Brancheneinstufung2/company-explorer/backend/lib/metric_parser.py

import re
import logging
from typing import Optional, Union

logger = logging.getLogger(__name__)

class MetricParser:
    """
    Robust parser for extracting numeric values from text, specialized for
    German formats and business metrics (Revenue, Employees).
    Reconstructs legacy logic to handle thousands separators and year-suffixes.
    """

    @staticmethod
    def extract_numeric_value(text: str, is_revenue: bool = False) -> Optional[float]:
        """
        Extracts a float value from a string, handling German locale and suffixes.

        Args:
            text: The raw text containing the number (e.g. "1.005 Mitarbeiter (2020)").
            is_revenue: If True, prioritizes currency logic (e.g. handling "Mio").

        Returns:
            The parsed float value or None if no valid number found.
        """
        if not text:
            return None
        
        # 1. Cleaning: Remove Citations [1], [note 2]
        clean_text = re.sub(r'\[.*?\]', '', text)
        
        # 2. Cleaning: Remove Year/Date in parentheses to prevent "80 (2020)" -> 802020
        # Matches (2020), (Stand 2021), (31.12.2022), etc.
        # We replace them with space to avoid merging numbers.
        clean_text = re.sub(r'\(\s*(?:Stand\s*|ab\s*)?(?:19|20)\d{2}.*?\)', ' ', clean_text)
        
        # 3. Identify Multipliers (Mio, Mrd)
        multiplier = 1.0
        lower_text = clean_text.lower().replace('.', '') # Remove dots for word matching (e.g. "Mio." -> "mio")
        
        if any(x in lower_text for x in ['mrd', 'milliarde', 'billion']): # German Billion = 10^12? Usually in business context here Mrd=10^9
            multiplier = 1_000_000_000.0
        elif any(x in lower_text for x in ['mio', 'million']):
            multiplier = 1_000_000.0
        
        # 4. Extract the number candidate
        # We look for the FIRST pattern that looks like a number.
        # Must contain at least one digit.
        # We iterate over matches to skip pure punctuation like "..."
        matches = re.finditer(r'[\d\.,]+', clean_text)
        
        for match in matches:
            candidate = match.group(0)
            # Check if it actually has a digit
            if not re.search(r'\d', candidate):
                continue
                
            # Clean trailing/leading punctuation (e.g. "80." -> "80")
            candidate = candidate.strip('.,')
            if not candidate:
                continue

            try:
                val = MetricParser._parse_german_number_string(candidate)
                return val * multiplier
            except Exception as e:
                # If this candidate fails (e.g. "1.2.3.4"), try the next one?
                # For now, let's assume the first valid-looking number sequence is the target.
                # But "Wolfra ... 80" -> "..." skipped. "80" matched.
                # "1.005 Mitarbeiter" -> "1.005" matched.
                logger.debug(f"Failed to parse number string '{candidate}': {e}")
                continue
        
        return None

    @staticmethod
    def _parse_german_number_string(s: str) -> float:
        """
        Parses a number string dealing with ambiguous separators.
        Logic based on Lessons Learned:
        - "1.005" -> 1005.0 (Dot followed by exactly 3 digits = Thousands)
        - "1,5" -> 1.5 (Comma = Decimal)
        - "1.234,56" -> 1234.56
        """
        # Count separators
        dots = s.count('.')
        commas = s.count(',')
        
        # Case 1: No separators
        if dots == 0 and commas == 0:
            return float(s)
        
        # Case 2: Mixed separators (Standard German: 1.000.000,00)
        if dots > 0 and commas > 0:
            # Assume . is thousands, , is decimal
            s = s.replace('.', '').replace(',', '.')
            return float(s)
        
        # Case 3: Only Dots
        if dots > 0:
            # Ambiguity: "1.005" (1005) vs "1.5" (1.5)
            # Rule: If dot is followed by EXACTLY 3 digits (and it's the last dot or multiple dots), likely thousands.
            # But "1.500" is 1500. "1.5" is 1.5.
            
            # Split by dot
            parts = s.split('.')
            
            # Check if all parts AFTER the first one have exactly 3 digits
            # E.g. 1.000.000 -> parts=["1", "000", "000"] -> OK -> Thousands
            # 1.5 -> parts=["1", "5"] -> "5" len is 1 -> Decimal
            
            all_segments_are_3_digits = all(len(p) == 3 for p in parts[1:])
            
            if all_segments_are_3_digits:
                # Treat as thousands separator
                return float(s.replace('.', ''))
            else:
                # Treat as decimal (US format or simple float)
                # But wait, German uses comma for decimal. 
                # If we are parsing strict German text, "1.5" might be invalid or actually mean 1st May? 
                # Usually in Wikipedia DE: "1.5 Mio" -> 1.5 Million.
                # So if it's NOT 3 digits, it's likely a decimal point (US style or just typo/format variation).
                # User Rule: "1.005" -> 1005.
                return float(s) # Python handles 1.5 correctly
        
        # Case 4: Only Commas
        if commas > 0:
            # German Decimal: "1,5" -> 1.5
            # Or English Thousands: "1,000" -> 1000?
            # User context is German Wikipedia ("Mitarbeiter", "Umsatz").
            # Assumption: Comma is ALWAYS decimal in this context, UNLESS followed by 3 digits AND likely English?
            # Safer bet for German data: Comma is decimal.
            return float(s.replace(',', '.'))
            
        return float(s)