How do i open files in python with variable as part of filename?

Question

something where the filenames have numbers 1-32 and i want to open them in order in a loop like:

i = 1
while i < 32:
filename = "C:\\Documents and Settings\\file[i].txt"
f = open(filename, 'r')
text = f.read()
f.close()

but this looks for the file "file[i].txt" instead of file1.txt, file2.txt and so on. how do i make the variable become a variable inside double quotes? and yes i know its not indented, please dont think i m that stupid.

I think this might work : Build the filename just like you'd build any other string that contains a variable:

filename = "C:\\Documents and Settings\\file" + str( i ) + ".txt"

or if you need more options for formatting the number:

filename = "C:\\Documents and Settings\\file%d.txt" % i

Looks like you answered your own question.

lurker
– lurker

2013-08-17 21:01:57 +00:00
Commented Aug 17, 2013 at 21:01 — lurker
– lurker, Commented Aug 17, 2013 at 21:01

Jason Pierrepont · Accepted Answer · 2013-08-17 21:10:36Z

5

First, change your loop to while i <= 32 or you'll exclude the file with 32 in it's name. Your second option filename = "C:\\Documents and Settings\\file%d.txt" % i should work.

If the numbers in your files are 0 padded, like 'file01.txt', 'file02.txt', you can use %.2d instead of plain old %d

answered Aug 17, 2013 at 21:10

Jason Pierrepont

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3 Comments

Arunava Ghosh Over a year ago

If files are 0 padded then we can also do as filename = "C:\\Documents and Settings\\file0%d.txt" % i. I think this should also work.

Jason Pierrepont Over a year ago

Don't do that. That would assume that there's a 0 before every number from 1 to 32. %.2d will add 0's if there's only one number. For every number from 10-32, it'll leave the numbers as they are.

Arunava Ghosh Over a year ago

If files are have four integers in the end then we can also do as filename = "C:\\Documents and Settings\\file%.4d.txt" % i. I think this should also work. Like filename is file1234.

alecxe · Accepted Answer · 2013-08-17 21:01:52Z

5

You've already provided an answer. Btw, use with context manager instead of manually calling close():

i = 1
while i < 32:
    filename = "C:\\Documents and Settings\\file%d.txt" % i
    with open(filename, 'r') as f:
        print(f.read())

answered Aug 17, 2013 at 21:01

alecxe

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Comments

Vaibhav Aggarwal · Accepted Answer · 2013-08-17 21:02:21Z

2

Yes, the options you gave would work, why not just test it out?

filename = "C:\\Documents and Settings\\file" + str( i ) + ".txt"

or

filename = "C:\\Documents and Settings\\file%d.txt" % i

answered Aug 17, 2013 at 21:02

Vaibhav Aggarwal

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