[FIXED] How to silent or redirect output of shutil.move() or another method to move files in silent mode?

Issue

Was using Python shutil.move() method but need to move million files from one location to another and I would like to redirect / silent the output of shutil.move().

if os.path.exists(logs_path_archive):
    for source_file in downloaded_files:
        destination_file = source_file.replace(logs_path,logs_path_archive)
        shutil.move(source_file,destination_file)

Any recommendation?

Solution

In your comment you specify, you are using a Jupyter Notebooks. So you can use a couple of options built into IPython/Jupyter for that.

If the move step is just one cell, at the start of the cell put the following:

%%capture out_stream
shutil.move(<original_filename>,<new_file_name>)

However, it sounds like the move step is in amongst many other steps in a cell, and so to just hush the move, you can use the with io.capture_output() as captured: to suppress output from only what is in the with block, like so:

from IPython.utils import io
with io.capture_output() as captured:
    shutil.move(<original_filename>,<new_file_name>)

See the page here for more about both of those. The %%capture cell magic will probably be above where that link sends you.

---

If you do need to use pure python you can use a with statement and contextlib (like a contextmanager to handle output) to redirect that, similar to the with io.capture_output() as captured idea above. Here I'm sending it to _ which is often used when you don't plan to use the object assigned later.

import contextlib
with contextlib.redirect_stdout(_):
    shutil.move(<original_filename>,<new_file_name>)

Answered By - Wayne