Choose a file starting with a given string

Question

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?

Answer 1

import os prefixed = [filename for filename in os.listdir('.') if filename.startswith("prefix")] 

Answer 2

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...