Python: Iterate through opening files

Question

If anyone could point me in the right direction I'd be really grateful. I am looking to replace the following:

file1 = open ('filepath')
file1.write(data1)
file2 = open ('filepath2')
file2.write(data2)
file3 = open ('filepath3')
file3.write(data3)

With something like this which can be iterated through:

file[i] = open ('filepath')
file[i].write(data[i])

The reason they all need different names is because all the files must be open at once without closing. This is just a requirement of the system.

Is there any way in which this can be done?

score 3 · Answer 1 · answered Aug 06 '15 at 17:48

3

open_files = [open(fname) for fname in ['filepath1', 'filepath2', 'filepath3']]

for fh in open_files:
    fh.write(...)

or

for i, fh in enumerate(open_files):
    fh.write(data[i])

answered Aug 06 '15 at 17:48

chepner

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Padraic Cunningham · Answer 2 · 2015-08-06T17:40:00.833

2

You can iterate over the file paths using enumerate:

for f in enumerate("fil1","file2","file3"):
    with open(f,"w") as fle:
        fle.write(data[i])

Or zip the file names and data:

for f,d in zip(("fil1","file2","file3",data)):
        with open(f,"w") as fle:
            fle.write(d)

If you want them to stay open store the file objects in a dict:

d = {}
for f,d in zip(("fil1","file2","file3",data)):
        d[f] = open(f,"w")
        f[f].write(d)

edited Aug 06 '15 at 17:40

answered Aug 06 '15 at 17:31

Padraic Cunningham

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While that's correct, that's not what the user needs, which was stated "...all the files must be open at once without closing." – Manuel J. Diaz Aug 06 '15 at 17:32
@ManuelJ.Diaz, it is trivial to store the file objects in a dict as I have provided – Padraic Cunningham Aug 06 '15 at 17:35
1

Yea, you added the last piece after I wrote the comment. That should set Robin up with what he needs. – Manuel J. Diaz Aug 06 '15 at 17:39

Morgan Thrapp · Answer 3 · 2015-08-06T17:40:17.847

0

You can use a dictionary.

files = {'filepath1': open('filepath1'), 'filepath2': open('filepath2')}

If you want to generate the dictionary in an iterative way, you can do something like this:

path = 'filepath{0}'
for i in range(10):
    filepath = path.format(i)
    files[filepath] = open(filepath)

edited Aug 06 '15 at 17:40

answered Aug 06 '15 at 17:34

Morgan Thrapp

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score 0 · Answer 4 · answered Aug 06 '15 at 18:01

0

Well You can do something like:-

filepaths=[List of all your file paths (Ex. "abc.txt", "\c\example\abc.txt")]
fileptr =[]
for file in filepaths :
    fileptr += open(file,'mode')

for fil in fileptr :
    fil.write(data[i])

answered Aug 06 '15 at 18:01

btechhb

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score 0 · Answer 5 · edited May 23 '17 at 12:03

Thank you for the suggestions made, I explored around the subject after having looked at the various ways of doing this and came up with a good solution (for my application anyhow).

The trouble I was having was that I'm writing data to a networked csv file with a raspberry pi. If the file is open elsewhere, the pi can no longer access it and data is lost as it can't be recorded on the csv file.

The initial solution which prompted this question was to keep all the files open at once, allowing other users to only open the csv file in read only mode which still allows the pi to write data.This however needs all files to be kept open in the python script and is rather memory intensive if I'm right in thinking,

Therefore, the solution I found was to make the file read only all the time I wasn't writing to it with the pi and then only making it writable just as I was adding information to the csv file.

I used the code explained here: How to remove read-only attrib directory with Python in Windows?

Many thanks

Python: Iterate through opening files

5 Answers5