One context manager instead of two for open file?

Question

My question looks similar to this one but not sure... I want to parse some log files that are sometimes compressed in gzip sometimes not.

I've got the following:

if file[-3:] == ".gz":
     with gzip.open(file, 'rb') as f:
          # do something
else:
    with open(file) as f:
          # do the same thing.

Is it possible to have only one with statement ?

it may be possible, but I think it would be more confusing. perhaps just put the `something` in a function so you don't duplicate your code. — user1558604, Dec 09 '19 at 14:44

score 2 · Answer 1 · answered Dec 09 '19 at 14:47

fn = gzip.open if file.endswith('.gz') else open

with fn(file, 'rb') as f:
    ...

Also note that the call to the function returning the context manager does not have to happen inside the with line:

if file.endswith('.gz'):
    ctx = gzip.open(file, 'rb')
else:
    ctx = open(file)

with ctx as f:
    ...

score 1 · Answer 2 · answered Dec 09 '19 at 14:46

1

Put your "Do Something" in a function

def processFile(f)
        Do Something...

if file[-3:] == ".gz":
     with gzip.open(file, 'rb') as f:
          processFile(f)
else:
    with open(file) as f:
          processFile(f)

answered Dec 09 '19 at 14:46

tomgalpin

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score 1 · Accepted Answer · answered Dec 09 '19 at 15:31

1

You can put the conditional statement in the with line:

with gzip.open(file, 'rb') if file[-3:] == '.gz' else open(file) as f:
   processFile(f)

answered Dec 09 '19 at 15:31

Jab

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One context manager instead of two for open file?

3 Answers3