CODEWITHSUNDEEP
pythonpython-3.xfile-iolist-comprehensioncontextmanager
kupgov
kupgov

Reputation: 393

How to use list comprehension with list of variable number of filenames?

Given the list of filenames filenames = [...].

Is it possibly rewrite the next list comprehension for I/O-safety: [do_smth(open(filename, 'rb').read()) for filename in filenames]? Using with statement, .close method or something else.

Another problem formulation: is it possibly to write I/O-safe list comprehension for the next code?

results = []
for filename in filenames:
   with open(filename, 'rb') as file:
      results.append(do_smth(file.read()))

Upvotes: 4

Views: 372

Answers (2)

Sven Marnach
Sven Marnach

Reputation: 602115

You can use the ExitStack introduced in Python 3.3 for this purpose:

with ExitStack() as stack:
    files = [stack.enter_context(open(name, "rb")) for name in filenames]
    results = [do_smth(file.read()) for file in files]

Note that this opens all the files at once, which is not necessary for this use case, and might not be a good idea if you have a big number of files.

Upvotes: 3

Eugene Yarmash
Eugene Yarmash

Reputation: 149963

You can put the with statement/block to a function and call that in the list comprehension:

def slurp_file(filename):
    with open(filename, 'rb') as f:
        return f.read()

results = [do_smth(slurp_file(f)) for f in filenames]

Upvotes: 9

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