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.