python-使用os.walk查找文件名中值最高的文件

Question

I have some folders and in each folder is a list of files like below:

file-Test01-summary.xlsm

file-Test02-summary.xlsm

file-Test03-summary.xlsm

the number of files can be different. in each folder could be 3 to 5 file! I need a code to find the file with highest test number.

It could be great if the code will be write with os.walk()

Answer 1

Try to get the name of the file and split it as below to filter the integers.

I didn't write the code in os.walk(), but i hope this gives you a start.

#List with all file names
from os import walk

FilenameList = ['file-Test01-summary.xlsm', 'file-Test02-summary.xlsm', 'file-Test03-summary.xlsm']

for (<dirpath>, <dirnames>, <filenames>) in walk (<mypath>):
    FilenameList.extend(filenames)
    break

#This list will contain all the test numbers
IntegerList = []

#This loop will go through all filenames and filter the testnumbers and appand them to Integerlist
for file in FilenameList:
      for item in File:
        try:
            IntegerList.append(float(t))
        except ValueError:
          pass

#This will print the highest test number
print(max(IntegerList))

hope this helps;)

Answer 2

One easy way is to iterate over the required files, than use sorted to sort the list ascendent or descendent.

Using ofunctions.file_utils comes in handy to list only files with .xlsm extension. Install the package using:

python -m pip install ofunctions.file_utils`

Here's the working code:

import ofunctions.file_utils

files = ofunctions.file_utils.get_files_recursive('.', ext_include_list='.xlsm')
files = sorted(files, reverse=True) # Remove reverse=True to achieve ASC ordering
print(files)

Result:

['.\\file-Test03-summary.xlsm', '.\\file-Test02-summary.xlsm', '.\\file-Test01-summary.xlsm']

Your highest number will then be files[0] .

You can add parameter depth=1 to get_files_recursive in order to prevent descending into subdirectories.

Disclaimer: I am the author of ofunctions