Read nth line of importlib.resources.files without loading whole file in memory

I thought I would be able to use a importlib.resources.files(...).open() object just like a file:

from importlib.resources import files


with files('data').joinpath('graphics.txt').open() as graphics:
    for i, line in graphics:
        if i == 4:
            print(line)

but I get a ValueError: too many values to unpack (expected 2)

In this kind of project structure of course:

.
├── __init__.py
└── data
    ├── __init__.py (empty)
    └── graphics.txt

sample graphics.txt:

line 0
line 1
line 2
line 3
line 4
line 5
line 6

I want to read a single line I know the number of, wasting minimal time getting to that line, and without loading the whole file into memory (it is quite large, but not too large)

Solution

It works just fine, you just forgot enumerate:

for i, line in graphics:

should be:

for i, line in enumerate(graphics):

Without enumerate, you're only getting the lines, not the line number.

If you want a small micro-optimization, you can use itertools.islice to avoid explicitly looping over and checking line numbers for the lines you don't care about:

from itertools import islice  # At top of file

with files('data').joinpath('graphics.txt').open() as graphics:
    line = next(islice(graphics, 4, None), None)
    if line is not None:
        print(line)

But to be clear, the preceding lines are still being read (and immediately discarded) under the hood; you can't skip variable lengths lines without scanning from the beginning.