寻求commandline工具：从CSV坐标列表中查找距离最近的点对

Hey there! I’ve put together a Python command-line script that exactly addresses your need—finding the closest pair of points (not just each point’s nearest neighbor) from a CSV of geocoordinates, perfect for matching your GEO-sourced power plant data.

How It Works #

This script uses the Haversine formula to calculate accurate great-circle distances between geographic points (Euclidean distance isn’t reliable for lat/long data). It checks all unique point pairs to pinpoint the minimum distance pair, with basic validation for messy CSV rows.

Full Script #

Save this as closest_point_pair.py:

import csv
import math
import argparse

def haversine(lat1, lon1, lat2, lon2):
    # Convert degrees to radians for trigonometric calculations
    lat1_rad = math.radians(lat1)
    lon1_rad = math.radians(lon1)
    lat2_rad = math.radians(lat2)
    lon2_rad = math.radians(lon2)

    # Haversine formula to compute great-circle distance
    dlon = lon2_rad - lon1_rad
    dlat = lat2_rad - lat1_rad
    a = math.sin(dlat/2)**2 + math.cos(lat1_rad) * math.cos(lat2_rad) * math.sin(dlon/2)**2
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))
    # Earth radius in kilometers (use 3956 for miles)
    radius = 6371
    return radius * c

def find_closest_pair(csv_file):
    points = []
    # Read and parse CSV data
    with open(csv_file, mode='r', newline='', encoding='utf-8') as f:
        reader = csv.reader(f)
        # Skip header (assumes CSV has columns: Name, Latitude, Longitude)
        next(reader)
        for row in reader:
            if len(row) >= 3:
                name = row[0].strip()
                try:
                    lat = float(row[1])
                    lon = float(row[2])
                    points.append( (name, lat, lon) )
                except ValueError:
                    print(f"Skipping invalid row (non-numeric coordinates): {row}")

    if len(points) < 2:
        print("Error: Need at least two points to find a closest pair.")
        return

    min_distance = float('inf')
    closest_pair = None

    # Check all unique point pairs to find the smallest distance
    for i in range(len(points)):
        name1, lat1, lon1 = points[i]
        for j in range(i+1, len(points)):
            name2, lat2, lon2 = points[j]
            distance = haversine(lat1, lon1, lat2, lon2)
            if distance < min_distance:
                min_distance = distance
                closest_pair = (name1, name2, lat1, lon1, lat2, lon2)

    if closest_pair:
        name1, name2, lat1, lon1, lat2, lon2 = closest_pair
        print("✅ Closest pair found:")
        print(f"- {name1} | Lat: {lat1}, Lon: {lon1}")
        print(f"- {name2} | Lat: {lat2}, Lon: {lon2}")
        print(f"Distance: {round(min_distance, 4)} kilometers")

if __name__ == "__main__":
    parser = argparse.ArgumentParser(description='Find the closest pair of geographic points from a CSV file.')
    parser.add_argument('csv_file', help='Path to your CSV file (columns: Name, Latitude, Longitude)')
    args = parser.parse_args()
    find_closest_pair(args.csv_file)

Usage Instructions #

1. Prepare your CSV: Ensure your file follows this structure (comma-delimited):

Name,Latitude,Longitude
Chicoasén Dam,16.941064,-93.100828
Tuxpan Oil Power Plant,21.014891,-97.3344
Coastal Generator,16.950000,-93.090000

2. Run the script: Open your terminal and run:

python closest_point_pair.py your_power_plants.csv

3. Adjust units: If you need miles instead of kilometers, change the radius value in the haversine function from 6371 to 3956.

Key Notes #

• Validation: The script skips rows with non-numeric coordinates and alerts you to invalid entries.
• Performance: This brute-force approach works great for most power plant datasets (hundreds to thousands of points). For extremely large datasets (10k+ points), you could upgrade to a divide-and-conquer algorithm, but that’s overkill for typical use cases here.
• Custom delimiters: If your CSV uses tabs or semicolons, modify the csv.reader call to include delimiter='\t' or delimiter=';'.

内容的提问来源于stack exchange，提问作者pickenpack