CODEWITHSUNDEEP
python
kkard
kkard

Reputation: 79

Opening multiple files using loop in python

Basically I am trying to open multiple files according to name of the list pass. I have files in Log/ with following name

here is the dir structure
script---myfile.py
       |
       |----Log/*.txt


following files are in Log/
parse_1d_30a.txt, 
parse_10d_60a.txt,
parse_20d_90a.txt


#!/usr/bin/python
deviation = ['1', '10', '20']
angle = ['30', '60', '90']

def openFile(dev, ang):
    p = open('Log/parse_%sd_%sa.txt'%(dev, ang), 'r')
    print "open file is", p.name
    p.close()

    print "file closed."

def main():
    for d, a in zip(deviation, angle):
        openFile(d, a)
main()

So, when I execute the code first file parse_1d_30a.txt opens but for other files it gives IOError: no such file or directory.

I think by using 'glob' it might work. I know how to open the files individually in python but not sure why I am wrong with above code and what is the alternative for the same.

Thanks

Upvotes: 2

Views: 2847

Answers (1)

heemayl
heemayl

Reputation: 41987

With glob you can not impose absolute preciseness like Regex. In glob, you need to use one of * (any number of characters) or ? (any single character), which makes it difficult to do strict matching.

The close i can get:

>>> import glob
>>> glob.glob(r'parse_[0-9]*d_[0-9][0-9]a.txt')
['parse_20d_90a.txt', 'parse_1d_30a.txt', 'parse_10d_60a.txt']

Here * can match any number of characters which might lead to wrong output based on the file names and your desired output.

Upvotes: 1

Related Questions