Choose a file starting with a given string

file, python

Solution

Try using `os.listdir`,`os.path.join` and `os.path.isfile`. In long form (with for loops),

import os
path = 'C:/'
files = []
for i in os.listdir(path):
    if os.path.isfile(os.path.join(path,i)) and '001_MN_DX' in i:
        files.append(i)

Code, with list-comprehensions is

import os
path = 'C:/'
files = [i for i in os.listdir(path) if os.path.isfile(os.path.join(path,i)) and \
         '001_MN_DX' in i]

Check here for the long explanation...

Problem

In a directory I have a lot of files, named more or less like this: ``` 001_MN_DX_1_M_32 001_MN_SX_1_M_33 012_BC_2_F_23 ... ... ``` In Python, I have to write a code that selects from the directory a file starting with a certain string. For example, if the string is `001_MN_DX`, Python selects the first file, and so on. How can I do it?

Original source