file
# So, after you find the number:
newfname = "Example %s%s" % (number, ext)
try:
    os.rename(fname, newfname)
except Exception, e:
    print "Error: couldn't rename the file '%s' to '%s' because:\n%s\nRenaming skipped" % (fname, newfname, str(e))

Now, the above code might seem as a little too much. You can use regular expressions to do it as well though. (if you know how :D ) Or any number of variants of above code(s). Put it in a function and then iterate over the folder using it to rename files. The code I provided is much more flexible than using eg formatting. It is usable on a file of any extension and it makes sure that what you are getting is really the number and not a part of some added text like: "thingy1.txt" with formatting solution would result in "exampley1.txt", which you do not want to happen. So, either use my code or regex.