How do I using with open() as f: ...
to write the file in a directory that doesn’t exist.
For example:
with open('/Users/bill/output/output-text.txt', 'w') as file_to_write: file_to_write.write("{}n".format(result))
Let’s say the /Users/bill/output/
directory doesn’t exist. If the directory doesn’t exist just create the directory and write the file there.
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Answer
You need to first create the directory.
The mkdir -p
implementation from this answer will do just what you want. mkdir -p
will create any parent directories as required, and silently do nothing if it already exists.
Here I’ve implemented a safe_open_w()
method which calls mkdir_p
on the directory part of the path, before opening the file for writing:
import os, os.path import errno # Taken from https://stackoverflow.com/a/600612/119527 def mkdir_p(path): try: os.makedirs(path) except OSError as exc: # Python >2.5 if exc.errno == errno.EEXIST and os.path.isdir(path): pass else: raise def safe_open_w(path): ''' Open "path" for writing, creating any parent directories as needed. ''' mkdir_p(os.path.dirname(path)) return open(path, 'w') with safe_open_w('/Users/bill/output/output-text.txt') as f: f.write(...)
Updated for Python 3:
import os, os.path def safe_open_w(path): ''' Open "path" for writing, creating any parent directories as needed. ''' os.makedirs(os.path.dirname(path), exist_ok=True) return open(path, 'w') with safe_open_w('/Users/bill/output/output-text.txt') as f: f.write(...)