graze 0.1.19

Cache (a tiny part of) the internet

graze

Cache (a tiny part of) the internet.

(For the technically inclined, graze is meant to enable the separation of the concerns of getting and caching data from the internet.)

install

pip install graze

Example

from graze import Graze
import os
rootdir = os.path.expanduser('~/graze')
g = Graze(rootdir)
list(g)

If this is your first time, you got nothing:

[]

So get something. For no particular reason let's be self-referential and get myself:

url = 'https://raw.githubusercontent.com/thorwhalen/graze/master/README.md'
content = g[url]
type(content), len(content)

Before I grew up, I had only 46 petty bytes:

(bytes, 46)

These were:

print(content.decode())

# graze

Cache (a tiny part of) the internet

But now, here's the deal. List your g keys now. Go ahead, don't be shy!

list(g)

['https://raw.githubusercontent.com/thorwhalen/graze/master/README.md']

What does that mean?

I means you have a local copy of these contents.

The file path isn't really https://..., it's rootdir/https/..., but you only have to care about that if you actually have to go get the file with something else than graze. Because graze will give it to you.

How? Same way you got it in the first place:

content_2 = g[url]
assert content_2 == content

But this time, it didn't ask the internet. It just got it's local copy.

And if you want a fresh copy?

No problem, just delete your local one. You guessed! The same way you would delete a key from a dict:

del g[url]

Q&A

The pages I need to slurp need to be rendered, can I use selenium of other such engines?

Sure!

We understand that sometimes you might have special slurping needs -- such as needing to let the JS render the page fully, and/or extract something specific, in a specific way, from the page.

Selenium is a popular choice for these needs.

graze doesn't install selenium for you, but if you've done that, you just need to specify a different Internet object for Graze to source from, and to make an internet object, you just need to specify what a url_to_contents function that does exactly what it says.

Note that the contents need to be returned in bytes for Graze to work.

If you want to use some of the default selenium url_to_contents functions to make an Internet (we got Chrome, Firefox, Safari, and Opera), you go ahead! here's an example using the default Chrome driver (again, you need to have the driver installed already for this to work; see https://selenium-python.readthedocs.io/):

from graze import Graze, url_to_contents, Internet

g = Graze(source=Internet(url_to_contents=url_to_contents.selenium_chrome))

And if you'll be using it often, just do:

from graze import Graze, url_to_contents, Internet
from functools import partial
my_graze =  partial(
    Graze,
    rootdir='a_specific_root_dir_for_your_project',
    source=Internet(url_to_contents=url_to_contents.selenium_chrome)
)

# and then you can just do
g = my_graze()
# and get on with the fun...

What if I want a fresh copy of the data?

Classic caching problem. You like the convenience of having a local copy, but then how do you keep in sync with the data source if it changes?

If you KNOW the source data changed and want to sync, it's easy. You delete the local copy (like deleting a key from a dict: del Graze()[url]) and you try to access it again. Since you don't have a local copy, it will get one from the url source.

What if you want this to happen automatically?

Well, there's several ways to do that.

If you have a way to know if the source and local are different (through modified dates, or hashes, etc.), then you can write a little function to keep things in sync. But that's context dependent; graze doesn't offer you any default way to do it.

Another way to do this is sometimes known as a TTL Cache (time-to-live cache). You get such functionality with the graze.GrazeWithDataRefresh store, or for most cases, simply getting your data through the graze function specifying a max_age value (in seconds):

from graze import graze

content_bytes = graze(url, max_age=in_seconds)

Can I make graze notify me when it gets a new copy of the data?

Sure! Just specify a key_ingress function when you make your Graze object, or call graze. This function will be called on the key (the url) just before contents are being downloaded from the internet. The typical function would be:

key_ingress = lambda key: print(f"Getting {key} from the internet")

Does graze work for dropbox links?

Yes it does, but you need to be aware that dropbox systematically send the data as a zip, even if there's only one file in it.

Here's some code that can help.

def zip_store_of_gropbox_url(dropbox_url: str):
    """Get a key-value perspective of the (folder) contents 
    of the zip a dropbox url gets you"""
    from graze import graze
    from py2store import FilesOfZip
    return FilesOfZip(graze(dropbox_url))
    
def filebytes_of_dropbox_url(dropbox_url: str, assert_only_one_file=True):
    """Get the bytes of the first file in a zip that a dropbox url gives you"""
    zip_store = zip_store_of_gropbox_url(dropbox_url)
    zip_filepaths = iter(zip_store)
    first_filepath = next(zip_filepaths)
    if assert_only_one_file:
        assert next(zip_filepaths, None) is None, f"More than one file in {dropbox_url}"
    return zip_store[first_filepath]

Notes

New url-to-path mapping

graze used to have a more straightforward url-to-local_filepath mapping, but it ended up being problematic: In a nutshell, if you slurp abc.com and it goes to a file of that name, where is abc.com/data.zip supposed to go (abc.com needs to be a folder in that case).
See issue.

It's with a heavy heart that I changed the mapping to one that was still straightforward, but has the disadvantage of mapping all files to the same file name, without extension.

Hopefully a better solution will show up soon.

If you already have graze files from the old way, you can use the change_files_to_new_url_to_filepath_format function to change these to the new format.