Open file knowing only a part of its name

Question

I’m currently reading a file and importing the data in it with the line:

# Read data from file.
data = np.loadtxt(join(mypath, 'file.data'), unpack=True)

where the variable mypath is known. The issue is that the file file.data will change with time assuming names like:

file_3453453.data
file_12324.data
file_987667.data
...

So I need a way to tell the code to open the file in that path that has a name like file*.data, assuming that there will always be only one file by that name in the path. Is there a way to do this in python?

Asked By: Gabriel

||

Source

Answer 1

Try

import os

[os.path.join(root, f) for root, _, files in os.walk(mypath)
                       for f in files
                       if f.startswith('file') and f.endswith('.data')]

It’ll return a list of all files file*.data, in case there are more than one. You can just iterate through them. If there is only one file, then just put [0] at then end of the list comprehension.

Answered By: wflynny

Answer 2

I simple solution would be to use the python modules “os” and “re”:

import os
import re
for file in os.listdir(mypath):
     if re.match("file_d+.data", file):
     ...

Answered By: go2

Answer 3

You can use the glob module. It allows pattern matching on filenames and does exactly what you’re asking

import glob

for fpath in glob.glob(mypath):
    print fpath

e.g I have a directory with files named google.xml, google.json and google.csv.

I can use glob like this:

>>> import glob
>>> glob.glob('g*gle*')
['google.json', 'google.xml', 'google.csv']

Note that glob uses the fnmatch module but it has a simpler interface and it matches paths instead of filenames only.

You can search relative paths and don’t have to use os.path.join. In the example above if I change to the parent directory and try to match file names, it returns the relative paths:

>>> import os
>>> import glob
>>> os.chdir('..')
>>> glob.glob('foo/google*')
['foo/google.json', 'foo/google.xml', 'foo/google.csv']

Answered By: MrD

Answer 4

Also check out fnmatch:

>>> import fnmatch
>>> import os
>>>
>>> fnmatch.filter(os.listdir('.'), 'file_*.data')
['file_3453453.data']
>>>

Answered By: user1129665

Open file knowing only a part of its name

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