Filter lines from a csv file inside a zip. It does this in an efficient way so to not waste time on doing this.

filter_lines_from_zip_csv.py

import pandas as pd
import zipfile
import io
import re

from typing import Iterator

def rows_with_index(pattern, sep, file) -> Iterator[str]:
    row_index = 0
    for row in io.TextIOWrapper(file):
        if pattern.match(row):
            yield f"{row_index}{sep}{row}"
        row_index += 1

def filter_lines_from_zipped_csv(path: str, pattern: re.Pattern[str], sep: str = ",", column_names: list[str] | None = None) -> pd.DataFrame:
    """
    Filter lines from a zipped csv file (with one file inside) using a regex pattern.
    The pattern needs to be compiled generating the re.Pattern object.
    If no lines are found, an empty dataframe is returned.
    Parameters
    ----------
        path: str
            path to the zipped csv file
        pattern: re.Pattern[str]
            compiled regex pattern
        sep: str, default :
            separator of the csv file
    """
    with zipfile.ZipFile(path, 'r') as zip_file:
        with zip_file.open(zip_file.namelist()[0]) as file:
            rows = rows_with_index(pattern, sep, file)

            return pd.read_csv(io.StringIO("\n".join(rows)), sep=sep, index_col=0, header=None, names=column_names, lineterminator='\n')